NCERT Solutions Class 12 Maths Chapter 9 Differential Equations

NCERT Solutions for Class 12 Maths Chapter 9 - Differential Equations

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Access NCERT Solutions for Class-12 Maths Chapter 9 – Differential Equation part-1

Access NCERT Solutions for Class-12 Maths Chapter 9 – Differential Equation

Exercise 9.1

1. Determine order and degree (if defined) of differential equation $\frac{{{\text{d}}^{\text{4}}}\text{y}}{\text{d}{{\text{x}}^{\text{4}}}}\text{+sin}\left( \text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{  }\!\!'\!\!\text{ } \right)\text{=0}.$

Ans: Rewrite the equation $\frac{{{\text{d}}^{\text{4}}}\text{y}}{\text{d}{{\text{x}}^{\text{4}}}}\text{+sin}\left( \text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{  }\!\!'\!\!\text{ } \right)\text{=0}$. as:

\[\Rightarrow \text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{  }\!\!'\!\!\text{  }\!\!'\!\!\text{ +sin}\left( y''' \right)=0\]  

The highest order between the two terms is of $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{  }\!\!'\!\!\text{  }\!\!'\!\!\text{ }$ which is four.

The differential equation contains a trigonometric derivative term and is not completely polynomial in its derivative, thus degree is not defined.


2. Determine order and degree (if defined) of differential equation $\text{y }\!\!'\!\!\text{ +5y=0}$.

Ans: The given differential equation is $\text{y }\!\!'\!\!\text{ +5y=0}$.

The highest order term is $\text{y }\!\!'\!\!\text{ }$, thus the order is one.

As the derivative is of completely polynomial nature is and highest power of derivative is of $\text{y }\!\!'\!\!\text{ }$ which is one. Thus degree is one.


3. Determine order and degree (if defined) of differential equation ${{\left( \frac{\text{ds}}{\text{dt}} \right)}^{\text{4}}}\text{+3s}\frac{{{\text{d}}^{\text{2}}}\text{s}}{\text{d}{{\text{t}}^{\text{2}}}}\text{=0}$.

Ans: The given differential equation is ${{\left( \frac{\text{ds}}{\text{dt}} \right)}^{\text{4}}}\text{+3s}\frac{{{\text{d}}^{\text{2}}}s}{\text{d}{{\text{t}}^{\text{2}}}}\text{=0}$.

The highest order term is $\frac{{{\text{d}}^{\text{2}}}\text{s}}{\text{d}{{\text{t}}^{\text{2}}}}$, thus the order is two.

As the derivative is of completely polynomial nature is and highest power of derivative term$\frac{{{\text{d}}^{\text{2}}}\text{s}}{\text{d}{{\text{t}}^{\text{2}}}}$ which is one. Thus the degree is one.


4. Determine order and degree (if defined) of differential equation ${{\left( \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}} \right)}^{\text{2}}}\text{+cos}\left( \frac{\text{dy}}{\text{dx}} \right)\text{=0}$.

Ans: The given differential equation is ${{\left( \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}} \right)}^{\text{2}}}\text{+cos}\left( \frac{\text{dy}}{\text{dx}} \right)\text{=0}$.

The highest order term is $\frac{{{\text{d}}^{\text{2}}}y}{\text{d}{{\text{x}}^{\text{2}}}}$, thus the order is two.

The differential equation contains a trigonometric derivative term and is not completely polynomial in its derivative, thus degree is not defined.


5. Determine order and degree (if defined) of differential equation ${{\left( \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}} \right)}}\text{-cos3x+sin3x=0}$.

Ans: The given differential equation is ${{\left( \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}} \right)}^{\text{2}}}\text{-cos3x+sin3x=0}$.

The highest order term is $\frac{{{\text{d}}^{\text{2}}}y}{\text{d}{{\text{x}}^{\text{2}}}}$, thus the order is two.

As the derivative is of completely polynomial nature is and highest power of derivative term $\frac{{{\text{d}}^{\text{2}}}y}{\text{d}{{\text{x}}^{\text{2}}}}$ which is one. Thus degree is one.


6. Determine order and degree (if defined) of differential equation ${{\left( \text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{  }\!\!'\!\!\text{ } \right)}^{\text{2}}}\text{+}{{\left( \text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ } \right)}^{\text{3}}}\text{+}{{\left( \text{y }\!\!'\!\!\text{ } \right)}^{\text{4}}}\text{+}{{\text{y}}^{\text{5}}}\text{=0}$.

Ans: The given differential equation is ${{\left( \text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{  }\!\!'\!\!\text{ } \right)}^{\text{2}}}\text{+}{{\left( \text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ } \right)}^{\text{3}}}\text{+}{{\left( \text{y }\!\!'\!\!\text{ } \right)}^{\text{4}}}\text{+}{{\text{y}}^{\text{5}}}\text{=0}$.

The highest order term is ${{\left( y''' \right)}^{2}}$, thus the order is three.

The differential equation is of the polynomial form and the power of highest order term $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{  }\!\!'\!\!\text{ }$ is two, thus the degree is two.


7. Determine order and degree (if defined) of differential equation $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{  }\!\!'\!\!\text{ +2y }\!\!'\!\!\text{  }\!\!'\!\!\text{ +y }\!\!'\!\!\text{ =0}$.

Ans: The given differential equation is $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{  }\!\!'\!\!\text{ +2y }\!\!'\!\!\text{  }\!\!'\!\!\text{ +y }\!\!'\!\!\text{ =0}$.

The highest order derivative in the differential equation is $y'''$. Thus its order is three.

The differential equation is polynomial with the highest order term $y'''$ having a degree one. Thus the degree is one.


8. Determine order and degree (if defined) of differential equation $\text{y }\!\!'\!\!\text{ +y=e }\!\!'\!\!\text{ }$.

Ans: The given differential equation is $\text{y }\!\!'\!\!\text{ +y=e }\!\!'\!\!\text{ }$. Therefore:

$\Rightarrow \text{y }\!\!'\!\!\text{ +y-e }\!\!'\!\!\text{ =0}$ 

The highest order derivative in the differential equation is $\text{y }\!\!'\!\!\text{ }$. Thus its order is one.

The given equation is of polynomial form with the highest order term $\text{y }\!\!'\!\!\text{ }$ with degree one. Thus the degree is one.


9. Determine order and degree (if defined) of differential equation $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ +}{{\left( \text{y }\!\!'\!\!\text{ } \right)}^{\text{2}}}\text{+2y=0}$.

Ans: The given differential equation is $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ +}{{\left( \text{y }\!\!'\!\!\text{ } \right)}^{\text{2}}}\text{+2y=0}$.

The highest order derivative in the differential equation is $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ }$. Thus its order is two.

The given equation is of polynomial form with the highest order term $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ }$ with highest degree one. Thus the degree is one.


10. Determine order and degree (if defined) of differential equation $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ +2y }\!\!'\!\!\text{ +siny=0}$ .

Ans: The given differential equation is $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ +2y }\!\!'\!\!\text{ +siny=0}$.

The highest order derivative in the differential equation is $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ }$. Thus its order is two.

The given equation is of polynomial form with the highest order term $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ }$ with the highest degree one. Thus the degree is one.


11. Find the degree of the differential equation ${{\left( \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}} \right)}^{\text{3}}}\text{+}{{\left( \frac{\text{dy}}{\text{dx}} \right)}^{\text{2}}}\text{+sin}\left( \frac{\text{dy}}{\text{dx}} \right)\text{+1=0}$.

(A)$\text{3}$

(B)$\text{2}$

(C)$\text{1}$  

(D)not defined

Ans: The given differential equation is ${{\left( \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}} \right)}^{\text{3}}}\text{+}{{\left( \frac{\text{dy}}{\text{dx}} \right)}^{\text{2}}}\text{+sin}\left( \frac{\text{dy}}{\text{dx}} \right)\text{+1=0}$.

The differential equation is not polynomial in its derivative because of the term $\text{sin}\left( \frac{\text{dy}}{\text{dx}} \right)$ thus its order is not defined.

The correct answer is (D).


12. Find the order of the differential equation $\text{2}{{\text{x}}^{\text{2}}}\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{-3}\frac{\text{dy}}{\text{dx}}\text{+y=0}$.

(A)$\text{2}$

(B)$\text{1}$

(C)$\text{0}$  

(D)not defined

Ans: The given differential equation is $\text{2}{{\text{x}}^{\text{2}}}\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{-3}\frac{\text{dy}}{\text{dx}}\text{+y=0}$.

The highest order term of the equation is $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}$, thus the order is two.

The correct answer is (A).


EXERCISE- 9.2

1. Verify the function $\text{y=}{{\text{e}}^{\text{x}}}\text{+1}$ is solution of differential equation $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ -y=0}$ .

Ans: The given function is $\text{y=}{{\text{e}}^{\text{x}}}+1$ .

Take its derivative:

$\frac{dy}{dx}=\frac{d}{dx}\left( {{e}^{x}}+1 \right)$ 

$\Rightarrow y'={{e}^{x}}$       ……(1)

Take the derivative of the above equation:

$\frac{d}{dx}\left( y' \right)=\frac{d}{dx}\left( {{e}^{x}} \right)$ 

$\Rightarrow y''={{e}^{x}}$ 

Using result from equation (1):

$y''-y'=0$ 

Thus the given function is solution of differential equation $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ -y }\!\!'\!\!\text{ =0}$.


 2. Verify the function $\text{y=}{{\text{x}}^{\text{2}}}\text{+2x+C}$ is solution of differential equation $\text{y }\!\!'\!\!\text{ -2x-2=0}$ .

Ans: The given function is $\text{y=}{{\text{x}}^{\text{2}}}\text{+2x+C}$ .

Take its derivative:

$\frac{dy}{dx}=\frac{d}{dx}\left( {{x}^{2}}+2x+C \right)$ 

$\Rightarrow y'=2x+2$

$\Rightarrow y'-2x-2=0$ 

Thus the given function is solution of differential equation $\text{y }\!\!'\!\!\text{ -2x-2=0}$.


3. Verify the function $\text{y=cos x +C}$ is solution of differential equation $\text{y }\!\!'\!\!\text{ +sin x =0}$ .

Ans: The given function is $\text{y=cos x + C}$ .

Take its derivative:

$\frac{dy}{dx}=\frac{d}{dx}\left( \cos x+C \right)$ 

$\Rightarrow y'=-\sin x$

$\Rightarrow y'+\sin x=0$ 

Thus the given function is solution of differential equation $\text{y }\!\!'\!\!\text{  + sin x = 0}$.


4. Verify the function $\text{y=}\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}$ is solution of differential equation $\text{y }\!\!'\!\!\text{ =}\frac{\text{xy}}{\text{1+}{{\text{x}}^{\text{2}}}}$ .

Ans: The given function is $\text{y = }\sqrt{1+{{x}^{2}}}$ .

Take its derivative:

$\frac{dy}{dx}=\frac{d}{dx}\left( \sqrt{1+{{x}^{2}}} \right)$ 

$y'=\frac{1}{2\sqrt{1+{{x}^{2}}}}\times \frac{d}{dx}\left( 1+{{x}^{2}} \right)$

$y'=\frac{2}{2\sqrt{1+{{x}^{2}}}}$ 

$\Rightarrow y'=\frac{1}{\sqrt{1+{{x}^{2}}}}$ 

Multiply numerator and denominator by $\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}$:

$y'=\frac{1}{\sqrt{1+{{x}^{2}}}}\times \frac{\sqrt{1+{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}}$

Substitute $\text{y  = }\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}$ 

$\Rightarrow y'=\frac{xy}{1+{{x}^{2}}}$ 

Thus the given function is solution of differential equation $\text{y }\!\!'\!\!\text{  =}\frac{xy}{1+{{x}^{2}}}$.


5. Verify the function $\text{y=Ax}$ is solution of differential equation $\text{xy }\!\!'\!\!\text{ =y}\left( x\ne 0 \right)$ .

Ans: The given function is $\text{y = Ax}$ .

Take its derivative:

$\frac{dy}{dx}=\frac{d}{dx}\left( Ax \right)$ 

$\Rightarrow y'=A$

Multiply by $\text{x}$ on both side:

$xy'=Ax$ 

Substitute $\text{y = Ax}$:

$\Rightarrow xy'=y$ 

Thus the given function is solution of differential equation $xy'=y\left( x\ne 0 \right)$.


6. Verify the function $\text{y=xsin x}$ is solution of differential equation $\text{xy }\!\!'\!\!\text{ =y+x}\sqrt{{{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}}\left( x\ne 0\,\text{and}\,x>y\,\text{or}\,x<-y \right)$ .

Ans: The given function is $\text{y = xsin x}$ .

Take its derivative:

$\frac{dy}{dx}=\frac{d}{dx}\left( x\sin x \right)$ 

$\Rightarrow y'=\sin x\frac{d}{dx}\left( x \right)+x\frac{d}{dx}\left( \sin x \right)$

$\Rightarrow y'=\sin x+x\cos x$ 

Multiply by $\text{x}$ on both side:

$xy'=x\left( \sin x+x\cos x \right)$ 

$xy'=x\sin x+{{x}^{2}}\cos x$ 

Substitute $\text{y = xsin x}$:

$\Rightarrow xy'=y+{{x}^{2}}\cos x$ 

Use $\text{sin x = }\frac{\text{y}}{\text{x}}$ and substitute $\text{cos x}$:

$xy'=y+{{x}^{2}}\sqrt{1-{{\sin }^{2}}x}$ 

$xy'=y+{{x}^{2}}\sqrt{1-{{\left( \frac{y}{x} \right)}^{2}}}$ 

$\Rightarrow xy'=y+x\sqrt{{{y}^{2}}-{{x}^{2}}}$ 

Thus the given function is solution of differential equation $\text{xy }\!\!'\!\!\text{  = y + x}\sqrt{{{\text{y}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}}}$.


7. Verify the function $\text{xy=log y + C}$ is solution of differential equation $\text{y }\!\!'\!\!\text{ =}\frac{{{\text{y}}^{\text{2}}}}{\text{1-xy}}\left( \text{xy}\ne \text{1} \right)$ .

Ans: The given function is $\text{xy = log y + C}$ .

Take derivative on both side:

$\frac{d}{dx}\left( xy \right)=\frac{d}{dx}\left( \log y \right)$ 

$\Rightarrow y\frac{d}{dx}\left( x \right)+x\frac{dy}{dx}=\frac{1}{y}\frac{dy}{dx}$

$\Rightarrow y+xy'=\frac{1}{y}y'$ 

$\Rightarrow {{y}^{2}}+xy\,y'=y'$ 

Shift the $\text{y }\!\!'\!\!\text{ }$ terms on one side and take it common. 

$\Rightarrow \left( xy-1 \right)y'=-{{y}^{2}}$ 

$\Rightarrow y'=\frac{{{y}^{2}}}{1-xy}$ 

Thus the given function is solution of differential equation $\text{y }\!\!'\!\!\text{  = }\frac{{{\text{y}}^{\text{2}}}}{\text{1-xy}}$.


8. Verify the function $\text{y-cosy=x}$ is solution of differential equation \[\left( \text{ysin y + cos y + x} \right)\text{y }\!\!'\!\!\text{ = 1}\] .

Ans: The given function is $\text{y - cos y = x}$ .

Take derivative on both side:

$\frac{dy}{dx}-\frac{d}{dx}\left( \cos y \right)=\frac{d}{dx}\left( x \right)$ 

$\Rightarrow y'+y'\sin y=1$

$\Rightarrow y'\left( 1+\sin y \right)=1$ 

$\Rightarrow y'=\frac{1}{1+\sin y}$ 

Multiply by \[\left( \text{ysin y + cos y + x} \right)\] on both side:

\[\left( \text{ysin y + cos y + x} \right)y'=\frac{\left( \text{ysin y + cos y + x} \right)}{1+\sin y}\] 

Substitute $y=\cos y+x$ in the numerator:

\[\left( \text{ysin y + cos y + x} \right)y'=\frac{\left( \text{ysin y + y} \right)}{1+\sin y}\] 

\[\left( \text{ysin y + cos y + x} \right)y'=\frac{y\left( \text{sin y + 1} \right)}{1+\sin y}\] 

$\Rightarrow \left( \text{ysin y + cos y + x} \right)y'=y$ 

Thus the given function is solution of differential equation$\left( \text{ysin y + cos y + x} \right)\text{y }\!\!'\!\!\text{  = y}$.


9. Verify the function $\text{x+y=ta}{{\text{n}}^{-1}}\text{y}$ is solution of differential equation \[{{\text{y}}^{\text{2}}}\text{y }\!\!'\!\!\text{ +}{{\text{y}}^{\text{2}}}\text{+1=0}\] .

Ans: The given function is $\text{x + y = ta}{{\text{n}}^{-1}}y$ .

Take derivative on both side:

$\frac{d}{dx}\left( x+y \right)=\frac{d}{dx}\left( ta{{n}^{-1}}y \right)$ 

$1+y'=\left( \frac{1}{1+{{y}^{2}}} \right)y'$

$\Rightarrow y'\left[ \frac{1}{1+{{y}^{2}}}-1 \right]=1$ 

$\Rightarrow y'\left[ \frac{1-\left( 1+{{y}^{2}} \right)}{1+{{y}^{2}}} \right]=1$ 

$\Rightarrow y'\left[ \frac{-{{y}^{2}}}{1+{{y}^{2}}} \right]=1$ 

$\Rightarrow -{{y}^{2}}y'=1+{{y}^{2}}$ 

$\Rightarrow {{y}^{2}}y'+{{y}^{2}}+1=0$ 

Thus the given function is solution of differential equation ${{y}^{2}}y'+{{y}^{2}}+1=0$.


10. Verify the function $\text{y=}\sqrt{{{\text{a}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}}}\text{x}\in \left( \text{-a,a} \right)$ is solution of differential equation \[\text{x+y}\frac{\text{dy}}{\text{dx}}\text{=0}\left( \text{y}\ne \text{0} \right)\] .

Ans: The given function is:

\[y\text{ }=\text{ }\sqrt{{{a}^{2}}-{{x}^{2}}}\,x\in \left( -a,a \right)\] .

Take derivative on both side:

$\frac{dy}{dx}=\frac{d}{dx}\left( \sqrt{{{a}^{2}}-{{x}^{2}}} \right)$ 

$\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{{{a}^{2}}-{{x}^{2}}}}\frac{d}{dx}\left( {{a}^{2}}-{{x}^{2}} \right)$

$\Rightarrow \frac{dy}{dx}=\frac{-2x}{2\sqrt{{{a}^{2}}-{{x}^{2}}}}$ 

$\Rightarrow \frac{dy}{dx}=\frac{-x}{\sqrt{{{a}^{2}}-{{x}^{2}}}}$ 

Substitute $\text{y = }\sqrt{{{\text{a}}^{\text{2}}}\text{ - }{{\text{x}}^{\text{2}}}}$ 

$\Rightarrow \frac{dy}{dx}=\frac{-x}{y}$ 

$\Rightarrow x+y\frac{dy}{dx}=0$ 

Thus the given function is solution of differential equation:

$x+y\frac{dy}{dx}=0\left( y\ne 0 \right)$.


11. Find the numbers of arbitrary constants in the general solution of a differential equation of fourth order.

(A)$\text{0}$

(B)$\text{2}$

(C)$\text{3}$  

(D)$\text{4}$ 

Ans: The number of arbitrary constants in the general solution of a differential equation is equal to its order. As the given differential equation is of fourth order, thus it has four arbitrary constants in its solution.

The correct answer is (D).


12. Find the numbers of arbitrary constants in the particular solution of a differential equation of third order.

(A)$\text{3}$

(B)$\text{2}$

(C)$\text{1}$  

(D)$\text{0}$ 

Ans: The particular solution of any differential equation does not have any arbitrary constants. Thus it has zero constants in its solution.

The correct answer is (D).



Exercise 9.3

1. Form a differential equation representing the family of the curve $\frac{\text{x}}{\text{a}}\text{+}\frac{\text{y}}{\text{b}}\text{= 1}$ by eliminating arbitrary constants.

Ans: The given differential equation is $\frac{\text{x}}{\text{a}}\text{+}\frac{\text{y}}{\text{b}}\text{= 1}$:

Take the derivative on both side:

$\frac{d}{dx}\left( \frac{x}{a}+\frac{y}{b} \right)=0$ 

$\Rightarrow \frac{1}{a}+\frac{1}{b}y'=0$ 

Take the derivative of the above equation to eliminate the constants.

$\Rightarrow 0+\frac{1}{b}y''=0$ 

$\Rightarrow y''=0$ 


The curve for given differential equation is $y''=0$.


2. Form a differential equation representing the family of the curve ${{\text{y}}^{\text{2}}}\text{= a}\left( {{\text{b}}^{\text{2}}}\text{- }{{\text{x}}^{\text{2}}} \right)$ by eliminating arbitrary constants.

Ans: The given differential equation is ${{\text{y}}^{\text{2}}}\text{= a}\left( {{\text{b}}^{\text{2}}}\text{- }{{\text{x}}^{\text{2}}} \right)$:

Take the derivative on both side:

$2y\frac{dy}{dx}=a\left( -2x \right)$ 

$\Rightarrow 2yy'=-2ax$ 

$\Rightarrow yy'=-ax$       ……(1)

Take the derivative of the above equation to eliminate the constants.

$\Rightarrow yy''+y'y'=-a$ 

$\Rightarrow {{\left( y' \right)}^{2}}+yy''=-a$ 

Substitute this in result (1):

$yy'=\left( {{\left( y' \right)}^{2}}+yy'' \right)x$ 

$\Rightarrow xyy''+x{{\left( y' \right)}^{2}}-yy'=0$ 


The curve for given differential equation is $xyy''+x{{\left( y' \right)}^{2}}-yy'=0$ 

.


3. Form a differential equation representing the family of the curve $\text{y = a}{{\text{e}}^{\text{3x}}}\text{+b}{{\text{e}}^{\text{-2x}}}$ by eliminating arbitrary constants.

Ans: The given differential equation is $\text{y = a}{{\text{e}}^{\text{3x}}}\text{+b}{{\text{e}}^{\text{-2x}}}$:

Take the derivative on both side:

$y'=3a{{e}^{3x}}-2b{{e}^{-2x}}$ 

Take derivative of above equation:

$y''=9a{{e}^{3x}}+4b{{e}^{-2x}}$ 

Compute $\text{y }\!\!'\!\!\text{ +2y}$:

$y'+2y=3a{{e}^{3x}}-2b{{e}^{-2x}}+2a{{e}^{3x}}+2b{{e}^{-2x}}$ 

$\Rightarrow 5{{a}^{3x}}=y'+2y$ 

$\Rightarrow {{a}^{3x}}=\frac{y'+2y}{5}$ 

Compute $\text{3y - y }\!\!'\!\!\text{ }$:

$3y-y'=3a{{e}^{3x}}+3b{{e}^{-2x}}-\left( 3a{{e}^{3x}}-2b{{e}^{-2x}} \right)$ 

\[\Rightarrow 5{{b}^{-2x}}=3y-y'\] 

\[\Rightarrow {{b}^{-2x}}=\frac{3y-y'}{5}\] 

Substitute the above results in  y'' :

$\Rightarrow y''=9\left( \frac{y'+2y}{5} \right)+4\left( \frac{3y-y'}{5} \right)$ 

$\Rightarrow y''=\frac{5y'+30y}{5}$ 

$\Rightarrow y''=y'+6y$ 

$\Rightarrow y''-y'-6y=0$ 

The curve for the given differential equation is $ y''\text{  - y }\!\!'\!\!\text{  - 6y = 0}$.


4. Form a differential equation representing the family of the curve $\text{y = }{{\text{e}}^{\text{2x}}}\left( \text{a+bx} \right)$ by eliminating arbitrary constants.

Ans: The given differential equation is $\text{y = }{{\text{e}}^{\text{2x}}}\left( \text{a+bx} \right)$:

Take the derivative on both side:

$y'=2{{e}^{2x}}(a+bx)+{{e}^{2x}}\left( b \right)$ 

$\Rightarrow y'={{e}^{2x}}\left( 2a+2bx+b \right)$ 

Compute $\text{y }\!\!'\!\!\text{  - 2y}$:

$y'-2y={{e}^{2x}}\left( 2a+2bx+b \right)-2{{e}^{2x}}\left( a+bx \right)$ 

$\Rightarrow y'-2y=b{{e}^{2x}}$       ……(1)

Take derivative of the above equation:

$\Rightarrow y''-2y'=2b{{e}^{2x}}$ 

Substitute the above results using result from equation (1):

$\Rightarrow y''-2y'=2\left( y'-2y \right)$ 

$\Rightarrow y''-4y'+4y=0$ 

The curve for given differential equation is \[\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{  - 4y }\!\!'\!\!\text{  + 4y = 0}\].


5. Form a differential equation representing the family of the curve $\text{y = }{{\text{e}}^{\text{x}}}\left( \text{acosx+bsinx} \right)$ by eliminating arbitrary constants.

Ans: The given differential equation is:

$y\text{ }=\text{ }{{e}^{x}}\left( a\cos x+b\sin x \right)$.

Take the derivative on both side:

$y'={{e}^{x}}\left( a\cos x+b\sin x \right)+{{e}^{x}}\left( -a\sin x+b\cos x \right)$ 

$\Rightarrow y'={{e}^{x}}\left[ \left( a+b \right)\cos x-\left( a-b \right)\sin x \right]$ 

Take the derivative of the above equation:

$y''={{e}^{x}}\left[ \left( a+b \right)\cos x-\left( a-b \right)\sin x \right]+{{e}^{x}}\left[ -\left( a+b \right)\sin x-\left( a-b \right)\cos x \right]$ 

$\Rightarrow y''={{e}^{x}}\left[ \left( a+b-a+b \right)\cos x-\left( a-b+a+b \right)\sin x \right]$    

$\Rightarrow y''={{e}^{x}}\left( 2b\cos x-2a\sin x \right)$  

$\Rightarrow y''=2{{e}^{x}}\left( b\cos x-a\sin x \right)$ 

$\Rightarrow \frac{y''}{2}={{e}^{x}}\left( b\cos x-a\sin x \right)$ 

Add $\text{y}$ on both side:

$y+\frac{y''}{2}={{e}^{x}}\left( a\cos x+b\sin x \right)+{{e}^{x}}\left( b\cos x-a\sin x \right)$ 

$\Rightarrow y+\frac{y''}{2}={{e}^{x}}\left( \left( a+b \right)\cos x-\left( a-b \right)\sin x \right)$ 

Back substitute $\text{y }\!\!'\!\!\text{ }$:

$y+\frac{y''}{2}=y'$ 

$\Rightarrow 2y+y''=2y'$ 

$\Rightarrow y''-2y'+2y=0$ 

The curve for given differential equation is \[\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{  - 2y }\!\!'\!\!\text{  + 2y = 0}\].


6. Form the differential equation of the family of circles touching the $\text{y}$ -axis at the origin.

Ans: Let a circle with radius $\text{a}$ touches $\text{y}$-axis at origin.


(Image will be uploaded soon)


Thus the given circle will have the center at $\left( \text{a,0} \right)$. So its equation will be:

${{\left( x-a \right)}^{2}}+{{y}^{2}}={{a}^{2}}$ 

$\Rightarrow {{x}^{2}}+{{y}^{2}}=2ax$       ……(1)

Take the derivative of the above equation:

$2x+2yy'=2a$ 

$\Rightarrow x+yy'=a$ 

Back substitute $\text{a}$ in equation (1):
${{x}^{2}}+{{y}^{2}}=2\left( x+yy' \right)x$ 

$\Rightarrow 2{{x}^{2}}+2yy'x={{x}^{2}}+{{y}^{2}}$ 

$\Rightarrow {{x}^{2}}+2yy'x={{y}^{2}}$ 

Thus the differential equation for the given family of the circle is ${{\text{x}}^{\text{2}}}\text{+ 2yy }\!\!'\!\!\text{ x = }{{\text{y}}^{\text{2}}}$ 


7. Form the differential equation of the family of parabolas having vertex at origin and axis along positive$\text{y}$ -axis.

Ans: Draw a general parabola with given properties.


(Image will be uploaded soon)


${{x}^{2}}=4ay$       ……(1)

Take the derivative of the above equation:

$2x=4ay'$ 

$\Rightarrow a=\frac{x}{2y'}$ 

Back substitute $\text{a}$ in equation (1):
${{x}^{2}}=4\left( \frac{x}{2y'} \right)y$ 

$\Rightarrow {{x}^{2}}=2\left( \frac{x}{y'} \right)y$ 

$\Rightarrow {{x}^{2}}y'-2xy=0$ 

Thus the differential equation for the given family of the parabolas is ${{\text{x}}^{\text{2}}}\text{y }\!\!'\!\!\text{  - 2xy = 0}$ .


8. Form the differential equation of the family of ellipses having foci on $\text{y}$ -axis and center at origin.

Ans: Draw a standard eclipse with given properties.


(Image will be uploaded soon)


$\frac{{{x}^{2}}}{{{b}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1$

Take the derivative of the above equation:

$\frac{2x}{{{b}^{2}}}+\frac{2yy'}{{{a}^{2}}}=0$

$\Rightarrow \frac{x}{{{b}^{2}}}+\frac{yy'}{{{a}^{2}}}=0$       ……(1)

Take the derivative of the above equation:

$\frac{1}{{{b}^{2}}}+\frac{y'y'+yy''}{{{a}^{2}}}=0$  

$\frac{1}{{{b}^{2}}}=-\frac{1}{{{a}^{2}}}\left( y{{'}^{2}}+yy'' \right)$ 

Back substitute in equation (1):
$x\left[ -\frac{1}{{{a}^{2}}}\left( y{{'}^{2}}+yy'' \right) \right]+\frac{yy'}{{{a}^{2}}}=0$ 

$\Rightarrow -xy{{'}^{2}}-xyy''+yy'=0$ 

Thus the differential equation for the given family of the parabolas is $\text{-xy}{{\text{ }\!\!'\!\!\text{ }}^{\text{2}}}\text{ - xyy }\!\!'\!\!\text{  }\!\!'\!\!\text{  + yy }\!\!'\!\!\text{  = 0}$ .


9. Form the differential equation of the family of hyperbolas having foci on $\text{x}$ -axis and center at origin.

Ans: Draw a standard hyperbola with given properties.


(Image will be uploaded soon)


$\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1$

Take the derivative of the above equation:

$\frac{2x}{{{a}^{2}}}-\frac{2yy'}{{{b}^{2}}}=0$

$\Rightarrow \frac{x}{{{b}^{2}}}+\frac{yy'}{{{a}^{2}}}=0$       ……(1)

Take the derivative of the above equation:

$\frac{1}{{{b}^{2}}}+\frac{y'y'+yy''}{{{a}^{2}}}=0$  

$\frac{1}{{{b}^{2}}}=-\frac{1}{{{a}^{2}}}\left( y{{'}^{2}}+yy'' \right)$ 

Back substitute in equation (1):
$x\left[ -\frac{1}{{{a}^{2}}}\left( y{{'}^{2}}+yy'' \right) \right]+\frac{yy'}{{{a}^{2}}}=0$ 

$\Rightarrow -xy{{'}^{2}}-xyy''+yy'=0$ 

$\Rightarrow xyy''+xy{{'}^{2}}-yy'=0$ 


Thus the differential equation for the given family of the parabolas is $xyy''+xy{{'}^{2}}-yy'=0$  .


10. Form the differential equation of the family of circles having center on $\text{y}$ -axis and radius $\text{3}$ units.

Ans: Draw the circle with given properties.


(Image will be uploaded soon)


${{x}^{2}}+{{\left( y-b \right)}^{2}}=9$

Take the derivative of the above equation:

$2x+2\left( y-b \right)y'=0$

$\Rightarrow x+\left( y-b \right)y'=0$

$\Rightarrow \left( y-b \right)=-\frac{x}{y'}$ 

Back substitute in the equation of circle:
${{x}^{2}}+{{\left( \frac{-x}{y'} \right)}^{2}}=9$ 

$\Rightarrow {{x}^{2}}y{{'}^{2}}+{{x}^{2}}=9y{{'}^{2}}$ 

$\Rightarrow \left( {{x}^{2}}-9 \right)y{{'}^{2}}+{{x}^{2}}=0$ 


Thus the differential equation for the given family of the parabolas is $\left( {{\text{x}}^{\text{2}}}\text{- 9} \right)\text{y}{{\text{ }\!\!'\!\!\text{ }}^{\text{2}}}\text{+ }{{\text{x}}^{\text{2}}}\text{= 0}$ .


11. Which of the following differential equations has

$\text{y=}{{\text{c}}_{\text{1}}}{{\text{e}}^{\text{x}}}\text{+}{{\text{c}}_{\text{2}}}{{\text{e}}^{\text{-x}}}$ as the general solution?

A.$\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{+y=0}$ 

B.$\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{- y=0}$ 

C.$\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{+1=0}$ 

D.$\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{-1=0}$ 

Ans: The given equation is $y={{c}_{1}}{{e}^{x}}+{{c}_{2}}{{e}^{-x}}$.

Differentiate the equation:

$\frac{dy}{dx}={{c}_{1}}{{e}^{x}}-{{c}_{2}}{{e}^{-x}}$ 

Differentiate the above equation:
$\frac{{{d}^{2}}y}{d{{x}^{2}}}={{c}_{1}}{{e}^{x}}+{{c}_{2}}{{e}^{-x}}$ 

Back substitute the $\text{y}$:

$\frac{{{d}^{2}}y}{d{{x}^{2}}}=y$ 

$\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}-y=0$ 

Thus the correct answer is Option B.


12. Which of the following differential equation has $\text{y = x}$ as one of its particular solutions?

A.$\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{-}{{\text{x}}^{\text{2}}}\frac{\text{dy}}{\text{dx}}\text{+xy=x}$ 

B.$\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{+x}\frac{\text{dy}}{\text{dx}}\text{+xy=x}$ 

C.$\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{-}{{\text{x}}^{\text{2}}}\frac{\text{dy}}{\text{dx}}\text{+xy=0}$ 

D.$\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{+x}\frac{\text{dy}}{\text{dx}}\text{+xy=0}$ 

Ans: The given equation is $y=x$.

Differentiate the equation:

$\frac{dy}{dx}=1$ 

Differentiate the above equation:
$\frac{{{d}^{2}}y}{d{{x}^{2}}}=0$ 

Deducing for the options:

$\frac{{{d}^{2}}y}{d{{x}^{2}}}-{{x}^{2}}\frac{dy}{dx}+xy=0-{{x}^{2}}\left( 1 \right)+x\left( x \right)$ 

$\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}-{{x}^{2}}\frac{dy}{dx}+xy=-{{x}^{2}}+{{x}^{2}}$ 

$\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}-{{x}^{2}}\frac{dy}{dx}+xy=0$ 

Thus the correct answer is Option C.


Exercise 9.4

1. Find the general solution for $\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{1- cos x}}{\text{1+ cos x}}$ .

Ans: The given differential equation is $\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{1- cos x}}{\text{1+ cos x}}$.

Use trigonometric half –angle identities to simplify:
$\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}$ 

$\Rightarrow \frac{dy}{dx}=\frac{2{{\sin }^{2}}\frac{x}{2}}{2{{\cos }^{2}}\frac{x}{2}}$ 

$\Rightarrow \frac{dy}{dx}={{\tan }^{2}}\frac{x}{2}$ 

$\Rightarrow \frac{dy}{dx}={{\sec }^{2}}\frac{x}{2}-1$ 

Separate the differentials and integrate:

$\int dy=\int \left( {{\sec }^{2}}\frac{x}{2}-1 \right)dx$ 

$\Rightarrow y=\int {{\sec }^{2}}\frac{x}{2}dx-\int dx$ 

$\Rightarrow y=2\tan \frac{x}{2}-x+C$ 

Thus the general solution of given differential equation is $\text{y = 2tan}\frac{\text{x}}{\text{2}}\text{ - x + C}$.


2. Find the general solution for $\frac{\text{dy}}{\text{dx}}\text{=}\sqrt{\text{4 - }{{\text{y}}^{\text{2}}}}\left( \text{-2  y  2} \right)$ .

Ans: The given differential equation is $\frac{\text{dy}}{\text{dx}}\text{=}\sqrt{\text{4 - }{{\text{y}}^{\text{2}}}}\left( \text{-2  y  2} \right)$.

Simplify the expression:

$\frac{dy}{dx}=\sqrt{4-{{y}^{2}}}$ 

$\Rightarrow \frac{dy}{\sqrt{4-{{y}^{2}}}}=dx$ 

Use standard integration:

$\int \frac{dy}{\sqrt{4-{{y}^{2}}}}=\int dx$ 

$\Rightarrow {{\sin }^{-1}}\frac{y}{2}=x+C$ 

$\Rightarrow \frac{y}{2}=\sin \left( x+C \right)$ 

$\Rightarrow y=2\sin \left( x+C \right)$ 

Thus the general solution of given differential equation is $\text{y = 2 sin}\left( \text{x + C} \right)$.


3. Find the general solution for $\frac{\text{dy}}{\text{dx}}\text{+ y =1}\left( \text{y}\ne \text{1} \right)$ .

Ans: The given differential equation is $\frac{\text{dy}}{\text{dx}}\text{+ y =1}\left( \text{y}\ne \text{1} \right)$.

Simplify the expression:

$\frac{dy}{dx}+y=1$ 

$\Rightarrow \frac{dy}{dx}=1-y$ 

$\Rightarrow \frac{dy}{1-y}=dx$ 

Use standard integration:

$\int \frac{dy}{1-y}=\int dx$ 

$\Rightarrow -\log \left( 1-y \right)=x+C$ 

$\Rightarrow \log \left( 1-y \right)=-\left( x+C \right)$ 

$\Rightarrow 1-y={{e}^{-\left( x+C \right)}}$ 

$y=1-A{{e}^{-x}}\,\left( A={{e}^{-C}} \right)$ 

Thus the general solution of given differential equation is $\text{y=1- A}{{\text{e}}^{\text{-x}}}\,$.


4. Find the general solution for $\text{se}{{\text{c}}^{\text{2}}}\text{x}\,\text{tany}\,\text{dx+se}{{\text{c}}^{\text{2}}}\text{y}\,\text{tanx}\,\text{dy=0}$.

Ans: The given differential equation is:

${{\sec }^{2}}x\tan ydx+{{\sec }^{2}}y\tan xdy=0$.

Divide both side by $\text{tan x tan y}$:

$\frac{{{\sec }^{2}}x\tan ydx+{{\sec }^{2}}y\tan xdy}{\tan x\tan y}=0$ 

\[\Rightarrow \frac{{{\sec }^{2}}x}{\tan x}dx+\frac{{{\sec }^{2}}y}{\tan y}dy=0\] 

Integrate both side:

$\int \frac{{{\sec }^{2}}x}{\tan x}dx+\int \frac{{{\sec }^{2}}y}{\tan y}dy=0$ 

$\Rightarrow \int \frac{{{\sec }^{2}}y}{\tan y}dy=-\int \frac{{{\sec }^{2}}x}{\tan x}dx$       ……(1)

Use a substitution method for integration. Substitute $\text{tan x = u}$:

For integral on RHS:

$\Rightarrow \tan x=u$ 

$\Rightarrow {{\sec }^{2}}xdx=du$ 

$\Rightarrow \int \frac{{{\sec }^{2}}x}{\tan x}dx=\int \frac{du}{u}$ 

$\Rightarrow \int \frac{{{\sec }^{2}}x}{\tan x}dx=\log u$ 

$\Rightarrow \int \frac{{{\sec }^{2}}x}{\tan x}dx=\log \left( \tan x \right)$ 

Thus evaluating result form (1):

$\Rightarrow \log \left( \tan y \right)=-\log \left( \tan x \right)+\log \left( C \right)$ 

$\Rightarrow \log \left( \tan y \right)=\log \left( \frac{C}{\tan x} \right)$ 

$\Rightarrow \tan x\tan y=C$ 

Thus the general solution of given differential equation is $\text{tan x tan y = C}$.


5. Find the general solution for $\left( {{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{\text{-x}}} \right)\text{dy-}\left( {{\text{e}}^{\text{x}}}\text{-}{{\text{e}}^{\text{-x}}} \right)\text{dx=0}$ .

Ans: The given differential equation is:

$\left( {{e}^{x}}+{{e}^{-x}} \right)dy-\left( {{e}^{x}}-{{e}^{-x}} \right)dx=0$.

Simplify the expression:

$dy=\left[ \frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}} \right]dx$ 

Integrate both side:

$\int dy=\int \left[ \frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}} \right]dx$ ……(1)

Use a substitution method for integration. Substitute \[{{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{\text{-x}}}\text{=t}\]:

For integral on RHS:

$\Rightarrow {{e}^{x}}+{{e}^{-x}}=t$ 

$\Rightarrow \left( {{e}^{x}}-{{e}^{-x}} \right)dx=dt$ 

$\Rightarrow \int \left[ \frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}} \right]dx=\int \frac{dt}{t}$ s

$\Rightarrow \int \left[ \frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}} \right]dx=\ln t+C$ 

$\Rightarrow \int \left[ \frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}} \right]dx=\log \left( {{e}^{x}}+{{e}^{-x}} \right)+C$ 

Thus evaluating result form (1):

$y=\log \left( {{e}^{x}}+{{e}^{-x}} \right)+C$ 

Thus the general solution of given differential equation is $\text{y=log}\left( {{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{\text{-x}}} \right)\text{+C}$.


6. Find the general solution for $\frac{\text{dy}}{\text{dx}}\text{=}\left( \text{1+}{{\text{x}}^{\text{2}}} \right)\left( \text{1+}{{\text{y}}^{\text{2}}} \right)$.

Ans: The given differential equation is:

$\frac{dy}{dx}=\left( 1+{{x}^{2}} \right)\left( 1+{{y}^{2}} \right)$.

Simplify the expression:

$\frac{dy}{1+{{y}^{2}}}=\left( 1+{{x}^{2}} \right)dx$ 

Integrate both side:

$\int \frac{dy}{1+{{y}^{2}}}=\int \left( 1+{{x}^{2}} \right)dx$

Use standard integration:

${{\tan }^{-1}}y=\int dx+\int {{x}^{2}}dx$ 

$\Rightarrow {{\tan }^{-1}}y=x+\frac{{{x}^{3}}}{3}+C$ 

Thus the general solution of given differential equation is $\text{ta}{{\text{n}}^{\text{-1}}}\text{y=x+}\frac{{{\text{x}}^{\text{3}}}}{\text{3}}\text{+C}$.


7. Find the general solution for $\text{ylog y dx-xdy=0}$.

Ans: The given differential equation is:

$y\log ydx-xdy=0$.

Simplify the expression:

$y\log ydx=xdy$ 

$\Rightarrow \frac{dx}{x}=\frac{dy}{y\log y}$ 

Integrate both side:

$\int \frac{dy}{y\log y}=\int \frac{dx}{x}$ ……(1)

Use substitution method for integration on LHS. Substitute \[\text{log y=t}\]:

$\log y=t$ 

$\Rightarrow \frac{1}{y}dy=dt$ 

$\Rightarrow \int \frac{dy}{y\log y}=\int \frac{dt}{t}$ 

$\Rightarrow \int \frac{dy}{y\log y}=\log t$ 

$\Rightarrow \int \frac{dy}{y\log y}=\log \left( \log y \right)$ 

Evaluating expression (1):

$\log \left( \log y \right)=\log x+\log C$ 

$\Rightarrow \log \left( \log y \right)=\log Cx$ 

$\Rightarrow \log y=Cx$ 

$\Rightarrow y={{e}^{Cx}}$ 

Thus the general solution of given differential equation is $\text{y=}{{\text{e}}^{\text{Cx}}}$.


8. Find the general solution for ${{\text{x}}^{\text{5}}}\frac{\text{dy}}{\text{dx}}\text{=-}{{\text{y}}^{\text{5}}}$.

Ans: The given differential equation is:

${{x}^{5}}\frac{dy}{dx}=-{{y}^{5}}$.

Simplify the expression:

$\frac{dy}{{{y}^{5}}}=-\frac{dx}{{{x}^{5}}}$ 

Integrate both side:

$\int \frac{dy}{{{y}^{5}}}=-\int \frac{dx}{{{x}^{-5}}}$ 

\[\Rightarrow \int {{y}^{-5}}dy=-\int {{x}^{-5}}dx\] 

$\Rightarrow \frac{{{y}^{-5+1}}}{-5+1}=-\frac{{{x}^{-5+1}}}{-5+1}+C$ 

$\Rightarrow \frac{{{y}^{-4}}}{-4}=-\frac{{{x}^{-4}}}{-4}+C$ 

$\Rightarrow {{x}^{-4}}+{{y}^{-4}}=-4C$ 

$\Rightarrow {{x}^{-4}}+{{y}^{-4}}=A\,\,\,\,\left( A=-4C \right)$ 

Thus the general solution of given differential equation is ${{\text{x}}^{\text{-4}}}\text{+}{{\text{y}}^{\text{-4}}}\text{=A}$.


9. Find the general solution for $\frac{\text{dy}}{\text{dx}}\text{=si}{{\text{n}}^{\text{-1}}}\text{x}$.

Ans: The given differential equation is:

$\frac{dy}{dx}={{\sin }^{-1}}x$.

Simplify the expression:

$dy={{\sin }^{-1}}xdx$ 

Integrate both side:

$\int dy=\int {{\sin }^{-1}}xdx$ 

\[\Rightarrow y=\int 1\times {{\sin }^{-1}}xdx\] 

Use product rule of integration:

$\int {{\sin }^{-1}}xdx={{\sin }^{-1}}x\int dx-\int \left( \frac{1}{\sqrt{1-{{x}^{2}}}}\int dx \right)dx$ 

$\Rightarrow \int {{\sin }^{-1}}xdx=x{{\sin }^{-1}}x-\int \frac{x}{\sqrt{1-{{x}^{2}}}}dx$ 

Substitute $\text{1-}{{\text{x}}^{\text{2}}}\text{=}{{\text{t}}^{\text{2}}}$ 

$1-{{x}^{2}}={{t}^{2}}$ 

$\Rightarrow -2xdx=2tdt$ 

$\Rightarrow -xdx=tdt$ 

Evaluating the integral:

$\frac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}=\frac{A}{x+1}+\frac{Bx+C}{{{x}^{2}}+1}$ 

$\Rightarrow \int {{\sin }^{-1}}xdx=x{{\sin }^{-1}}x+\int \frac{tdt}{\sqrt{{{t}^{2}}}}$ 

$\Rightarrow \int {{\sin }^{-1}}xdx=x{{\sin }^{-1}}x+t+C$ 

$\Rightarrow \int {{\sin }^{-1}}xdx=x{{\sin }^{-1}}x+\sqrt{1-{{x}^{2}}}+C$ 

$\Rightarrow y=x{{\sin }^{-1}}x+\sqrt{1-{{x}^{2}}}+C$ 

Thus the general solution of given differential equation is $\text{y=xsi}{{\text{n}}^{\text{-1}}}\text{x+}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}\text{+C}$,


10. Find the general solution for ${{\text{e}}^{\text{x}}}\text{tanydx+}\left( \text{1-}{{\text{e}}^{\text{x}}} \right)\text{se}{{\text{c}}^{\text{2}}}\text{ydy=0}$.

Ans: The given differential equation is:

${{e}^{x}}\tan ydx+\left( 1-{{e}^{x}} \right){{\sec }^{2}}ydy=0$.

Simplify the expression:

$\left( 1-{{e}^{x}} \right){{\sec }^{2}}ydy=-{{e}^{x}}\tan ydx$ 

$\Rightarrow \frac{{{\sec }^{2}}y}{\tan y}dy=-\frac{{{e}^{x}}}{\left( 1-{{e}^{x}} \right)}dx$ 

Integrate both side:

$\int \frac{{{\sec }^{2}}y}{\tan y}dy=-\int \frac{{{e}^{x}}}{\left( 1-{{e}^{x}} \right)}dx$ ……(1)

Substitute $\text{tan y=u}$ 

$\tan y=u$ 

\[\Rightarrow {{\sec }^{2}}y=du\] 

Evaluating the LHS integral of (1):

$\Rightarrow \int \frac{{{\sec }^{2}}y}{\tan y}dy=\int \frac{du}{u}$ 

$\Rightarrow \int \frac{{{\sec }^{2}}y}{\tan y}dy=\log u$ 

$\Rightarrow \int \frac{{{\sec }^{2}}y}{\tan y}dy=\log \left( \tan y \right)$ 

Substitute $\text{1-}{{\text{e}}^{x}}\text{=v}$ 

$1-{{e}^{x}}=v$ 

$\Rightarrow -{{e}^{x}}dx=dv$ 

Evaluating the RHS integral of (1):

$\Rightarrow -\int \frac{{{e}^{x}}}{\left( 1-{{e}^{x}} \right)}dx=\int \frac{dv}{v}$ 

$\Rightarrow -\int \frac{{{e}^{x}}}{\left( 1-{{e}^{x}} \right)}dx=\log v$ 

$\Rightarrow -\int \frac{{{e}^{x}}}{\left( 1-{{e}^{x}} \right)}dx=\log \left( 1-{{e}^{x}} \right)$ 

Therefore the integral (1) will be:

$\log \left( \tan y \right)=\log \left( 1-{{e}^{x}} \right)+\log C$ 

$\Rightarrow \log \left( \tan y \right)=\log C\left( 1-{{e}^{x}} \right)$ 

$\Rightarrow \tan y=C\left( 1-{{e}^{x}} \right)$ 

Thus the general solution of given differential equation is $\text{tany=C}\left( \text{1-}{{\text{e}}^{\text{x}}} \right)$,


11. Find the particular solution of $\left( {{\text{x}}^{\text{3}}}\text{+}{{\text{x}}^{\text{2}}}\text{+x+1} \right)\frac{\text{dy}}{\text{dx}}\text{=2}{{\text{x}}^{\text{2}}}\text{+x;}\,\,\text{y=1,}\,\text{x=0}$ to satisfy the given condition.

Ans: The given differential equation is:

$\left( {{x}^{3}}+{{x}^{2}}+x+1 \right)\frac{dy}{dx}=2{{x}^{2}}+x;\,\,y=1,\,x=0$.

Simplify the expression:

$\frac{dy}{dx}=\frac{2{{x}^{2}}+x}{\left( {{x}^{3}}+{{x}^{2}}+x+1 \right)}$ 

$\Rightarrow \frac{dy}{dx}=\frac{2{{x}^{2}}+x}{\left( {{x}^{3}}+x+{{x}^{2}}+1 \right)}$ 

$\Rightarrow \frac{dy}{dx}=\frac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}$ 

$\Rightarrow dy=\frac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}dx$ 

Integrate both side:

$\int dy=\int \frac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}dx$ ……(1)

Use partial fraction method to simplify the RHS:


$\frac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}=\frac{A}{x+1}+\frac{Bx+C}{{{x}^{2}}+1}$

$\Rightarrow 2{{x}^{2}}+x=A\left( {{x}^{2}}+1 \right)+\left( Bx+C \right)\left( x+1 \right)$ 

\[\Rightarrow 2{{x}^{2}}+x=\left( A+B \right){{x}^{2}}+\left( B+C \right)x+\left( A+C \right)\] 

By comparing coefficients:

$A+B=2$ 

$B+C=1$ 

$A+C=0$ 

Solving this we get:

$\frac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}=\frac{\left( \frac{1}{2} \right)}{x+1}+\frac{\left( \frac{3}{2} \right)x+\left( -\frac{1}{2} \right)}{{{x}^{2}}+1}$ 

$\Rightarrow \frac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}=\frac{1}{2}\left( \frac{1}{x+1}+\frac{3x-1}{{{x}^{2}}+1} \right)$ 

Rewriting the  integral(1):

$y=\frac{1}{2}\int \left( \frac{1}{x+1}+\frac{3x-1}{{{x}^{2}}+1} \right)dx$ 

\[\Rightarrow y=\frac{1}{2}\int \frac{1}{x+1}dx+\frac{1}{2}\int \frac{3x-1}{{{x}^{2}}+1}dx\] 

\[\Rightarrow y=\frac{1}{2}\log \left( x+1 \right)+\frac{3}{2}\int \frac{x}{{{x}^{2}}+1}dx-\frac{1}{2}\int \frac{1}{{{x}^{2}}+1}dx\] 

\[\Rightarrow y=\frac{1}{2}\log \left( x+1 \right)+\frac{3}{4}\int \frac{2x}{{{x}^{2}}+1}dx-\frac{1}{2}{{\tan }^{-1}}x\] 

\[\Rightarrow y=\frac{1}{2}\log \left( x+1 \right)+\frac{3}{4}\log \left( {{x}^{2}}+1 \right)-\frac{1}{2}{{\tan }^{-1}}x+C\] 

For $y=1$ when $\text{x=0}$. 

\[1=\frac{1}{2}\log \left( 0+1 \right)+\frac{3}{4}\log \left( 0+1 \right)-\frac{1}{2}{{\tan }^{-1}}0+C\] 

\[\Rightarrow C=1\] 

Thus the required particular solution is :

$\Rightarrow y=\frac{1}{2}\log \left( x+1 \right)+\frac{3}{4}\log \left( {{x}^{2}}+1 \right)-\frac{1}{2}{{\tan }^{-1}}x+1$.


12. Find the particular solution of $\text{x}\left( {{\text{x}}^{\text{2}}}\text{-1} \right)\frac{\text{dy}}{\text{dx}}\text{=1}$; $\text{y=0}$ when $\text{x=2}$ to satisfy the given condition.

Ans: The given differential equation is:

$x\left( {{x}^{2}}-1 \right)\frac{dy}{dx}=1$; $y=0$ when $x=2$ 

Simplify the expression:

$x\left( {{x}^{2}}-1 \right)\frac{dy}{dx}=1$ 

$\Rightarrow dy=\frac{dx}{x\left( {{x}^{2}}-1 \right)}$ 

$\Rightarrow dy=\frac{dx}{x\left( x-1 \right)\left( x+1 \right)}$ 

Integrate both side:

$\int dy=\int \frac{dx}{x\left( x-1 \right)\left( x+1 \right)}$ ……(1)

Use partial fraction method to simplify the RHS:


$\frac{1}{x\left( x-1 \right)\left( x+1 \right)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$

$\Rightarrow 1=A\left( {{x}^{2}}-1 \right)+Bx\left( x+1 \right)+Cx\left( x-1 \right)$ 

\[\Rightarrow 1=\left( A+B+C \right){{x}^{2}}+\left( B-C \right)x-A\] 

By comparing coefficients:

$A+B+C=0$ 

$B-C=0$ 

$-A=1$ 

Solving this we get:

$\frac{1}{x\left( x-1 \right)\left( x+1 \right)}=\frac{\left( -1 \right)}{x}+\frac{\left( \frac{1}{2} \right)}{x-1}+\frac{\left( \frac{1}{2} \right)}{x+1}$ 

$\Rightarrow \frac{1}{x\left( x-1 \right)\left( x+1 \right)}=-\frac{1}{x}+\frac{1}{2}\left( \frac{1}{x-1}+\frac{1}{x+1} \right)$ 

Rewriting the  integral(1):

$y=\int \left( -\frac{1}{x}+\frac{1}{2}\left( \frac{1}{x-1}+\frac{1}{x+1} \right) \right)dx$ 

\[\Rightarrow y=-\int \frac{1}{x}dx+\frac{1}{2}\int \frac{1}{x-1}dx+\frac{1}{2}\int \frac{1}{x+1}dx\] 

\[\Rightarrow y=-\log x+\frac{1}{2}\log \left( x-1 \right)+\frac{1}{2}\log \left( x+1 \right)+\log C\] 

\[\Rightarrow y=-\frac{2}{2}\log x+\frac{1}{2}\log \left( x-1 \right)+\frac{1}{2}\log \left( x+1 \right)+\frac{2}{2}\log C\] 

\[\Rightarrow y=\frac{1}{2}\left( -\log {{x}^{2}}+\log \left( x-1 \right)+\log \left( x+1 \right)+\log {{C}^{2}} \right)\] 

\[\Rightarrow y=\frac{1}{2}\log \left[ \frac{{{C}^{2}}\left( {{x}^{2}}-1 \right)}{{{x}^{2}}} \right]\] 

For $y=0$ when $\text{x=2}$. 

\[0=\frac{1}{2}\log \left[ \frac{{{C}^{2}}\left( {{2}^{2}}-1 \right)}{{{2}^{2}}} \right]\] 

\[\Rightarrow 0=\log \left[ \frac{3{{C}^{2}}}{4} \right]\] 

$\Rightarrow \frac{3{{C}^{2}}}{4}=1$ 

$\Rightarrow {{C}^{2}}=\frac{4}{3}$ 

Thus the required particular solution is :

$y=\frac{1}{2}\log \left[ \frac{4\left( {{x}^{2}}-1 \right)}{3{{x}^{2}}} \right]$.


13. Find the particular solution of $\text{cos}\left( \frac{\text{dy}}{\text{dx}} \right)\text{=a}\,\,\left( \text{a}\in \text{R} \right)$; $\text{y=1}$ when $\text{x=0}$ to satisfy the given condition.

Ans: The given differential equation is:

$\cos \left( \frac{dy}{dx} \right)=a\,\,\left( a\in R \right)$; $y=1$ when $x=0$

Simplify the expression:

$\cos \left( \frac{dy}{dx} \right)=a\,$ 

$\Rightarrow \frac{dy}{dx}={{\cos }^{-1}}a$ 

$\Rightarrow dy={{\cos }^{-1}}adx$ 

Integrate both side:

$\int dy=\int {{\cos }^{-1}}adx$ 

$\Rightarrow y={{\cos }^{-1}}a\int dx$ 

\[\Rightarrow y=x{{\cos }^{-1}}a+C\] 

For $y=1$ when $\text{x=0}$. 

\[1=0{{\cos }^{-1}}a+C\] 

\[\Rightarrow C=1\] 

Thus the required particular solution is:

\[y=x{{\cos }^{-1}}a+1\].

\[\Rightarrow \frac{y-1}{x}={{\cos }^{-1}}a\]

\[\Rightarrow \cos \left( \frac{y-1}{x} \right)= a\]


14. Find the particular solution of $\frac{\text{dy}}{\text{dx}}\text{=ytanx}$; $\text{y=1}$ when $\text{x=0}$ to satisfy the given condition.

Ans: The given differential equation is:

$\frac{dy}{dx}=y\tan x$; $y=1$ when $x=0$

Simplify the expression:

$\frac{dy}{dx}=y\tan x$

$\Rightarrow \frac{dy}{y}=\tan xdx$ 

Integrate both side:

$\int \frac{dy}{y}=\int \tan xdx$ 

$\Rightarrow \log y=\log \left( \sec x \right)+\log C$ 

\[\Rightarrow \log y=\log \left( C\sec x \right)\] 

$\Rightarrow y=C\sec x$ 

For $\text{y=1}$ when $\text{x=0}$. 

\[1=C\sec 0\] 

\[\Rightarrow C=1\] 

Thus the required particular solution is :

\[y=\sec x\]. 


15. Find the equation of a curve passing through the point (0, 0) and whose differential equation is $\text{y }\!\!'\!\!\text{ =}{{\text{e}}^{\text{x}}}\text{sin x}$.

Ans: The given differential equation is:

$y'={{e}^{x}}\sin x$

The curve passes through $\left( 0,0 \right)$.

Simplify the expression:

$\Rightarrow \frac{dy}{dx}={{e}^{x}}\sin x$

$\Rightarrow dy={{e}^{x}}\sin xdx$ 

Integrate both side:

$\int dy=\int {{e}^{x}}\sin xdx$ 

Use product rules for integration of RHS. Let:

$I=\int {{e}^{x}}\sin xdx$ 

$\Rightarrow I=\sin x\int {{e}^{x}}dx-\int \left( \cos x\int {{e}^{x}}dx \right)dx$ 

$\Rightarrow I={{e}^{x}}\sin x-\int {{e}^{x}}\cos xdx$ 

$\Rightarrow I={{e}^{x}}\sin x-\left( \cos x\int {{e}^{x}}dx+\int \left( \sin x\int {{e}^{x}}dx \right)dx \right)$ 

$\Rightarrow I={{e}^{x}}\sin x-{{e}^{x}}\cos x-\int \left( {{e}^{x}}\sin x \right)dx$ 

$\Rightarrow I={{e}^{x}}\sin x-{{e}^{x}}\cos x-I$ 

$\Rightarrow I=\frac{{{e}^{x}}}{2}\left( \sin x-\cos x \right)$ 

Thus integral will be:

$y=\frac{{{e}^{x}}}{2}\left( \sin x-\cos x \right)+C$ 

Thus as the curve passes through $\left( \text{0,0} \right)$ 

$0=\frac{{{e}^{0}}}{2}\left( \sin 0-\cos 0 \right)+C$

$\Rightarrow 0=\frac{1}{2}\left( 0-1 \right)+C$ 

$\Rightarrow C=\frac{1}{2}$ 

Thus the equation of the curve will be:

$y=\frac{{{e}^{x}}}{2}\left( \sin x-\cos x \right)+\frac{1}{2}$ 

$\Rightarrow y=\frac{{{e}^{x}}}{2}\left( \sin x-\cos x+1 \right)$ 


16. For the differential equation $\text{xy}\frac{\text{dy}}{\text{dx}}\text{=}\left( \text{x+2} \right)\left( \text{y+2} \right)$ find the solution curve passing through the point $\left( \text{1,-1} \right)$.

Ans: The given differential equation is:

$xy\frac{dy}{dx}=\left( x+2 \right)\left( y+2 \right)$

The curve passes through $\left( 1,-1 \right)$.

Simplify the expression:

\[\Rightarrow \left( \frac{y}{y+2} \right)dy=\frac{\left( x+2 \right)}{x}dx\]

$\Rightarrow \left( 1-\frac{2}{y+2} \right)dy=\frac{\left( x+2 \right)}{x}dx$ 

Integrate both side:

$\int \left( 1-\frac{2}{y+2} \right)dy=\int \frac{\left( x+2 \right)}{x}dx$ 

$\Rightarrow \int dy-2\int \frac{1}{y+2}dy=\int \frac{x}{x}dx+\int \frac{2}{x}dx$ 

$\Rightarrow y-2\log \left( y+2 \right)=x+2\log x+C$ 

$\Rightarrow y-x=2\log \left( y+2 \right)+2\log x+C$ 

$\Rightarrow \Rightarrow y-x=2\log \left[ x\left( y+2 \right) \right]+C$ 

$\Rightarrow y-x=\log \left[ {{x}^{2}}{{\left( y+2 \right)}^{2}} \right]+C$ 

Thus as the curve passes through $\left( \text{1,-1} \right)$ 

$\Rightarrow -1-1=\log \left[ {{\left( 1 \right)}^{2}}{{\left( -1+2 \right)}^{2}} \right]+C$

$\Rightarrow -2=\log 1+C$ 

$\Rightarrow C=-2$ 

Thus the equation of the curve will be:

$y-x=\log \left[ {{x}^{2}}{{\left( y+2 \right)}^{2}} \right]-2$ 

$\Rightarrow y-x+2=\log \left( {{x}^{2}}{{\left( y+2 \right)}^{2}} \right)$ 


17. Find the equation of a curve passing through the point $\left( \text{0,-2} \right)$ given that at any point $\left( \text{x,y} \right)$ on the curve, the product of the slope of its tangent and $\text{y}$-coordinate of the point is equal to the $\text{x}$ -coordinate of the point.

Ans: According to , the equation is given by:

$y\frac{dy}{dx}=x$ 

The curve passes through $\left( 0,-2 \right)$.

Simplify the expression:

\[\Rightarrow ydy=xdx\] 

Integrate both side:

$\int ydy=\int xdx$ 

$\Rightarrow \frac{{{y}^{2}}}{2}=\frac{{{x}^{2}}}{2}+C$ 

$\Rightarrow {{y}^{2}}-{{x}^{2}}=2C$ 

Thus as the curve passes through $\left( \text{0,-2} \right)$ 

$\Rightarrow {{\left( -2 \right)}^{2}}-{{0}^{2}}=2C$

$\Rightarrow 4=2C$ 

$\Rightarrow C=2$ 

Thus the equation of the curve will be:

${{y}^{2}}-{{x}^{2}}=2\left( 2 \right)$ 

$\Rightarrow {{y}^{2}}-{{x}^{2}}=4$ .


18. At any point $\left( \text{x,y} \right)$ of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point$\left( \text{-4,-3} \right)$ . Find the equation of the curve given that it passes through $\left( \text{-2,1} \right)$.

Ans: Let the point of contact of the tangent be $\left( \text{x,y} \right)$.Then the slope of the segment joining point of contact and $\left( \text{-4,-3} \right)$:

$m=\frac{y+3}{x+4}$ 

According to  the for the slope of tangent $\frac{\text{dy}}{\text{dx}}$ it follows:

$\frac{dy}{dx}=2m$ 

$\Rightarrow \frac{dy}{dx}=2\left( \frac{y+3}{x+4} \right)$

Simplify the expression:

$\frac{dy}{dx}=2\left( \frac{y+3}{x+4} \right)$ 

$\frac{dy}{y+3}=\frac{2}{x+4}dx$ 

Integrate both side:

$\int \frac{dy}{y+3}=\int \frac{2}{x+4}dx$ 

$\Rightarrow \log \left( y+3 \right)=2\log \left( x+4 \right)+\log C$ 

\[\Rightarrow \log \left( y+3 \right)=\log {{\left( x+4 \right)}^{2}}+\log C\] 

\[\Rightarrow \log \left( y+3 \right)=\log C{{\left( x+4 \right)}^{2}}\] 

\[\Rightarrow y+3=C{{\left( x+4 \right)}^{2}}\] 

Thus as the curve passes through $\left( \text{-2,1} \right)$ 

\[1+3=C{{\left( -2+4 \right)}^{2}}\]

$\Rightarrow 4=4C$ 

$\Rightarrow C=1$ 

Thus the equation of the curve will be:

\[y+3={{\left( x+4 \right)}^{2}}\] .


19. The volume of spherical balloons being inflated changes at a constant rate. If initially its radius is $\text{3}$ units and after \[\text{3}\] seconds it is $\text{6}$ units. Find the radius of balloon after $\text{t}$ seconds.

Ans: Let the volume of spherical balloon be $\text{V}$ and its radius $\text{r}$. Let the rate of change of volume be $\text{k}$.

$\frac{dV}{dt}=k$ 

$\Rightarrow \frac{d}{dt}\left( \frac{4}{3}\pi {{r}^{3}} \right)=k$ 

$\Rightarrow \frac{4}{3}\pi \frac{d}{dt}\left( {{r}^{3}} \right)=k$ 

$\Rightarrow \frac{4}{3}\pi \left( 3{{r}^{2}} \right)\frac{dr}{dt}=k$ 

$\Rightarrow 4\pi {{r}^{2}}\frac{dr}{dt}=k$ 

$\Rightarrow 4\pi {{r}^{2}}dr=kdt$ 

Integrate both side:
$\int 4\pi {{r}^{2}}dr=\int kdt$ 

$\Rightarrow 4\pi \int {{r}^{2}}dr=kt+C$ 

$\Rightarrow \frac{4}{3}\pi {{r}^{3}}=kt+C$ 

At initial time, $\text{t=0}$ and $\text{r=3}$:

$\frac{4}{3}\pi {{3}^{3}}=k\left( 0 \right)+C$ 

$\Rightarrow C=36\pi $

At $\text{t=3}$ the radius $\text{r=6}$:

$\frac{4}{3}\pi \left( {{6}^{3}} \right)=k\left( 3 \right)+36\pi $ 

$\Rightarrow 3k=288\pi -36\pi $ 

$\Rightarrow k=84\pi $ 

Thus the radius-time relation can be given by:

$\frac{4}{3}\pi {{r}^{3}}=84\pi t+36\pi$ 

$\Rightarrow {{r}^{3}}=63t+27$ 

$\Rightarrow r={{\left( 63t+27 \right)}^{\frac{1}{3}}}$

The radius of balloon after $\text{t}$ seconds is given by: $r={{\left( 63t+27 \right)}^{\frac{1}{3}}}$ .

.

20. In a bank, principal increases continuously at the rate of $\text{r}$% per year. Find the value of $\text{r}$ if Rs. $\text{100}$doubles itself in $\text{10}$ years $\left( \text{lo}{{\text{g}}_{\text{e}}}\text{2=0}\text{.6931} \right)$.

Ans: Let the principal be $\text{p}$, according to :

$\frac{dp}{dt}=\left( \frac{r}{100} \right)p$ 

Simplify the expression:

$\frac{dp}{p}=\left( \frac{r}{100} \right)dt$ 

Integrate both side:
$\int \frac{dp}{p}=\int \left( \frac{r}{100} \right)dt$ 

$\Rightarrow \log p=\frac{rt}{100}+c$ 

\[\Rightarrow p={{e}^{\frac{rt}{100}+c}}\] 

\[\Rightarrow p=A{{e}^{\frac{rt}{100}}}\,\,\,\left( A={{e}^{c}} \right)\] 

At $\text{t=0}$, $\text{p=100}$:

\[100=A{{e}^{\frac{r\left( 0 \right)}{100}}}\] 

$\Rightarrow A=100$ 

Thus the principle and rate of interest relation:

$p=100{{e}^{\frac{rt}{100}}}$ 

At $\text{t=10}$, $\text{p=2}\times \text{100=200}$:

\[200=100{{e}^{\frac{r\left( 10 \right)}{100}}}\] 

$\Rightarrow 2={{e}^{\frac{r}{10}}}$ 

Take logarithm on both side:

$\log \left( {{e}^{\frac{r}{10}}} \right)=\log \left( 2 \right)$ 

$\Rightarrow \frac{r}{10}=0.6931$ 

$\Rightarrow r=6.931$ 

Thus the rate of interest $\text{r=6}\text{.931 }\!\!%\!\!\text{ }$. 


21. In a bank, principal increases continuously at the rate of $\text{5}$% per year. An amount of Rs $\text{1000}$ is deposited with this bank, how much will it worth after $\text{10}$  years $\left( {{\text{e}}^{\text{0}\text{.5}}}\text{=1}\text{.648} \right)$.

Ans: Let the principal be $\text{p}$, according to  principle increases at the rate of $\text{5}$% per year.

$\frac{dp}{dt}=\left( \frac{5}{100} \right)p$ 

Simplify the expression:

$\frac{dp}{dt}=\frac{p}{20}$ 

$\Rightarrow \frac{dp}{p}=\frac{1}{20}dt$ 

Integrate both side:

$\int \frac{dp}{p}=\int \frac{1}{20}dt$ 

$\Rightarrow \log p=\frac{t}{20}+C$ 

$\Rightarrow p={{e}^{\frac{t}{20}+C}}$ 

$\Rightarrow p=A{{e}^{\frac{t}{20}}}\,\,\left( A={{e}^{C}} \right)$
At $\text{t=0}$, $\text{p=1000}$: 

$1000=A{{e}^{\frac{0}{20}}}$ 

$\Rightarrow A=1000$ 

Thus the relation of principal and time relation:

$\Rightarrow p=1000{{e}^{\frac{t}{20}}}$ 

At $\text{t=10}$: 

$p=1000{{e}^{\frac{10}{20}}}$ 

$\Rightarrow p=1000{{e}^{0.5}}$

$\Rightarrow p=1000\times 1.648$ 

$\Rightarrow p=1648$  

Thus after $\text{10}$ this year the amount will become Rs. $\text{1648}$.


22. In a culture, the bacteria count is $\text{100000}$ . The number is increased by $\text{10 }\!\!%\!\!\text{ }$  in $\text{2}$ hours. In how many hours will the count reach $\text{200000}$, if the rate of growth of bacteria is proportional to the number present?

Ans: Let the number of bacteria be $\text{y}$ at time $\text{t}$. According to :

$\frac{dy}{dt}\propto y$ 

$\frac{dy}{dt}=cy$ 

Here $\text{c}$ is constant.

Simplify the expression:
$\frac{dy}{y}=cdt$ 

Integrate both side:
$\int \frac{dy}{y}=\int cdt$ 

$\Rightarrow \log y=ct+D$ 

$\Rightarrow y={{e}^{ct+D}}$ 

$\Rightarrow y=A{{e}^{ct}}\,\,\left( A={{e}^{D}} \right)$ 

At $\text{t=0}$, $\text{y=100000}$: 

$100000=A{{e}^{c\left( 0 \right)}}$ 

$\Rightarrow A=100000$ 

At $\text{t=2}$, $\text{y=}\frac{11}{10}\left( \text{100000} \right)=110000$: 

$y=100000{{e}^{ct}}$ 

$\Rightarrow 110000=100000{{e}^{c\left( 2 \right)}}$ 

$\Rightarrow {{e}^{2c}}=\frac{11}{10}$ 

$\Rightarrow 2c=\log \left( \frac{11}{10} \right)$ 

$\Rightarrow c=\frac{1}{2}\log \left( \frac{11}{10} \right)$ ……(1)

For $\text{y=200000}$:

$200000=100000{{e}^{ct}}$ 

$\Rightarrow {{e}^{ct}}=2$ 

$\Rightarrow ct=\log 2$ 

$\Rightarrow t=\frac{\log 2}{c}$ 

Back substituting using expression (1):
\[t=\frac{\log 2}{\frac{1}{2}\log \left( \frac{11}{10} \right)}\] 

$\Rightarrow t=\frac{2\log 2}{\log \left( \frac{11}{10} \right)}$ 

Thus time required for bacteria to reach $\text{200000}$ is \[\text{t=}\frac{\text{log2}}{\frac{\text{1}}{\text{2}}\text{log}\left( \frac{\text{11}}{\text{10}} \right)}\] hrs.


23. Find the general solution of the differential equation $\frac{\text{dy}}{\text{dx}}\text{=}{{\text{e}}^{\text{x+y}}}$.

(A)${{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{\text{-y}}}\text{=C}$ 

(B) ${{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{\text{y}}}\text{=C}$

(C) ${{\text{e}}^{\text{-x}}}\text{+}{{\text{e}}^{\text{y}}}\text{=C}$

(D) ${{\text{e}}^{\text{-x}}}\text{+}{{\text{e}}^{\text{-y}}}\text{=C}$

Ans: The given differential equation is $\frac{\text{dy}}{\text{dx}}\text{=}{{\text{e}}^{\text{x+y}}}$. Simplify the expression:

$\frac{dy}{dx}={{e}^{x}}{{e}^{y}}$ 

$\Rightarrow \frac{dy}{{{e}^{y}}}={{e}^{x}}dx$ 

$\Rightarrow {{e}^{-y}}dy={{e}^{x}}dx$ 

Integrate both side:

$\int {{e}^{-y}}dy=\int {{e}^{x}}dx$ 

$\Rightarrow -{{e}^{-y}}={{e}^{x}}+D$ 

$\Rightarrow {{e}^{x}}+{{e}^{-y}}=-D$ 

$\Rightarrow {{e}^{x}}+{{e}^{-y}}=C\,\,\left( C=-D \right)$ 

Thus the general solution of given differential equation is $ {{e}^{x}}+{{e}^{-y}}=C$ 

Thus the correct option is (A).


Exercise 9.5

1. Show that, differential equation $\left( {{\text{x}}^{\text{2}}}\text{+xy} \right)\text{dy=}\left( {{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}} \right)\text{dx}$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$\frac{dy}{dx}=\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}+xy}$ 

Checking for homogeneity:

$F\left( x,y \right)=\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}+xy}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{{{\left( \lambda x \right)}^{2}}+{{\left( \lambda y \right)}^{2}}}{{{\left( \lambda x \right)}^{2}}+\left( \lambda x \right)\left( \lambda y \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{{{\lambda }^{2}}\left( {{x}^{2}}+{{y}^{2}} \right)}{{{\lambda }^{2}}\left( {{x}^{2}}+xy \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}+xy}$ 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=\frac{{{x}^{2}}+{{\left( vx \right)}^{2}}}{{{x}^{2}}+x\left( vx \right)}$ 

$\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{{{x}^{2}}\left( 1+{{v}^{2}} \right)}{{{x}^{2}}\left( 1+v \right)}$ 

$\Rightarrow v+x\frac{dv}{dx}=\frac{1+{{v}^{2}}}{1+v}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{1+{{v}^{2}}}{1+v}-v$ 

$\Rightarrow x\frac{dv}{dx}=\frac{1+{{v}^{2}}-v-{{v}^{2}}}{1+v}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{1-v}{1+v}$ 

Separate the differentials:

$\frac{1+v}{1-v}dv=\frac{dx}{x}$ 

Integrate both side:

$\int \frac{1+v}{1-v}dv=\int \frac{dx}{x}$ 

$\Rightarrow \int \frac{2-\left( 1-v \right)}{1-v}dv=\log x-\log k$ 

$\Rightarrow \int \frac{2}{1-v}dv-\int \frac{1-v}{1-v}dv=\log \frac{x}{k}$ 

$\Rightarrow -2\log \left( 1-v \right)-\int dv=\log \frac{x}{k}$ 

$\Rightarrow -2\log \left( 1-v \right)-v=\log \frac{x}{k}$ 

$\Rightarrow v=-\log \frac{x}{k}-2\log \left( 1-v \right)$ 

$\Rightarrow v=\log \left( \frac{k}{x{{\left( 1-v \right)}^{2}}} \right)$ 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

$\Rightarrow \frac{y}{x}=\log \left( \frac{k}{x{{\left( 1-\frac{y}{x} \right)}^{2}}} \right)$ 

$\Rightarrow \frac{y}{x}=\log \left( \frac{kx}{{{\left( x-y \right)}^{2}}} \right)$ 

$\Rightarrow {{e}^{\frac{y}{x}}}=\frac{kx}{{{\left( x-y \right)}^{2}}}$

$\Rightarrow {{\left( x-y \right)}^{2}}=kx{{e}^{-\frac{y}{x}}}$  

 The solution of the given differential equation ${{\left( \text{x-y} \right)}^{\text{2}}}\text{=kx}{{\text{e}}^{\text{-}\frac{\text{y}}{\text{x}}}}$ .


2. Show that, differential equation $\text{y }\!\!'\!\!\text{ =}\frac{\text{x+y}}{\text{x}}$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$\frac{dy}{dx}=\frac{x+y}{x}$ 

Checking for homogeneity:

.$F\left( x,y \right)=\frac{x+y}{x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda x+\lambda y}{\lambda x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda \left( x+y \right)}{\lambda \left( x \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{x+y}{x}$ 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=\frac{x+\left( vx \right)}{x}$ 

$\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{x\left( 1+v \right)}{x}$ 

$\Rightarrow v+x\frac{dv}{dx}=1+v$ 

$\Rightarrow x\frac{dv}{dx}=1$ 

Separate the differentials:

$dv=\frac{dx}{x}$ 

Integrate both side:

$\int dv=\int \frac{dx}{x}$ 

$\Rightarrow \int dv=\log x+\log k$ 

$\Rightarrow v=\log kx$ 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

$\Rightarrow \frac{y}{x}=\log kx$ 

$\Rightarrow y=x\log kx$  

 The solution of the given differential equation $\text{y=x log kx}$ 


3. Show that, differential equation $\left( \text{x-y} \right)\text{dy-}\left( \text{x+y} \right)\text{dx=0}$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$\frac{dy}{dx}=\frac{x+y}{x-y}$ 

Checking for homogeneity:

$F\left( x,y \right)=\frac{x+y}{x-y}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda x+\lambda y}{\lambda x-\lambda y}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda \left( x+y \right)}{\lambda \left( x-y \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{x+y}{x-y}$ 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=\frac{x+\left( vx \right)}{x-vx}$ 

$\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{x\left( 1+v \right)}{x\left( 1-v \right)}$ 

$\Rightarrow v+x\frac{dv}{dx}=\frac{1+v}{1-v}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{1+v}{1-v}-v$

$\Rightarrow x\frac{dv}{dx}=\frac{1+v-v+{{v}^{2}}}{1-v}$  

$\Rightarrow x\frac{dv}{dx}=\frac{1+{{v}^{2}}}{1-v}$ 

Separate the differentials:

$\frac{1-v}{1+{{v}^{2}}}dv=\frac{dx}{x}$ 

Integrate both side:

$\int \frac{1-v}{1+{{v}^{2}}}dv=\int \frac{dx}{x}$ 

$\Rightarrow \int \frac{1-v}{1+{{v}^{2}}}dv=\int \frac{dx}{x}$ 

$\Rightarrow \int \frac{1}{1+{{v}^{2}}}dv-\int \frac{v}{1+{{v}^{2}}}dv=\log x+C$ 

$\Rightarrow {{\tan }^{-1}}v-\frac{1}{2}\int \frac{2v}{1+{{v}^{2}}}dv=\log x+C$ 

$\Rightarrow {{\tan }^{-1}}v-\frac{1}{2}\log \left( 1+{{v}^{2}} \right)=\log x+C$ 

$\Rightarrow {{\tan }^{-1}}v=\log x+\frac{1}{2}\log \left( 1+{{v}^{2}} \right)+C$ 

$\Rightarrow {{\tan }^{-1}}v=\frac{1}{2}\log {{x}^{2}}+\frac{1}{2}\log \left( 1+{{v}^{2}} \right)+C$ 

$\Rightarrow {{\tan }^{-1}}v=\frac{1}{2}\log \left[ {{x}^{2}}\left( 1+{{v}^{2}} \right) \right]+C$ 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

$\Rightarrow {{\tan }^{-1}}\frac{y}{x}=\frac{1}{2}\log \left[ {{x}^{2}}\left( 1+\frac{{{y}^{2}}}{{{x}^{2}}} \right) \right]+C$ 

$\Rightarrow {{\tan }^{-1}}\frac{y}{x}=\frac{1}{2}\log \left( {{x}^{2}}+{{y}^{2}} \right)+C$  

The solution of the given differential equation $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{y}}{\text{x}}\text{=}\frac{\text{1}}{\text{2}}\text{log}\left( {{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}} \right)\text{+C}$.


4. Show that, differential equation $\left( {{\text{x}}^{2}}\text{-}{{\text{y}}^{2}} \right)\text{dx+2xy dy=0}$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$\frac{dy}{dx}=-\frac{{{x}^{2}}-{{y}^{2}}}{2xy}$ 

Checking for homogeneity:

$F\left( x,y \right)=-\frac{{{x}^{2}}-{{y}^{2}}}{2xy}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{{{\left( \lambda x \right)}^{2}}-{{\left( \lambda y \right)}^{2}}}{2\left( \lambda x \right)\left( \lambda y \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{{{\lambda }^{2}}{{x}^{2}}-{{\lambda }^{2}}{{y}^{2}}}{2{{\lambda }^{2}}xy}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{{{\lambda }^{2}}\left( {{x}^{2}}-{{y}^{2}} \right)}{{{\lambda }^{2}}\left( 2xy \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{{{x}^{2}}-{{y}^{2}}}{2xy}$ 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=-\frac{{{x}^{2}}-{{\left( vx \right)}^{2}}}{2x\left( vx \right)}$ 

$\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{{{x}^{2}}\left( {{v}^{2}}-1 \right)}{{{x}^{2}}\left( 2v \right)}$ 

$\Rightarrow v+x\frac{dv}{dx}=\frac{{{v}^{2}}-1}{2v}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{{{v}^{2}}-1}{2v}-v$

$\Rightarrow x\frac{dv}{dx}=\frac{{{v}^{2}}-1-2{{v}^{2}}}{2v}$  

$\Rightarrow x\frac{dv}{dx}=-\frac{1+{{v}^{2}}}{2v}$ 

Separate the differentials:

$\frac{2v}{1+{{v}^{2}}}dv=-\frac{dx}{x}$ 

Integrate both side:

$\int \frac{2v}{1+{{v}^{2}}}dv=-\int \frac{dx}{x}$ 

$\Rightarrow \log \left( 1+{{v}^{2}} \right)=-\log x+C$ 

$\Rightarrow \log \left( 1+{{v}^{2}} \right)+\log x=C$ 

$\Rightarrow \log \left[ x\left( 1+{{v}^{2}} \right) \right]=C$

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

$\Rightarrow \log \left[ x\left( 1+\frac{{{y}^{2}}}{{{x}^{2}}} \right) \right]=C$ 

$\Rightarrow \left( \frac{{{x}^{2}}+{{y}^{2}}}{x} \right)=k\,\,k={{e}^{C}}$  

$\Rightarrow {{x}^{2}}+{{y}^{2}}=kx$ 

 The solution of the given differential equation ${{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}}\text{=kx}$.


5. Show that, differential equation ${{\text{x}}^{\text{2}}}\frac{\text{dy}}{\text{dx}}\text{=}{{\text{x}}^{\text{2}}}\text{-2}{{\text{y}}^{\text{2}}}\text{+xy}$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

\[\frac{dy}{dx}=\frac{{{x}^{2}}-2{{y}^{2}}+xy}{{{x}^{2}}}\] 

Checking for homogeneity:

$F\left( x,y \right)=\frac{{{x}^{2}}-2{{y}^{2}}+xy}{{{x}^{2}}}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{{{\lambda }^{2}}{{x}^{2}}-2{{\lambda }^{2}}{{y}^{2}}+\left( \lambda x \right)\left( \lambda y \right)}{{{\lambda }^{2}}{{x}^{2}}}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{{{\lambda }^{2}}\left( {{x}^{2}}-2{{y}^{2}}+xy \right)}{{{\lambda }^{2}}\left( {{x}^{2}} \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{{{x}^{2}}-2{{y}^{2}}+xy}{{{x}^{2}}}$ 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=\frac{{{x}^{2}}-2{{\left( vx \right)}^{2}}+x\left( vx \right)}{{{x}^{2}}}$ 

$\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{{{x}^{2}}\left( 1-2{{v}^{2}}+v \right)}{{{x}^{2}}}$ 

$\Rightarrow v+x\frac{dv}{dx}=1-2{{v}^{2}}+v$ 

$\Rightarrow x\frac{dv}{dx}=1-2{{v}^{2}}$ 

Separate the differentials:

$\frac{1}{1-2{{v}^{2}}}dv=\frac{dx}{x}$ 

Integrate both side:

$\int \frac{1}{1-2{{v}^{2}}}dv=\int \frac{dx}{x}$ 

$\Rightarrow \frac{1}{2}\int \frac{1}{\frac{1}{2}-{{v}^{2}}}dv=\log x+C$ 

$\Rightarrow \frac{1}{2}\int \frac{1}{{{\left( \frac{1}{\sqrt{2}} \right)}^{2}}-{{v}^{2}}}dv=\log x+C$ 

$\Rightarrow \frac{1}{2}\left( \frac{1}{2\times \frac{1}{\sqrt{2}}} \right)\log \frac{\left| \frac{1}{\sqrt{2}}+v \right|}{\left| \frac{1}{\sqrt{2}}-v \right|}=\log x+C$

$\Rightarrow \frac{1}{2\sqrt{2}}\log \frac{\left| 1+\sqrt{2}v \right|}{\left| 1-\sqrt{2}v \right|}=\log x+C$ 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:s

$\Rightarrow \frac{1}{2\sqrt{2}}\log \frac{\left| 1+\sqrt{2}\left( \frac{y}{x} \right) \right|}{\left| 1-\sqrt{2}\left( \frac{y}{x} \right) \right|}=\log x+C$ 

$\Rightarrow \frac{1}{2\sqrt{2}}\log \frac{\left| x+\sqrt{2}y \right|}{\left| x-\sqrt{2}y \right|}=\log x+C$ 

 The solution of the given differential equation $\frac{\text{1}}{\text{2}\sqrt{\text{2}}}\text{log}\frac{\left| \text{x+}\sqrt{\text{2}}\text{y} \right|}{\left| \text{x-}\sqrt{\text{2}}\text{y} \right|}\text{=logx+C}$.


6. Show that, differential equation $\text{xdy-ydx=}\sqrt{{{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}}}\text{dx}$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$xdy=\sqrt{{{x}^{2}}+{{y}^{2}}}dx+ydx$ 

$\Rightarrow \frac{dy}{dx}=\frac{\sqrt{{{x}^{2}}+{{y}^{2}}}+y}{x}$ 

Checking for homogeneity:

$F\left( x,y \right)=\frac{\sqrt{{{x}^{2}}+{{y}^{2}}}+y}{x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\sqrt{{{\lambda }^{2}}{{x}^{2}}+{{\lambda }^{2}}{{y}^{2}}}+\lambda y}{\lambda x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\sqrt{{{\lambda }^{2}}\left( {{x}^{2}}+{{y}^{2}} \right)}+\lambda y}{\lambda x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda \left( \sqrt{{{x}^{2}}+{{y}^{2}}}+y \right)}{\lambda \left( x \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\sqrt{{{x}^{2}}+{{y}^{2}}}+y}{x}$ 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=\frac{\sqrt{{{x}^{2}}+{{\left( vx \right)}^{2}}}+vx}{x}$ 

$\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{x\left( \sqrt{1+{{v}^{2}}}+v \right)}{x}$ 

$\Rightarrow v+x\frac{dv}{dx}=\sqrt{1+{{v}^{2}}}+v$ 

$\Rightarrow x\frac{dv}{dx}=\sqrt{1+{{v}^{2}}}$ 

Separate the differentials:

$\frac{1}{\sqrt{1+{{v}^{2}}}}dv=\frac{dx}{x}$ 

Integrate both side:

$\int \frac{1}{\sqrt{1+{{v}^{2}}}}dv=\int \frac{dx}{x}$ 

$\Rightarrow \log \left| v+\sqrt{1+{{v}^{2}}} \right|=\log x+\log C$ 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:s

$\Rightarrow \log \left| \frac{y}{x}+\sqrt{1+\frac{{{y}^{2}}}{{{x}^{2}}}} \right|=\log Cx$ 

$\Rightarrow \log \left| \frac{y+\sqrt{{{y}^{2}}+{{x}^{2}}}}{x} \right|=\log Cx$ 

$y+\sqrt{{{y}^{2}}+{{x}^{2}}}=C{{x}^{2}}$ 

The solution of the given differential equation $\text{y+}\sqrt{{\text{y}}^{\text{2}}}+{{\text{x}}^{\text{2}}}}\text{=C}{{\text{x}}^{\text{2}}}$.


7. Show that, differential equation $\left\{ \text{xcos}\left( \frac{\text{y}}{\text{x}} \right)\text{+ysin}\left( \frac{\text{y}}{\text{x}} \right) \right\}\text{ydx=}\left\{ \text{ysin}\left( \frac{\text{y}}{\text{x}} \right)\text{-xcos}\left( \frac{\text{y}}{\text{x}} \right) \right\}\text{xdy}$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$\left\{ x\cos \left( \frac{y}{x} \right)+y\sin \left( \frac{y}{x} \right) \right\}ydx=\left\{ y\sin \left( \frac{y}{x} \right)-x\cos \left( \frac{y}{x} \right) \right\}xdy$ 

$\frac{dy}{dx}=\frac{\left\{ x\cos \left( \frac{y}{x} \right)+y\sin \left( \frac{y}{x} \right) \right\}y}{\left\{ y\sin \left( \frac{y}{x} \right)-x\cos \left( \frac{y}{x} \right) \right\}x}$ 

Checking for homogeneity:

$F\left( x,y \right)=\frac{\left\{ x\cos \left( \frac{y}{x} \right)+y\sin \left( \frac{y}{x} \right) \right\}y}{\left\{ y\sin \left( \frac{y}{x} \right)-x\cos \left( \frac{y}{x} \right) \right\}x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\left\{ \lambda x\cos \left( \frac{\lambda y}{\lambda x} \right)+y\sin \left( \frac{\lambda y}{\lambda x} \right) \right\}\lambda y}{\left\{ \lambda y\sin \left( \frac{\lambda y}{\lambda x} \right)-\lambda x\cos \left( \frac{\lambda y}{\lambda x} \right) \right\}\lambda x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{{{\lambda }^{2}}\left\{ x\cos \left( \frac{y}{x} \right)+y\sin \left( \frac{y}{x} \right) \right\}y}{{{\lambda }^{2}}\left\{ y\sin \left( \frac{y}{x} \right)-x\cos \left( \frac{y}{x} \right) \right\}x}$ 

\[\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\left\{ x\cos \left( \frac{y}{x} \right)+y\sin \left( \frac{y}{x} \right) \right\}y}{\left\{ y\sin \left( \frac{y}{x} \right)-x\cos \left( \frac{y}{x} \right) \right\}x}\] 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

\[\frac{d\left( vx \right)}{dx}=\frac{\left\{ x\cos \left( \frac{xv}{x} \right)+vx\sin \left( \frac{vx}{x} \right) \right\}vx}{\left\{ vx\sin \left( \frac{vx}{x} \right)-x\cos \left( \frac{vx}{x} \right) \right\}x}\] 

$\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{{{x}^{2}}\left\{ \cos v+v\sin v \right\}v}{{{x}^{2}}\left\{ v\sin v-\cos v \right\}}$ 

$\Rightarrow v+x\frac{dv}{dx}=\frac{v\cos v+{{v}^{2}}\sin v}{v\sin v-\cos v}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{v\cos v+{{v}^{2}}\sin v}{v\sin v-\cos v}-v$ 

$\Rightarrow x\frac{dv}{dx}=\frac{v\cos v+{{v}^{2}}\sin v-{{v}^{2}}\sin v+v\cos v}{v\sin v-\cos v}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{2v\cos v}{v\sin v-\cos v}$ 

Separate the differentials:

$\frac{v\sin v-\cos v}{v\cos v}dv=2\frac{dx}{x}$ 

Integrate both side:

$\int \frac{v\sin v-\cos v}{v\cos v}dv=2\int \frac{dx}{x}$ 

$\Rightarrow \int \frac{v\sin v}{v\cos v}dv-\int \frac{\cos v}{v\cos v}dv=2\int \frac{dx}{x}$

$\Rightarrow \int \tan vdv-\int \frac{1}{v}dv=2\log \left| x \right|+\log C$ 

$\Rightarrow \log \left| \sec v \right|-\log \left| v \right|=\log C{{\left| x \right|}^{2}}$

$\Rightarrow \log \frac{\left| \sec v \right|}{\left| v \right|}=\log C{{\left| x \right|}^{2}}$ 

$\Rightarrow \sec v=Cv{{x}^{2}}$   

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:s

$\Rightarrow \sec \frac{y}{x}=C\left( \frac{y}{x} \right){{x}^{2}}$ 

$\Rightarrow \cos \frac{y}{x}=\frac{k}{xy}\,\,k=\frac{1}{C}$ 

The solution of the given differential equation $\text{cos}\frac{\text{y}}{\text{x}}\text{=}\frac{\text{k}}{\text{xy}}$.


8. Show that, differential equation $\text{x}\frac{\text{dy}}{\text{dx}}\text{-y+xsin}\left( \frac{\text{y}}{\text{x}} \right)\text{=0}$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$x\frac{dy}{dx}-y+x\sin \left( \frac{y}{x} \right)=0$ 

$\frac{dy}{dx}=\frac{y-x\sin \left( \frac{y}{x} \right)}{x}$ 

Checking for homogeneity:

$F\left( x,y \right)=\frac{y-x\sin \left( \frac{y}{x} \right)}{x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda y-\lambda x\sin \left( \frac{\lambda y}{\lambda x} \right)}{\lambda x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda \left( y-x\sin \left( \frac{y}{x} \right) \right)}{\lambda x}$ 

\[\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{y-x\sin \left( \frac{y}{x} \right)}{x}\] 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

\[\frac{d\left( vx \right)}{dx}=\frac{vx-x\sin \left( \frac{vx}{x} \right)}{x}\] 

$\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{x\left( v-\sin v \right)}{x}$ 

$\Rightarrow v+x\frac{dv}{dx}=v-\sin v$ 

$\Rightarrow x\frac{dv}{dx}=-\sin v$ 

Separate the differentials:

$\frac{1}{\sin v}dv=-\frac{dx}{x}$ 

Integrate both side:

$\int \text{cosec}\,vdv=-\int \frac{dx}{x}$ 

$\Rightarrow \log \left| \text{cosec}\,v-\cot v \right|=-\log x+\log C$

$\Rightarrow \log \left| \text{cosec}\,v-\cot v \right|=\log \frac{C}{x}$ 

$\Rightarrow \text{cosec}\,v-\cot v=\frac{C}{x}$

$\Rightarrow \frac{1}{\sin v}-\frac{\cos v}{\sin v}=\frac{C}{x}$ 

$\Rightarrow 1-\cos v=\frac{C}{x}\sin v$   

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

$\Rightarrow 1-\cos \frac{y}{x}=\frac{C}{x}\sin \frac{y}{x}$ 

$\Rightarrow x\left( 1-\cos \frac{y}{x} \right)=C\sin \left( \frac{y}{x} \right)$ 

The solution of the given differential equation $\text{x}\left( \text{1-cos}\frac{\text{y}}{\text{x}} \right)\text{=Csin}\left( \frac{\text{y}}{\text{x}} \right)$.


9. Show that, differential equation $\text{ydx+xlog}\left( \frac{\text{y}}{\text{x}} \right)\text{dy-2xdy=0}$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$ydx=2xdy-x\log \left( \frac{y}{x} \right)dy$ 

$\frac{dy}{dx}=\frac{y}{2x-x\log \left( \frac{y}{x} \right)}$ 

Checking for homogeneity:

$F\left( x,y \right)=\frac{y}{2x-x\log \left( \frac{y}{x} \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda y}{2\lambda x-\lambda x\log \left( \frac{\lambda y}{\lambda x} \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda y}{\lambda \left( 2x-x\log \left( \frac{y}{x} \right) \right)}$ 

\[\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{y}{2x-x\log \left( \frac{y}{x} \right)}\] 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

\[\frac{d\left( vx \right)}{dx}=\frac{vx}{2x-x\log \left( \frac{vx}{x} \right)}\] 

$\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{xv}{x\left( 2-\log \left( v \right) \right)}$ 

$\Rightarrow v+x\frac{dv}{dx}=\frac{v}{2-\log \left( v \right)}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{v}{2-\log v}-v$

$\Rightarrow x\frac{dv}{dx}=\frac{v-2v+v\log v}{2-\log v}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{v\log v-v}{2-\log v}$  

Separate the differentials:

$\frac{2-\log v}{v\log v-v}dv=\frac{dx}{x}$ 

Integrate both side:

$\int \frac{2-\log v}{v\log v-v}dv=\int \frac{dx}{x}$ 

$\int \frac{1+1-\log v}{v\left( \log v-v \right)}dv=\log x+\log C$

$\Rightarrow \int \frac{1}{v\left( \log v-1 \right)}dv+\int \frac{1-\log v}{v\left( \log v-1 \right)}dv=\log x+\log C$ 

$\Rightarrow \int \frac{1}{v\left( \log v-1 \right)}dv-\int \frac{1}{v}dv=\log Cx$ ……(1)

Solving :

$\int \frac{1}{v\left( \log v-1 \right)}dv$

Substitute $\text{log v-1=t}$:

$\log v-1=t$   

$\Rightarrow \frac{1}{v}dv=dt$   

Thus the integral will be:

\[\Rightarrow \int \frac{1}{v\left( \log v-1 \right)}dv=\int \frac{dt}{t}\] 

\[\Rightarrow \int \frac{1}{v\left( \log v-1 \right)}dv=\log t\] 

\[\Rightarrow \int \frac{1}{v\left( \log v-1 \right)}dv=\log \left( \log v-1 \right)\] 

Using above result for solving (1):

$\Rightarrow \log \left( \log v-1 \right)-\log v=\log Cx$ 

$\Rightarrow \log \frac{\log v-1}{v}=\log Cx$ 

$\Rightarrow \frac{\log v-1}{v}=Cx$ 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

$\Rightarrow \frac{\log \frac{y}{x}-1}{\frac{y}{x}}=Cx$ 

$\Rightarrow \log \frac{y}{x}-1=Cx\left( \frac{y}{x} \right)$ 

$\Rightarrow \log \frac{y}{x}-1=Cy$ 

 The solution of the given differential equation $\text{log}\frac{\text{y}}{\text{x}}\text{-1=Cy}$.


10. Show that, differential equation $\left( \text{1+}{{\text{e}}^{\frac{\text{x}}{\text{y}}}} \right)\text{dx+}{{\text{e}}^{\frac{\text{x}}{\text{y}}}}\left( \text{1-}\frac{\text{x}}{\text{y}} \right)\text{dy=0}$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$\left( 1+{{e}^{\frac{x}{y}}} \right)dx=-{{e}^{\frac{x}{y}}}\left( 1-\frac{x}{y} \right)dy$ 

$\frac{dx}{dy}=-\frac{{{e}^{\frac{x}{y}}}\left( 1-\frac{x}{y} \right)}{\left( 1+{{e}^{\frac{x}{y}}} \right)}$ 

Checking for homogeneity:

$F\left( x,y \right)=-\frac{{{e}^{\frac{x}{y}}}\left( 1-\frac{x}{y} \right)}{\left( 1+{{e}^{\frac{x}{y}}} \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{{{e}^{\frac{\lambda x}{\lambda y}}}\left( 1-\frac{\lambda x}{\lambda y} \right)}{\left( 1+{{e}^{\frac{\lambda x}{\lambda y}}} \right)}$ 

\[\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{{{e}^{\frac{x}{y}}}\left( 1-\frac{x}{y} \right)}{\left( 1+{{e}^{\frac{x}{y}}} \right)}\] 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{x=vy}$:

\[\frac{d\left( vy \right)}{dy}=-\frac{{{e}^{\frac{vy}{y}}}\left( 1-\frac{vy}{y} \right)}{1+{{e}^{\frac{vy}{y}}}}\] 

$\Rightarrow v\frac{dy}{dy}+y\frac{dv}{dy}=-\frac{{{e}^{v}}\left( 1-v \right)}{1+{{e}^{v}}}$ 

$\Rightarrow v+y\frac{dv}{dy}=-\frac{{{e}^{v}}\left( 1-v \right)}{1+{{e}^{v}}}$ 

$\Rightarrow y\frac{dv}{dy}=-\frac{{{e}^{v}}\left( 1-v \right)}{1+{{e}^{v}}}-v$

$\Rightarrow y\frac{dv}{dy}=\frac{-{{e}^{v}}+v{{e}^{v}}-v-v{{e}^{v}}}{1+{{e}^{v}}}$ 

$\Rightarrow y\frac{dv}{dy}=\frac{-\left( {{e}^{v}}+v \right)}{1+{{e}^{v}}}$  

Separate the differentials:

$\frac{1+{{e}^{v}}}{{{e}^{v}}+v}dv=-\frac{dy}{y}$ 

Integrate both side:

$\int \frac{1+{{e}^{v}}}{{{e}^{v}}+v}dv=-\int \frac{dy}{y}$ 

\[\int \frac{{{e}^{v}}+1}{{{e}^{v}}+v}dv=-\log y+\log C\] ……(1)

Solving the LHS integral. Substitute ${{\text{e}}^{\text{v}}}\text{+v=t}$:

\[{{e}^{v}}+v=t\] 

$\Rightarrow \left( {{e}^{v}}+1 \right)dv=dt$ 

Solving the expression (1):

\[\Rightarrow \int \frac{1}{t}dt=\log \frac{C}{y}\] 

\[\Rightarrow \log \left( t \right)=\log \frac{C}{y}\]

\[\Rightarrow \log \left( {{e}^{v}}+v \right)=\log \frac{C}{y}\] 

\[\Rightarrow {{e}^{v}}+v=\frac{C}{y}\] 

Back substitute $\text{v=}\frac{\text{x}}{\text{y}}$:

\[\Rightarrow {{e}^{\frac{x}{y}}}+\frac{x}{y}=\frac{C}{y}\] 

\[\Rightarrow y{{e}^{\frac{x}{y}}}+x=C\] 

The solution of the given differential equation \[\text{y}{{\text{e}}^{\frac{\text{x}}{\text{y}}}}\text{+x=C}\].


11. For the differential equation $\left( \text{x+y} \right)\text{dy+}\left( \text{x-y} \right)\text{dx=0}$.Find the particular solution for the condition $\text{y=1}$ when $\text{x=1}$.

Ans: Given differential equation is:

$\left( x+y \right)dy+\left( x-y \right)dx=0$

$\Rightarrow \frac{dy}{dx}=-\frac{x-y}{x+y}$ 

Checking for homogeneity:

$F\left( x,y \right)=-\frac{x-y}{x+y}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{\lambda x-\lambda y}{\lambda x+\lambda y}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{\lambda \left( x-y \right)}{\lambda \left( x+y \right)}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{x-y}{x+y}$ 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=-\frac{x-\left( vx \right)}{x+\left( vx \right)}$ 

$\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{x\left( v-1 \right)}{x\left( v+1 \right)}$ 

$\Rightarrow v+x\frac{dv}{dx}=\frac{v-1}{v+1}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{v-1}{v+1}-v$ 

$\Rightarrow x\frac{dv}{dx}=\frac{v-1-{{v}^{2}}-v}{v+1}$ 

$\Rightarrow x\frac{dv}{dx}=-\frac{1+{{v}^{2}}}{v+1}$ 

Separate the differentials:

$\frac{v+1}{1+{{v}^{2}}}dv=-\frac{dx}{x}$ 

Integrate both side:

$\int \frac{v+1}{1+{{v}^{2}}}dv=-\int \frac{dx}{x}$ 

$\Rightarrow \int \frac{v}{1+{{v}^{2}}}dv+\int \frac{1}{1+{{v}^{2}}}dv=-\log x+k$ 

$\Rightarrow \frac{1}{2}\log \left( 1+{{v}^{2}} \right)+{{\tan }^{-1}}v+\log x=k$ 

$\Rightarrow \frac{1}{2}\log \left[ x\left( 1+{{v}^{2}} \right) \right]+{{\tan }^{-1}}v=k$ 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

$\Rightarrow \frac{1}{2}\log \left[ x\left( 1+\frac{{{y}^{2}}}{{{x}^{2}}} \right) \right]+{{\tan }^{-1}}\frac{y}{x}=k$ 

$\Rightarrow \frac{1}{2}\log \left[ \frac{{{x}^{2}}+{{y}^{2}}}{x} \right]+{{\tan }^{-1}}\frac{y}{x}=k$  

 Now $\text{y=1}$ and $\text{x=1}$:

$\Rightarrow \frac{1}{2}\log \left[ \frac{{{1}^{2}}+{{1}^{2}}}{1} \right]+{{\tan }^{-1}}\frac{1}{1}=k$  

$k=\frac{1}{2}\log 2+\frac{\pi }{4}$ 

The required particular solution:

$\Rightarrow \frac{1}{2}\log \left[ \frac{{{x}^{2}}+{{y}^{2}}}{x} \right]+{{\tan }^{-1}}\frac{y}{x}=\frac{1}{2}\log 2+\frac{\pi }{4}$ 


12. For the differential equation ${{\text{x}}^{2}}\text{dy+}\left( \text{xy+}{{\text{y}}^{2}} \right)\text{dx=0}$.Find the particular solution for the condition $\text{y=1}$ when $\text{x=1}$.

Ans: Given differential equation is:

${{x}^{2}}dy+\left( xy+{{y}^{2}} \right)dx=0$

$\Rightarrow \frac{dy}{dx}=-\frac{xy+{{y}^{2}}}{{{x}^{2}}}$ 

Checking for homogeneity:

$F\left( x,y \right)=-\frac{xy+{{y}^{2}}}{{{x}^{2}}}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{\left( \lambda x \right)\left( \lambda y \right)+{{\lambda }^{2}}{{y}^{2}}}{{{\lambda }^{2}}{{x}^{2}}}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{{{\lambda }^{2}}\left( xy+{{y}^{2}} \right)}{{{\lambda }^{2}}{{x}^{2}}}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{xy+{{y}^{2}}}{{{x}^{2}}}$ 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=-\frac{x\left( vx \right)+{{\left( vx \right)}^{2}}}{{{x}^{2}}}$ 

\[\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=-\frac{v{{x}^{2}}+{{v}^{2}}{{x}^{2}}}{{{x}^{2}}}\] 

$\Rightarrow v+x\frac{dv}{dx}=-\frac{{{x}^{2}}\left( v+{{v}^{2}} \right)}{{{x}^{2}}}$ 

$\Rightarrow v+x\frac{dv}{dx}=-v-{{v}^{2}}$ 

$\Rightarrow x\frac{dv}{dx}=-{{v}^{2}}-2v$ 

Separate the differentials:

$\frac{1}{{{v}^{2}}+2v}dv=-\frac{dx}{x}$ 

Integrate both side:

$\int \frac{1}{{{v}^{2}}+2v}dv=-\int \frac{dx}{x}$ 

$\Rightarrow \frac{1}{2}\int \frac{v+2-v}{v\left( v+2 \right)}dv=-\log x+\log C$ 

$\Rightarrow \frac{1}{2}\int \frac{v+2}{v\left( v+2 \right)}dv-\frac{1}{2}\int \frac{v}{v\left( v+2 \right)}dv=-\log x+\log C$ 

$\Rightarrow \frac{1}{2}\int \frac{1}{v}dv-\frac{1}{2}\int \frac{1}{v+2}dv=-\log x+\log C$ 

$\Rightarrow \frac{1}{2}\log v-\frac{1}{2}\log \left( v+2 \right)=\log \frac{C}{x}$ 

$\Rightarrow \frac{1}{2}\log \frac{v}{v+2}=\log \frac{C}{x}$ 

\[\Rightarrow \frac{v}{v+2}={{\left( \frac{C}{x} \right)}^{2}}\] 

\[\Rightarrow \frac{v}{v+2}=\frac{{{C}^{2}}}{{{x}^{2}}}\] 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

\[\Rightarrow \frac{\frac{y}{x}}{\frac{y}{x}+2}=\frac{{{C}^{2}}}{{{x}^{2}}}\] 

\[\Rightarrow \frac{{{x}^{2}}y}{y+2x}={{C}^{2}}\]  

 Now $\text{y=1}$ and $\text{x=1}$:

\[\Rightarrow \frac{{{1}^{2}}\left( 1 \right)}{1+2\left( 1 \right)}={{C}^{2}}\]  

\[\Rightarrow {{C}^{2}}=\frac{1}{3}\] 

The required particular solution:

\[\Rightarrow \frac{{{x}^{2}}y}{y+2x}=\frac{1}{3}\] 

\[\Rightarrow y+2x=3{{x}^{2}}y\] 


13. For the differential equation $\left[ \text{xsi}{{\text{n}}^{\text{2}}}\left( \frac{\text{x}}{\text{y}} \right)\text{-y} \right]\text{dx+xdy=0}$.Find the particular solution for the condition $\text{y=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$ when $\text{x=1}$.

Ans: Given differential equation is:

$\left[ x{{\sin }^{2}}\left( \frac{x}{y} \right)-y \right]dx+xdy=0$

$\Rightarrow \frac{dy}{dx}=-\frac{\left[ x{{\sin }^{2}}\left( \frac{x}{y} \right)-y \right]}{x}$ 

Checking for homogeneity:

$F\left( x,y \right)=-\frac{x{{\sin }^{2}}\left( \frac{x}{y} \right)-y}{x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{\lambda x{{\sin }^{2}}\left( \frac{\lambda x}{\lambda y} \right)-\lambda y}{\lambda x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{\lambda \left( x{{\sin }^{2}}\left( \frac{y}{x} \right)-y \right)}{\lambda x}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=-\frac{x{{\sin }^{2}}\left( \frac{y}{x} \right)-y}{x}$ 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=-\frac{x{{\sin }^{2}}\left( \frac{vx}{x} \right)-vx}{x}$ 

\[\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{-x{{\sin }^{2}}\left( v \right)+vx}{x}\] 

$\Rightarrow v+x\frac{dv}{dx}=-{{\sin }^{2}}v+v$ 

$\Rightarrow x\frac{dv}{dx}=-{{\sin }^{2}}v$ 

Separate the differentials:

$\text{cose}{{\text{c}}^{2}}vdv=-\frac{dx}{x}$ 

Integrate both side:

$\int \text{cose}{{\text{c}}^{2}}vdv=-\int \frac{dx}{x}$ 

$\Rightarrow -\cot v=-\log x-\log C$ 

$\Rightarrow \cot v=\log Cx$ 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

$\Rightarrow \cot \frac{y}{x}=\log Cx$ 

 Now $\text{y=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$ and $\text{x=1}$:

$\cot \frac{\frac{\pi }{4}}{1}=\log C\left( 1 \right)$  

\[\Rightarrow \log C=\cot \frac{\pi }{4}\] 

\[\Rightarrow \log C=1\] 

\[\Rightarrow C=e\] 

The required particular solution:

\[\Rightarrow \cot \frac{y}{x}=\log \left| ex \right|\] 


14. For the differential equation $\frac{\text{dy}}{\text{dx}}\text{-}\frac{\text{y}}{\text{x}}\text{+cosec}\left( \frac{\text{y}}{\text{x}} \right)\text{=0}$.Find the particular solution for the condition $\text{y=0}$ when $\text{x=1}$.

Ans: Given differential equation is:

$\frac{dy}{dx}-\frac{y}{x}+\text{cosec}\left( \frac{y}{x} \right)=0$

\[\Rightarrow \frac{dy}{dx}=\frac{y}{x}-\text{cosec}\left( \frac{y}{x} \right)\] 

Checking for homogeneity:

$F\left( x,y \right)=\frac{y}{x}-\text{cosec}\left( \frac{y}{x} \right)$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{\lambda y}{\lambda x}-\text{cosec}\left( \frac{\lambda y}{\lambda x} \right)$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{y}{x}-\text{cosec}\left( \frac{y}{x} \right)$ 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=v-\text{cosec}\left( v \right)$ 

\[\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=v-\text{cosec}\left( v \right)\] 

$\Rightarrow v+x\frac{dv}{dx}=v-\text{cosec}\left( v \right)$ 

$\Rightarrow x\frac{dv}{dx}=-\text{cosec}\left( v \right)$ 

Separate the differentials:

$\sin vdv=-\frac{dx}{x}$ 

Integrate both side:

$\int \sin vdv=-\int \frac{dx}{x}$ 

\[\Rightarrow -\cos v=-\log \left| x \right|-\log C\] 

\[\cos v=\log \left| Cx \right|\] 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

$\Rightarrow \cos \frac{y}{x}=\log \left| Cx \right|$ 

 Now $\text{y=0}$ and $\text{x=1}$:

$\Rightarrow \cos \frac{0}{1}=\log \left| C1 \right|$  

\[\Rightarrow \log C=1\] 

\[\Rightarrow C=e\] 

The required particular solution:

\[\Rightarrow \cos \frac{y}{x}=\log \left| ex \right|\].


15. For the differential equation $\text{2xy+}{{\text{y}}^{\text{2}}}\text{-2}{{\text{x}}^{\text{2}}}\frac{\text{dy}}{\text{dx}}\text{=0}$.Find the particular solution for the condition $\text{y=2}$ when $\text{x=1}$.

Ans: Given differential equation is:

$2xy+{{y}^{2}}-2{{x}^{2}}\frac{dy}{dx}=0$

\[\Rightarrow \frac{dy}{dx}=\frac{2xy+{{y}^{2}}}{2{{x}^{2}}}\] 

Checking for homogeneity:

$F\left( x,y \right)=\frac{2xy+{{y}^{2}}}{2{{x}^{2}}}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{2\left( \lambda x \right)\left( \lambda y \right)+{{\lambda }^{2}}{{y}^{2}}}{2{{\lambda }^{2}}{{x}^{2}}}$ 

$\Rightarrow F\left( \lambda x,\lambda y \right)=\frac{2xy+{{y}^{2}}}{2{{x}^{2}}}$ 

$\Rightarrow F\left( x,y \right)=F\left( \lambda x,\lambda y \right)$ 

Thus it is an homogenous equation.

Let $\text{y=vx}$:

$\frac{d\left( vx \right)}{dx}=\frac{2x\left( vx \right)+{{\left( vx \right)}^{2}}}{2{{x}^{2}}}$ 

\[\Rightarrow v\frac{dx}{dx}+x\frac{dv}{dx}=\frac{2v+{{v}^{2}}}{2}\] 

$\Rightarrow v+x\frac{dv}{dx}=v+\frac{{{v}^{2}}}{2}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{{{v}^{2}}}{2}$ 

Separate the differentials:

$\frac{dv}{{{v}^{2}}}=\frac{1}{2}\left( \frac{dx}{x} \right)$ 

Integrate both side:

$2\int \frac{dv}{{{v}^{2}}}=\int \left( \frac{dx}{x} \right)$ 

\[\Rightarrow \frac{{{v}^{-2+1}}}{-2+1}=\log \left| x \right|+C\] 

\[-\frac{2}{v}=\log \left| x \right|+C\] 

Back substitute $\text{v=}\frac{\text{y}}{\text{x}}$:

$\Rightarrow -\frac{2x}{y}=\log \left| x \right|+C$ 

 Now $\text{y=2}$ and $\text{x=1}$:

$\Rightarrow -\frac{2\left( 1 \right)}{2}=\log \left| 1 \right|+C$  

$\Rightarrow C=-1$ 

The required particular solution:

$\Rightarrow -\frac{2x}{y}=\log \left| x \right|-1$.

$\Rightarrow y=\frac{2x}{1-\log \left| x \right|}\,\,\left( x\ne 0,e \right)$ 


16. What substitution should be used for solving homogeneous differential equation $\frac{\text{dx}}{\text{dy}}\text{=h}\left( \frac{\text{x}}{\text{y}} \right)$.

A.$\text{y=vx}$ 

B. $\text{v=yx}$

C. $\text{x=vy}$

D. $\text{x=v}$

Ans: The required substitution will be: 

$\frac{x}{y}=v$ 

$\Rightarrow x=vy$ 

The correct answer is (C).


17. Which of the following equation is homogeneous:

  1. $\left( \text{4x+6y+5} \right)\text{dy-}\left( \text{3y+2x+4} \right)\text{dx=0}$ 

  2. $\left( \text{xy} \right)\text{dx-}\left( {{\text{x}}^{3}}\text{+}{{\text{y}}^{3}} \right)\text{dy=0}$ 

  3. $\left( {{\text{x}}^{\text{3}}}\text{+2}{{\text{y}}^{\text{2}}} \right)\text{dx+2xydy=0}$ 

  4. ${{\text{y}}^{\text{2}}}\text{dx+}\left( {{\text{x}}^{\text{2}}}\text{-xy-}{{\text{y}}^{\text{2}}} \right)\text{dy=0}$ 

Ans: For option (A):

$F\left( x,y \right)=\frac{3y+2x+4}{4x+6y+5}$

$F\left( \lambda x,\lambda y \right)=\frac{3\lambda y+2\lambda x+4}{4\lambda x+6\lambda y+5}$ 

$F\left( \lambda x,\lambda y \right)\ne F\left( x,y \right)$  


For option (B):

$F\left( x,y \right)=\frac{xy}{{{x}^{3}}+{{y}^{3}}}$

$F\left( \lambda x,\lambda y \right)=\frac{\left( \lambda x \right)\left( \lambda y \right)}{{{\left( \lambda x \right)}^{3}}+{{\left( \lambda y \right)}^{3}}}$ 

$F\left( \lambda x,\lambda y \right)=\frac{xy}{\lambda \left( x+y \right)}$ 

$F\left( \lambda x,\lambda y \right)\ne F\left( x,y \right)$  


For option (C):

$F\left( x,y \right)=-\frac{{{x}^{3}}+2{{y}^{2}}}{2xy}$

$F\left( \lambda x,\lambda y \right)=-\frac{{{\lambda }^{3}}{{x}^{3}}+2{{\lambda }^{2}}{{y}^{2}}}{2\left( \lambda x \right)\left( \lambda y \right)}$ 

$F\left( \lambda x,\lambda y \right)=-\frac{\lambda {{x}^{3}}+2{{y}^{2}}}{2xy}$ 

$F\left( \lambda x,\lambda y \right)\ne F\left( x,y \right)$  


For option (D):

$F\left( x,y \right)=-\frac{{{y}^{2}}}{{{x}^{2}}-xy-{{y}^{2}}}$

$F\left( \lambda x,\lambda y \right)=-\frac{{{\lambda }^{2}}{{y}^{2}}}{{{\lambda }^{2}}{{x}^{2}}-\left( \lambda x \right)\left( \lambda y \right)-{{\lambda }^{2}}{{y}^{2}}}$ 

$F\left( \lambda x,\lambda y \right)=-\frac{{{y}^{2}}}{{{x}^{2}}-xy-{{y}^{2}}}$ 

$F\left( \lambda x,\lambda y \right)=F\left( x,y \right)$  

Thus the correct answer is option (D).


Exercise 9.6

1. Find the general solution for the differential equation $\frac{\text{dy}}{\text{dx}}\text{+2y=sinx}$.

Ans: The given differential equation is:

$\frac{dy}{dx}+2y=\sin x$

It is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=2$ 

$Q=\sin x$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int 2dx}}$ 

$\Rightarrow I.F={{e}^{2x}}$ 

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$y{{e}^{2x}}=\int \sin x\left( {{e}^{2x}} \right)dx+C$ 

\[\Rightarrow y{{e}^{2x}}=I+C\,\,\,\left( I=\int \sin x\left( {{e}^{2x}} \right)dx \right)\] ……(1)

\[I=\int \sin x\left( {{e}^{2x}} \right)dx\] 

$\Rightarrow I=\left( \sin x \right)\int {{e}^{2x}}dx-\int \left( \left( \sin x \right)'\int {{e}^{2x}}dx \right)dx$ 

$\Rightarrow I=\frac{{{e}^{2x}}}{2}\sin x-\int \left( \cos x\left( \frac{{{e}^{2x}}}{2} \right) \right)dx$ 

$\Rightarrow I=\frac{{{e}^{2x}}}{2}\sin x-\frac{1}{2}\left[ \cos x\int {{e}^{2x}}dx-\int \left( \left( \cos x \right)'\left( \int {{e}^{2x}}dx \right) \right)dx \right]$ 

$\Rightarrow I=\frac{{{e}^{2x}}}{2}\sin x-\frac{1}{2}\left[ \frac{{{e}^{2x}}}{2}\cos x+\frac{1}{2}\int {{e}^{2x}}\left( \sin x \right)dx \right]$ 

$\Rightarrow I=\frac{{{e}^{2x}}}{2}\sin x-\frac{1}{2}\left[ \frac{{{e}^{2x}}}{2}\cos x+\frac{1}{2}I \right]$ 

$\Rightarrow I=\frac{{{e}^{2x}}}{2}\sin x-\frac{{{e}^{2x}}}{4}\cos x-\frac{1}{4}I$ 

$\Rightarrow \frac{5}{4}I=\frac{{{e}^{2x}}}{2}\sin x-\frac{{{e}^{2x}}}{4}\cos x$ 

\[\Rightarrow I=\frac{2{{e}^{2x}}}{5}\sin x-\frac{{{e}^{2x}}}{5}\cos x\] 

\[\Rightarrow I=\frac{{{e}^{2x}}}{5}\left[ 2\sin x-\cos x \right]\] 

Back substituting $\text{I}$ in expression (1):

\[\Rightarrow y{{e}^{2x}}=\frac{{{e}^{2x}}}{5}\left[ 2\sin x-\cos x \right]+C\,\] 

\[\Rightarrow y=\frac{1}{5}\left( 2\sin x-\cos x \right)+C{{e}^{-2x}}\] 

The general solution for given differential equation is \[\text{y=}\frac{\text{1}}{\text{5}}\left( \text{2sinx-cosx} \right)\text{+C}{{\text{e}}^{\text{-2x}}}\].


2. Find the general solution for the differential equation $\frac{\text{dy}}{\text{dx}}\text{+3y=}{{\text{e}}^{\text{-2x}}}$.

Ans: The given differential equation is:

$\frac{dy}{dx}+3y={{e}^{-2x}}$

It is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=3$ 

$Q={{e}^{-2x}}$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int 3dx}}$ 

$\Rightarrow I.F={{e}^{3x}}$ 

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow y{{e}^{3x}}=\int {{e}^{-2x}}\left( {{e}^{3x}} \right)dx+C$ 

$\Rightarrow y{{e}^{3x}}=\int {{e}^{-2x+3x}}dx+C$ 

$\Rightarrow y{{e}^{3x}}=\int {{e}^{x}}dx+C$ 

$\Rightarrow y{{e}^{3x}}={{e}^{x}}+C$ 

$\Rightarrow y={{e}^{-2x}}+C{{e}^{-3x}}$ 

The general solution for given differential equation is $\text{y=}{{\text{e}}^{\text{-2x}}}\text{+C}{{\text{e}}^{\text{-3x}}}$.


3. Find the general solution for the differential equation $\frac{\text{dy}}{\text{dx}}\text{+}\frac{\text{y}}{\text{x}}\text{=}{{\text{x}}^{\text{2}}}$.

Ans: The given differential equation is:

$\frac{dy}{dx}+\frac{y}{x}={{x}^{2}}$

It is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=\frac{1}{x}$ 

$Q={{x}^{2}}$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int \frac{1}{x}dx}}$ 

$\Rightarrow I.F={{e}^{\log x}}$

$\Rightarrow I.F=x$  

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow yx=\int {{x}^{2}}\left( x \right)dx+C$ 

$\Rightarrow xy=\int {{x}^{3}}dx+C$ 

$\Rightarrow xy=\frac{{{x}^{3+1}}}{3+1}+C$ 

$\Rightarrow xy=\frac{{{x}^{4}}}{4}+C$ 

The general solution for given differential equation is $\text{xy=}\frac{{{\text{x}}^{\text{4}}}}{\text{4}}\text{+C}$.


4. Find the general solution for the differential equation $\frac{\text{dy}}{\text{dx}}\text{+}\left( \text{secx} \right)\text{y=tanx}\,\,\left( \text{0}\le \text{x}\le \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \right)$.

Ans: The given differential equation is:

$\frac{dy}{dx}+\left( \sec x \right)y=\tan x$

It is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=\sec x$ 

$Q=\tan x$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int \sec xdx}}$ 

$\Rightarrow I.F={{e}^{\log \left( \sec x+\tan x \right)}}$

$\Rightarrow I.F=\left( \sec x+\tan x \right)$  

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow y\left( \sec x+\tan x \right)=\int \tan x\left( \sec x+\tan x \right)dx+C$ 

$\Rightarrow y\left( \sec x+\tan x \right)=\int \tan x\sec xdx+\int {{\tan }^{2}}xdx+C$ 

$\Rightarrow y\left( \sec x+\tan x \right)=\sec x+\int \left( {{\sec }^{2}}x-1 \right)dx+C$ 

$\Rightarrow y\left( \sec x+\tan x \right)=\sec x+\int {{\sec }^{2}}xdx-\int dx+C$ 

$\Rightarrow y\left( \sec x+\tan x \right)=\sec x+\tan x-x+C$ 

The general solution for given differential equation is $\text{y}\left( \text{secx+tanx} \right)\text{=secx+tanx-x+C}$.


5. Find the general solution for the differential equation $\text{co}{{\text{s}}^{\text{2}}}\text{x}\frac{\text{dy}}{\text{dx}}\text{+y=tanx}\,\,\left( \text{0}\le \text{x}\le \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \right)$.

Ans: The given differential equation is:

${{\cos }^{2}}x\frac{dy}{dx}+y=\tan x\,$

$\frac{dy}{dx}+\left( {{\sec }^{2}}x \right)y={{\sec }^{2}}x\tan x\,$ 

It is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p={{\sec }^{2}}x$ 

$Q={{\sec }^{2}}x\tan x$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int {{\sec }^{2}}xdx}}$ 

$\Rightarrow I.F={{e}^{\tan x}}$ 

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow y{{e}^{\tan x}}=\int {{e}^{\tan x}}\left( {{\sec }^{2}}x\tan x \right)dx+C$ 

$\Rightarrow y{{e}^{\tan x}}=I+C\,\,\left( I=\int {{e}^{\tan x}}\left( {{\sec }^{2}}x\tan x \right)dx \right)$ ……(1)

Solving the integral $\text{I}$:

$I=\int {{e}^{\tan x}}\left( {{\sec }^{2}}x\tan x \right)dx$ 

Substitute $\text{tanx=t}$ :
$\tan x=t$ 

$\Rightarrow {{\sec }^{2}}xdx=dt$ 

$\Rightarrow I=\int {{e}^{t}}tdt$ 

$\Rightarrow I=t\int {{e}^{t}}dt-\int \left( \left( t \right)'\int {{e}^{t}}dt \right)dt$ 

$\Rightarrow I=t{{e}^{t}}-\int \left( {{e}^{t}} \right)dt$ 

$\Rightarrow I=t{{e}^{t}}-{{e}^{t}}$ 

Back substitute $\text{t}$:

$I=\tan x{{e}^{\tan x}}-{{e}^{\tan x}}$ 

Back substitute $\text{I}$ in expression (1):

$\Rightarrow y{{e}^{\tan x}}=I+C\,\,$ 

$\Rightarrow y{{e}^{\tan x}}=\tan x{{e}^{\tan x}}-{{e}^{\tan x}}+C\,\,$ 

The general solution for given differential equation is $\text{y}{{\text{e}}^{\text{tanx}}}\text{=tanx}{{\text{e}}^{\text{tanx}}}\text{-}{{\text{e}}^{\text{tanx}}}\text{+C}\,\,$.


6. Find the general solution for the differential equation $\text{x}\frac{\text{dy}}{\text{dx}}\text{+2y=}{{\text{x}}^{2}}\text{logx}$.

Ans: The given differential equation is:

$x\frac{dy}{dx}+2y={{x}^{2}}\log x$

\[\frac{dy}{dx}+\frac{2}{x}y=x\log x\] 

It is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=\frac{2}{x}$ 

$Q=x\log x$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int \frac{2}{x}dx}}$ 

$\Rightarrow I.F={{e}^{2\log x}}$

$\Rightarrow I.F={{e}^{\log {{x}^{2}}}}$ 

$\Rightarrow I.F={{x}^{2}}$  

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow y{{x}^{2}}=\int {{x}^{2}}\left( x\log x \right)dx+C$ 

\[\Rightarrow y{{x}^{2}}=\int {{x}^{3}}\log xdx+C\] 

\[\Rightarrow y{{x}^{2}}=I+C\,\,\left( I=\int {{x}^{3}}\log xdx \right)\] ……(1)

Solving the integral $\text{I}$:

\[I=\int {{x}^{3}}\log xdx\] 

\[\Rightarrow I=\log x\int {{x}^{3}}dx-\int \left( \left( \log x \right)'\int {{x}^{3}}dx \right)dx\] 

\[\Rightarrow I=\frac{{{x}^{4}}}{4}\log x-\int \left( \frac{1}{x}\left( \frac{{{x}^{4}}}{4} \right) \right)dx\] 

\[\Rightarrow I=\frac{{{x}^{4}}}{4}\log x-\frac{1}{4}\int {{x}^{3}}dx\] 

\[\Rightarrow I=\frac{{{x}^{4}}}{4}\log x-\frac{1}{4}\left( \frac{{{x}^{4}}}{4} \right)\] 

\[\Rightarrow I=\frac{{{x}^{4}}}{4}\log x-\frac{{{x}^{4}}}{16}\] 

Back substitute $\text{I}$ in expression (1):

\[\Rightarrow y{{x}^{2}}=I+C\] 

\[\Rightarrow y{{x}^{2}}=\frac{{{x}^{4}}}{4}\log x-\frac{{{x}^{4}}}{16}+C\] 

\[\Rightarrow y=\frac{{{x}^{2}}}{16}\left( 4\log x-1 \right)+C{{x}^{-2}}\] 

The general solution for given differential equation is \[\text{y=}\frac{{{\text{x}}^{\text{2}}}}{\text{16}}\left( \text{4logx-1} \right)\text{+C}{{\text{x}}^{\text{-2}}}\].


7. Find the general solution for the differential equation $\text{xlogx}\frac{\text{dy}}{\text{dx}}\text{+y=}\frac{\text{2}}{\text{x}}\text{logx}$.

Ans: The given differential equation is:

$x\log x\frac{dy}{dx}+y=\frac{2}{x}\log x$

$\frac{dy}{dx}+\frac{y}{x\log x}=\frac{2}{{{x}^{2}}}$ 

It is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=\frac{1}{x\log x}$ 

$Q=\frac{2}{{{x}^{2}}}$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int \frac{1}{x\log x}dx}}$ 

Substitute $\text{logx=t}$:
$\log x=t$ 

$\Rightarrow \frac{1}{x}dx=dt$ 

$\Rightarrow I.F={{e}^{\int \frac{1}{t}dt}}$

$\Rightarrow I.F={{e}^{\log t}}$ 

$\Rightarrow I.F=t$

$\Rightarrow I.F=\log x$   

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow y\log x=\int \frac{2}{{{x}^{2}}}\left( \log x \right)dx+C$ 

$\Rightarrow y\log x=I+C\,\,\left( I=\int \frac{2}{{{x}^{2}}}\left( \log x \right)dx \right)$ ……(1)

Solving the integral $\text{I}$:

$I=\int \frac{2}{{{x}^{2}}}\left( \log x \right)dx$ 

\[I=2\left[ \log x\int \frac{1}{{{x}^{2}}}dx-\int \left( \left( \log x \right)'\int \frac{1}{{{x}^{2}}}dx \right)dx \right]\] 

\[I=2\left[ \log x\left( \frac{-1}{x} \right)-\int \left( \frac{1}{x}\left( \frac{-1}{x} \right) \right)dx \right]\] 

\[I=2\left[ -\frac{\log x}{x}+\int \left( \frac{1}{{{x}^{2}}} \right)dx \right]\] 

\[I=2\left[ -\frac{\log x}{x}-\frac{1}{x} \right]\] 

Back substitute $\text{I}$ in expression (1):

$y\log x=I+C$ 

\[\Rightarrow y\log x=2\left[ -\frac{\log x}{x}-\frac{1}{x} \right]+C\] 

\[\Rightarrow y\log x=-\frac{2}{x}\left( \log x+1 \right)+C\] 

The general solution for given differential equation is \[\text{ylogx=-}\frac{\text{2}}{\text{x}}\left( \text{logx+1} \right)\text{+C}\].


8. Find the general solution for the differential equation $\left( \text{1+}{{\text{x}}^{\text{2}}} \right)\text{dy+2xydx=cotxdx}\,\,\left( \text{x}\ne \text{0} \right)$.

Ans: The given differential equation is:

$\left( 1+{{x}^{2}} \right)dy+2xydx=\cot xdx\,\,\left( x\ne 0 \right)$

$\Rightarrow \frac{dy}{dx}+\frac{2xy}{1+{{x}^{2}}}=\frac{\cot x}{1+{{x}^{2}}}$ 

It is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=\frac{2x}{1+{{x}^{2}}}$ 

$Q=\frac{\cot x}{1+{{x}^{2}}}$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int \frac{2x}{1+{{x}^{2}}}dx}}$ 

Substitute $\text{logx=t}$:
$1+{{x}^{2}}=t$ 

$\Rightarrow 2xdx=dt$ 

$\Rightarrow I.F={{e}^{\int \frac{1}{t}dt}}$

$\Rightarrow I.F={{e}^{\log t}}$ 

$\Rightarrow I.F=t$

$\Rightarrow I.F=1+{{x}^{2}}$   

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow y\left( 1+{{x}^{2}} \right)=\int \frac{\cot x}{1+{{x}^{2}}}\left( 1+{{x}^{2}} \right)dx+C$ 

$\Rightarrow y\left( 1+{{x}^{2}} \right)=\int \cot xdx+C$

$\Rightarrow y\left( 1+{{x}^{2}} \right)=\log \left| \sin x \right|+C$ 

The general solution for given differential equation is $\text{y}\left( \text{1+}{{\text{x}}^{\text{2}}} \right)\text{=log}\left| \text{sinx} \right|\text{+C}$.


9. Find the general solution for the differential equation $\text{x}\frac{\text{dy}}{\text{dx}}\text{+y-x+xycotx=0}\,\,\left( \text{x}\ne \text{0} \right)$.

Ans: The given differential equation is:

$x\frac{dy}{dx}+y-x+xy\cot x=0\,$

$\Rightarrow \frac{dy}{dx}+\frac{y}{x}-1+y\cot x=0$ 

$\Rightarrow \frac{dy}{dx}+\left( \frac{1}{x}+\cot x \right)y=1$ 

It is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=\left( \frac{1}{x}+\cot x \right)$ 

$Q=1$ 

Calculate the integrating factor:

$I.F={{e}^{\int pdx}}$ 

$\Rightarrow I.F={{e}^{\int \left( \frac{1}{x}+\cot x \right)dx}}$ 

$\Rightarrow I.F={{e}^{\int \frac{1}{x}dx+\int \cot xdx}}$

$\Rightarrow I.F={{e}^{\log x+\log \left( \sin x \right)}}$ 

$\Rightarrow I.F={{e}^{\log \left( x\sin x \right)}}$

$\Rightarrow I.F=x\sin x$   

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow y\left( x\sin x \right)=\int \left( 1 \right)\left( x\sin x \right)dx+C$ 

$\Rightarrow xy\sin x=I+C\,\,\left( I=\int x\sin xdx \right)$ ……(1) 

Solving the integral $\text{I}$:

$I=\int x\sin xdx$ 

\[\Rightarrow I=x\int \sin xdx-\int \left( \left( x \right)'\int \sin xdx \right)dx\] 

\[\Rightarrow I=x\left( -\cos x \right)+\int \left( \cos x \right)dx\] 

\[\Rightarrow I=-x\cos x+\sin x\] 

Back substitute $\text{I}$ in expression (1):

$xy\sin x=I+C\,$ 

\[\Rightarrow xy\sin x=-x\cos x+\sin x+C\,\] 

\[\Rightarrow y=-\cot x+\frac{1}{x}+\frac{C}{x\sin x}\,\] 

The general solution for given differential equation is \[\text{y=-cotx+}\frac{\text{1}}{\text{x}}\text{+}\frac{\text{C}}{\text{xsinx}}\,\].


10. Find the general solution for the differential equation $\left( \text{x+y} \right)\frac{\text{dy}}{\text{dx}}=1$.

Ans: The given differential equation is:

$\left( x+y \right)\frac{dy}{dx}=1$

$\Rightarrow \frac{dy}{dx}=\frac{1}{x+y}$ 

$\Rightarrow \frac{dx}{dy}=x+y$ 

$\Rightarrow \frac{dx}{dy}-x=y$ 

It is differential equation of the form $\frac{\text{dx}}{\text{dy}}\text{+px=Q}$, with:

$p=-1$ 

$Q=y$ 

Calculate the integrating factor:

\[I.F={{e}^{\int pdy}}\] 

$\Rightarrow I.F={{e}^{\int -1dy}}$ 

$\Rightarrow I.F={{e}^{-y}}$

General solution is of the form:

$x\left( I.F \right)=\int \left( Q\times I.F \right)dy+C$ 

$\Rightarrow x\left( {{e}^{-y}} \right)=\int \left( y \right)\left( {{e}^{-y}} \right)dy+C$ 

$\Rightarrow x{{e}^{-y}}=I+C\,\,\left( I=\int y{{e}^{-y}}dy \right)$ ……(1) 

Solving the integral $\text{I}$:

$I=\int y{{e}^{-y}}dy$ 

$\Rightarrow I=y\int {{e}^{-y}}dy-\int \left( \left( y \right)'\int {{e}^{-y}}dy \right)dy$ 

$\Rightarrow I=-y{{e}^{-y}}+\int \left( \left( 1 \right){{e}^{-y}} \right)dy$ 

$\Rightarrow I=-y{{e}^{-y}}+\int {{e}^{-y}}dy$

$\Rightarrow I=-y{{e}^{-y}}-{{e}^{-y}}$  

Back substitute $\text{I}$ in expression (1):

$x{{e}^{-y}}=I+C$ 

\[\Rightarrow x{{e}^{-y}}=-y{{e}^{-y}}-{{e}^{-y}}+C\] 

\[\Rightarrow x=-y-1+C{{e}^{y}}\]

\[\Rightarrow x+y+1=C{{e}^{y}}\]  

The general solution for given differential equation is \[\text{x+y+1=C}{{\text{e}}^{\text{y}}}\].


11. Find the general solution for the differential equation $\text{ydx+}\left( \text{x-}{{\text{y}}^{\text{2}}} \right)\text{dy=0}$.

Ans: The given differential equation is:

$ydx+\left( x-{{y}^{2}} \right)dy=0$

$\Rightarrow \frac{dx}{dy}=\frac{{{y}^{2}}-x}{y}$ 

$\Rightarrow \frac{dx}{dy}+\left( \frac{1}{y} \right)x=y$ 

It is differential equation of the form $\frac{\text{dx}}{\text{dy}}\text{+px=Q}$, with:

$p=\frac{1}{y}$ 

$Q=y$ 

Calculate the integrating factor:

\[I.F={{e}^{\int pdy}}\] 

$\Rightarrow I.F={{e}^{\int \frac{1}{y}dy}}$ 

$\Rightarrow I.F={{e}^{\log y}}$

$\Rightarrow I.F=y$ 

General solution is of the form:

$x\left( I.F \right)=\int \left( Q\times I.F \right)dy+C$ 

$\Rightarrow x\left( y \right)=\int \left( y \right)\left( y \right)dy+C$ 

$\Rightarrow xy=\int {{y}^{2}}dy+C$ 

$\Rightarrow xy=\frac{{{y}^{3}}}{3}+C$ 

$\Rightarrow x=\frac{{{y}^{2}}}{3}+\frac{C}{y}$ 

The general solution for given differential equation is $\text{x=}\frac{{{\text{y}}^{\text{2}}}}{\text{3}}\text{+}\frac{\text{C}}{\text{y}}$.


12. Find the general solution for the differential equation $\left( \text{x+3}{{\text{y}}^{\text{2}}} \right)\frac{\text{dy}}{\text{dx}}\text{=y}\,\,\left( \text{y0} \right)$.

Ans: The given differential equation is:

$\left( x+3{{y}^{2}} \right)\frac{dy}{dx}=y$

$\Rightarrow \frac{dx}{dy}=\frac{x+3{{y}^{2}}}{y}$ 

$\Rightarrow \frac{dx}{dy}-\left( \frac{1}{y} \right)x=3y$ 

It is differential equation of the form $\frac{\text{dx}}{\text{dy}}\text{+px=Q}$, with:

$p=-\frac{1}{y}$ 

$Q=3y$ 

Calculate the integrating factor:

\[I.F={{e}^{\int pdy}}\] 

$\Rightarrow I.F={{e}^{-\int \frac{1}{y}dy}}$ 

$\Rightarrow I.F={{e}^{-\log y}}$

$\Rightarrow I.F={{e}^{\log {{y}^{-1}}}}$ 

$\Rightarrow I.F=\frac{1}{y}$ 

General solution is of the form:

$x\left( I.F \right)=\int \left( Q\times I.F \right)dy+C$ 

$\Rightarrow x\left( \frac{1}{y} \right)=\int \left( 3y \right)\left( \frac{1}{y} \right)dy+C$ 

$\Rightarrow \frac{x}{y}=3\int dy+C$ 

$\Rightarrow \frac{x}{y}=3y+C$ 

$\Rightarrow x=3{{y}^{2}}+Cy$ 

The general solution for given differential equation is $\text{x=3}{{\text{y}}^{\text{2}}}\text{+Cy}$.


13. Find particular solution for $\frac{\text{dy}}{\text{dx}}\text{+2ytanx=sinx}$ satisfying $\text{y=0}$ when $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$.

Ans: The given differential equation is:

$\frac{dy}{dx}+2y\tan x=\sin x$ 

$\frac{dy}{dx}+\left( 2\tan x \right)y=\sin x$ 

It is differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=2\tan x$ 

$Q=\sin x$ 

Calculate the integrating factor:

\[I.F={{e}^{\int pdx}}\] 

$\Rightarrow I.F={{e}^{2\int \tan xdx}}$ 

$\Rightarrow I.F={{e}^{2\log \left| \sec x \right|}}$

\[\Rightarrow I.F={{e}^{\log {{\left( \sec x \right)}^{2}}}}\] 

\[\Rightarrow I.F={{\sec }^{2}}x\] 

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow y{{\sec }^{2}}x=\int \left( \sin x \right)\left( {{\sec }^{2}}x \right)dx+C$ 

$\Rightarrow y{{\sec }^{2}}x=\int \tan x\sec xdx+C$ 

$\Rightarrow y{{\sec }^{2}}x=\sec x+C$ 

$\Rightarrow y=\cos x+C{{\cos }^{2}}x$ 

Given $\text{y=0}$ when $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$:

$0=\cos \left( \frac{\pi }{3} \right)+C{{\cos }^{2}}\left( \frac{\pi }{3} \right)$ 

$\Rightarrow 0=\frac{1}{2}+C{{\left( \frac{1}{2} \right)}^{2}}$ 

$\Rightarrow C=-2$ 

Therefore the particular solution will be:

$\Rightarrow y=\cos x-2{{\cos }^{2}}x$ 

The particular solution for given differential equation satisfying the given conditions is $\text{y=cosx-2co}{{\text{s}}^{\text{2}}}\text{x}$.


14. Find particular solution for $\left( \text{1+}{{\text{x}}^{\text{2}}} \right)\frac{\text{dy}}{\text{dx}}\text{+2xy=}\frac{\text{1}}{\text{1+}{{\text{x}}^{\text{2}}}}$ satisfying $\text{y=0}$ when $\text{x=1}$.

Ans: The given differential equation is:

$\left( 1+{{x}^{2}} \right)\frac{dy}{dx}+2xy=\frac{1}{1+{{x}^{2}}}$ 

\[\frac{dy}{dx}+\left( \frac{2x}{1+{{x}^{2}}} \right)y=\frac{1}{{{\left( 1+{{x}^{2}} \right)}^{2}}}\] 

It is differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=\frac{2x}{1+{{x}^{2}}}$ 

$Q=\frac{1}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$ 

Calculate the integrating factor:

\[I.F={{e}^{\int pdx}}\] 

$\Rightarrow I.F={{e}^{\int \frac{2x}{1+{{x}^{2}}}dx}}$ 

$\Rightarrow I.F={{e}^{2\log \left| \sec x \right|}}$

$\Rightarrow I.F={{e}^{\log \left( 1+{{x}^{2}} \right)}}$ 

\[\Rightarrow I.F=1+{{x}^{2}}\] 

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow y\left( 1+{{x}^{2}} \right)=\int \left( \frac{1}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \right)\left( 1+{{x}^{2}} \right)dx+C$ 

$\Rightarrow y\left( 1+{{x}^{2}} \right)=\int \frac{1}{1+{{x}^{2}}}dx+C$ 

$\Rightarrow y\left( 1+{{x}^{2}} \right)={{\tan }^{-1}}x+C$ 

Given $\text{y=0}$ when $\text{x=1}$:

$0\left( 1+1 \right)={{\tan }^{-1}}\left( 1 \right)+C$ 

$\Rightarrow C+\frac{\pi }{4}=0$ 

$\Rightarrow C=-\frac{\pi }{4}$ 

Therefore the particular solution will be:

$\Rightarrow y\left( 1+{{x}^{2}} \right)={{\tan }^{-1}}x-\frac{\pi }{4}$ 

The particular solution for given differential equation satisfying the given conditions is $\text{y}\left( \text{1+}{{\text{x}}^{\text{2}}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\text{x-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$.


15. Find particular solution for $\frac{\text{dy}}{\text{dx}}\text{-3ycotx=sin2x}$ satisfying $\text{y=2}$ when $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$.

Ans: The given differential equation is:

$\frac{dy}{dx}-3y\cot x=\sin 2x$ 

\[\frac{dy}{dx}+\left( -3\cot x \right)y=\sin 2x\] 

It is differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=-3\cot x$ 

$Q=\sin 2x$ 

Calculate the integrating factor:

\[I.F={{e}^{\int pdx}}\] 

$\Rightarrow I.F={{e}^{-3\int \cot xdx}}$ 

$\Rightarrow I.F={{e}^{-3\log \left| \sin x \right|}}$

$\Rightarrow I.F={{e}^{\log \left( \frac{1}{{{\sin }^{3}}x} \right)}}$ 

\[\Rightarrow I.F=\frac{1}{{{\sin }^{3}}x}\] 

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow y\left( \frac{1}{{{\sin }^{3}}x} \right)=\int \left( \sin 2x \right)\left( \frac{1}{{{\sin }^{3}}x} \right)dx+C$ 

$\Rightarrow y\left( \frac{1}{{{\sin }^{3}}x} \right)=2\int \left( \sin x\cos x \right)\left( \frac{1}{{{\sin }^{3}}x} \right)dx+C$ 

$\Rightarrow \frac{y}{{{\sin }^{3}}x}=2\int \left( \frac{\cos x}{{{\sin }^{2}}x} \right)dx+C$ 

$\Rightarrow \frac{y}{{{\sin }^{3}}x}=2\int \cot x\text{cosec}xdx+C$ 

$\Rightarrow \frac{y}{{{\sin }^{3}}x}=-2\text{cosec}\,x+C$ 

$\Rightarrow y=-2{{\sin }^{2}}x+C{{\sin }^{3}}x$ 

Given $\text{y=2}$ when $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$:

$2=-2{{\sin }^{2}}\left( \frac{\pi }{2} \right)+C{{\sin }^{3}}\left( \frac{\pi }{2} \right)$ 

$\Rightarrow C-2=2$ 

$\Rightarrow C=4$ 

Therefore the particular solution will be:

$\Rightarrow y=-2{{\sin }^{2}}x+4{{\sin }^{3}}x$ 

The particular solution for given differential equation satisfying the given conditions is $\text{y=-2si}{{\text{n}}^{\text{2}}}\text{x+4si}{{\text{n}}^{\text{3}}}\text{x}$.


16. Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point $\left( \text{x,y} \right)$ is equal to the sum of the coordinates of the point.

Ans: According to  the slope of tangent $\frac{\text{dy}}{\text{dx}}$ is equal to sum of the coordinates:

$\frac{dy}{dx}=x+y$ 

The given differential equation is:

$\frac{dy}{dx}=x+y$ 

\[\frac{dy}{dx}+\left( -1 \right)y=x\] 

It is differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=-1$ 

$Q=x$ 

Calculate the integrating factor:

\[I.F={{e}^{\int pdx}}\] 

$\Rightarrow I.F={{e}^{-1\int dx}}$ 

$\Rightarrow I.F={{e}^{-x}}$ 

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow y\left( {{e}^{-x}} \right)=\int x\left( {{e}^{-x}} \right)dx+C$ 

$\Rightarrow y{{e}^{-x}}=x\int {{e}^{-x}}dx-\int \left( \left( x \right)'\int {{e}^{-x}}dx \right)dx+C$ 

$\Rightarrow y{{e}^{-x}}=-x{{e}^{-x}}+\int \left( {{e}^{-x}} \right)dx+C$ 

$\Rightarrow y{{e}^{-x}}=-x{{e}^{-x}}-{{e}^{-x}}+C$ 

$\Rightarrow y=-x-1+C{{e}^{x}}$ 

$\Rightarrow y+x+1=C{{e}^{x}}$ 

Given $\text{y=0}$ when $\text{x=0}$ as it passes through origin:

$0+0+1=C{{e}^{0}}$ 

$\Rightarrow C=1$ 

Therefore the equation of the required curve is $\text{y+x+1=}{{\text{e}}^{\text{x}}}$.


17. Find the equation of a curve passing through the point $\left( \text{0,2} \right)$  given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by $\text{5}$.

Ans: Let the slope of tangent be $\frac{\text{dy}}{\text{dx}}$.

According to :

$x+y=\frac{dy}{dx}+5$ 

The given differential equation is:

$x+y=\frac{dy}{dx}+5$ 

\[\frac{dy}{dx}+\left( -1 \right)y=x-5\] 

It is differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$, with:

$p=-1$ 

$Q=x-5$ 

Calculate the integrating factor:

\[I.F={{e}^{\int pdx}}\] 

$\Rightarrow I.F={{e}^{-1\int dx}}$ 

$\Rightarrow I.F={{e}^{-x}}$ 

General solution is of the form:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$ 

$\Rightarrow y\left( {{e}^{-x}} \right)=\int \left( x-5 \right)\left( {{e}^{-x}} \right)dx+C$ 

$\Rightarrow y\left( {{e}^{-x}} \right)=\int x\left( {{e}^{-x}} \right)dx-5\int {{e}^{-x}}dx+C$ 

$\Rightarrow y{{e}^{-x}}=x\int {{e}^{-x}}dx-\int \left( \left( x \right)'\int {{e}^{-x}}dx \right)dx-5\int {{e}^{-x}}dxC$ 

$\Rightarrow y{{e}^{-x}}=-x{{e}^{-x}}+\int \left( {{e}^{-x}} \right)dx+5{{e}^{-x}}+C$ 

$\Rightarrow y{{e}^{-x}}=-x{{e}^{-x}}-{{e}^{-x}}+5{{e}^{-x}}+C$ 

$\Rightarrow y{{e}^{-x}}=-x{{e}^{-x}}+4{{e}^{-x}}+C$ 

$\Rightarrow y=-x+4+C{{e}^{x}}$ 

$\Rightarrow y+x-4=C{{e}^{x}}$ 

Given as it passes through $\left( \text{0,2} \right)$ :

$2+0-4=C{{e}^{0}}$ 

$\Rightarrow C=-2$ 

Therefore the equation of the required curve is:

 $y+x-4=-2{{e}^{x}}$.


18. Find the integrating factor of the differential equation $\text{x}\frac{\text{dy}}{\text{dx}}\text{-y=2}{{\text{x}}^{\text{2}}}$.

A. ${{e}^{-x}}$ 

B. ${{e}^{-y}}$ 

C. $\frac{1}{x}$ 

D. $x$ 

Ans: Given differential equation is:

$x\frac{dy}{dx}-y=2{{x}^{2}}$ 

$\Rightarrow \frac{dy}{dx}-\left( \frac{1}{x} \right)y=2{{x}^{2}}$ 

Thus it is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$:

$p=-\frac{1}{x}$ 

$I.F={{e}^{\int pdx}}$ 

\[\Rightarrow I.F={{e}^{-\int \frac{1}{x}dx}}\]

$\Rightarrow I.F={{e}^{-\log \left| x \right|}}$  

$\Rightarrow I.F={{e}^{\log {{x}^{-1}}}}$

$\Rightarrow I.F=\frac{1}{x}$

Therefore integrating factor is $\frac{\text{1}}{\text{x}}$. Thus the correct option is (C).


19. Find the integrating factor of the differential equation $\left( \text{1-}{{\text{y}}^{\text{2}}} \right)\frac{\text{dx}}{\text{dy}}\text{+yx=ay}\,\left( \text{-1y1} \right)$.

A. $\frac{1}{{{y}^{2}}-1}$ 

B. $\frac{1}{\sqrt{{{y}^{2}}-1}}$ 

C. $\frac{1}{1-{{y}^{2}}}$ 

D. $\frac{1}{\sqrt{1-{{y}^{2}}}}$ 

Ans: Given differential equation is:

$\left( 1-{{y}^{2}} \right)\frac{dx}{dy}+yx=ay$ 

$\Rightarrow \frac{dx}{dy}+\left( \frac{y}{1-{{y}^{2}}} \right)x=\frac{ay}{\left( 1-{{y}^{2}} \right)}$ 

Thus it is a linear differential equation of the form $\frac{\text{dx}}{\text{dy}}\text{+px=Q}$:

$p=\frac{y}{1-{{y}^{2}}}$ 

\[I.F={{e}^{\int pdy}}\] 

\[\Rightarrow I.F={{e}^{\int \frac{y}{1-{{y}^{2}}}dy}}\]

\[\Rightarrow I.F={{e}^{\frac{1}{-2}\int \frac{-2y}{1-{{y}^{2}}}dy}}\]  

\[\Rightarrow I.F={{e}^{\frac{1}{-2}\log \left( 1-{{y}^{2}} \right)}}\]

\[\Rightarrow I.F={{e}^{\log {{\left( 1-{{y}^{2}} \right)}^{-\frac{1}{2}}}}}\]

\[\Rightarrow I.F=\frac{1}{\sqrt{1-{{y}^{2}}}}\] 

Therefore integrating factor is $\frac{\text{1}}{\sqrt{\text{1-}{{\text{y}}^{\text{2}}}}}$. Thus the correct option is (D).


Miscellaneous Exercise on Chapter 9

1. For each of the differential equations given below, indicate its order and degree (if defined).

(i)$\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{+5x}{{\left( \frac{\text{dy}}{\text{dx}} \right)}^{\text{2}}}\text{-6y=logx}$ 

Ans: The given differential equation is:

 $\frac{{{d}^{2}}y}{d{{x}^{2}}}+5x{{\left( \frac{dy}{dx} \right)}^{2}}-6y-\log x=0$ 

The highest order derivative in the equation is of the term $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}$, thus the order of the equation is $\text{2}$ and its highest power is $\text{1}$. Therefore its degree is $\text{1}$.


(ii)${{\left( \frac{\text{dy}}{\text{dx}} \right)}^{3}}-4{{\left( \frac{\text{dy}}{\text{dx}} \right)}^{2}}\text{+7y=sin}\,\text{x}$ 

Ans: The given differential equation is:

 ${{\left( \frac{dy}{dx} \right)}^{3}}-4{{\left( \frac{dy}{dx} \right)}^{2}}+7y-\sin x=0$ 

The highest order derivative in the equation is of the term ${{\left( \frac{\text{dy}}{\text{dx}} \right)}^{3}}$, thus the order of the equation is $1$ and its highest power is $3$. Therefore its degree is $3$.


(iii)$\frac{{{\text{d}}^{\text{4}}}\text{y}}{\text{d}{{\text{x}}^{\text{4}}}}\text{-sin}\left( \frac{{{\text{d}}^{\text{3}}}\text{y}}{\text{d}{{\text{x}}^{\text{3}}}} \right)\text{=0}$ 

Ans: The given differential equation is:

 $\frac{{{d}^{4}}y}{d{{x}^{4}}}-\sin \left( \frac{{{d}^{3}}y}{d{{x}^{3}}} \right)=0$ 

The highest order derivative in the equation is of the term $\frac{{{\text{d}}^{4}}\text{y}}{\text{d}{{\text{x}}^{4}}}$, thus the order of the equation is $4$.

As the differential equation is not polynomial in its derivative, therefore its degree is not defined.


2. For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

  1. $\text{xy=a}{{\text{e}}^{\text{x}}}\text{+b}{{\text{e}}^{\text{-x}}}\text{+}{{\text{x}}^{\text{2}}}\,\,\,\text{:x}\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{+2}\frac{\text{dy}}{\text{dx}}\text{-xy+}{{\text{x}}^{\text{2}}}\text{-2=0}$ 

Ans: The given function is:
$xy=a{{e}^{x}}+b{{e}^{-x}}+{{x}^{2}}$ 

Take derivative on both side:

$\Rightarrow y+x\frac{dy}{dx}=a{{e}^{x}}-b{{e}^{-x}}+2x$ 

Take derivative on both side:

$\Rightarrow \frac{dy}{dx}+x\frac{{{d}^{2}}y}{d{{x}^{2}}}+\frac{dy}{dx}=a{{e}^{x}}+b{{e}^{-x}}+2$ 

$\Rightarrow x\frac{{{d}^{2}}y}{d{{x}^{2}}}+2\frac{dy}{dx}=a{{e}^{x}}+b{{e}^{-x}}+2$ ……(1)

The given differential equation is:
$x\frac{{{d}^{2}}y}{d{{x}^{2}}}+2\frac{dy}{dx}-xy+{{x}^{2}}-2=0$ 

Solving LHS:

Substitute $\text{x}\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{+2}\frac{\text{dy}}{\text{dx}}$ from the result (1) and $\text{xy}$:

$\Rightarrow \left( x\frac{{{d}^{2}}y}{d{{x}^{2}}}+2\frac{dy}{dx} \right)-xy+{{x}^{2}}-2$ 

$\Rightarrow \left( a{{e}^{x}}+b{{e}^{-x}}+2 \right)-\left( a{{e}^{x}}+b{{e}^{-x}}+{{x}^{2}} \right)+{{x}^{2}}-2$ 

$\Rightarrow 2-{{x}^{2}}+{{x}^{2}}-2$ 

$\Rightarrow 0$ 

Thus LHS=RHS, the given function is the solution of the given differential equation. 


  1. $\text{y=}{{\text{e}}^{\text{x}}}\left( \text{acosx+bsinx} \right)\,\,\text{:}\,\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{-2}\frac{\text{dy}}{\text{dx}}\text{+2y=0}$ 

Ans: The given function is:
$y={{e}^{x}}\left( a\cos x+b\sin x \right)$ 

Take derivative on both side:

$\Rightarrow \frac{dy}{dx}={{e}^{x}}\left( a\cos x+b\sin x \right)+{{e}^{x}}\left( -a\sin x+b\cos x \right)$ 

$\Rightarrow \frac{dy}{dx}={{e}^{x}}\left( \left( a+b \right)\cos x+\left( b-a \right)\sin x \right)$ 

Take derivative on both side:

$\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}={{e}^{x}}\left( \left( a+b \right)\cos x+\left( b-a \right)\sin x \right)+{{e}^{x}}\left( -\left( a+b \right)\sin x+\left( b-a \right)\cos x \right)$

$\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}={{e}^{x}}\left( \left( a+b+b-a \right)\cos x+\left( b-a-a-b \right)\sin x \right)$ 

$\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}={{e}^{x}}\left( 2b\cos x-2a\sin x \right)$  

The given differential equation is:
$\frac{{{d}^{2}}y}{d{{x}^{2}}}-2\frac{dy}{dx}+2y=0$ 

Solving LHS:

$\Rightarrow {{e}^{x}}\left( 2b\cos x-2a\sin x \right)-2{{e}^{x}}\left( \left( a+b \right)\cos x+\left( b-a \right)\sin x \right)+2y$ 

$\Rightarrow {{e}^{x}}\left( \left( 2b-2a-2b \right)\cos x+\left( -2a-2b+2a \right)\sin x \right)-2y$ 

$\Rightarrow {{e}^{x}}\left( -2a\cos x-2b\sin x \right)-2y$ 

$\Rightarrow -2{{e}^{x}}\left( a\cos x+b\sin x \right)-2y$

$\Rightarrow 0$ 

Thus LHS=RHS, the given function is the solution of the given differential equation. 


  1. $\text{y=x}\,\text{sin3x}\,\,\text{:}\,\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{+9y-6cos3x=0}$ 

Ans: The given function is:
$y=x\sin 3x$ 

Take derivative on both side:

$\Rightarrow \frac{dy}{dx}=\sin 3x+3x\cos 3x$ 

Take derivative on both side:

$\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=3\cos 3x+3\left( \cos 3x+x\left( -3\sin 3x \right) \right)$

\[\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=3\cos 3x+3\cos 3x-9x\sin 3x\] 

\[\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=6\cos 3x-9x\sin 3x\]  

The given differential equation is:
$\frac{{{d}^{2}}y}{d{{x}^{2}}}+9y-6\cos 3x=0$ 

Solving LHS:

$\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}+9y-6\cos 3x$ 

$\Rightarrow \left( 6\cos 3x-9x\sin 3x \right)+9\left( x\sin 3x \right)-6\cos 3x$ 

$\Rightarrow 6\cos 3x-9x\sin 3x+9x\sin 3x-6\cos 3x$ 

$\Rightarrow 0$ 

Thus LHS=RHS, the given function is the solution of the given differential equation. 


  1. ${{\text{x}}^{\text{2}}}\text{=2}{{\text{y}}^{\text{2}}}\,\text{log y}\,\,\text{:}\left( {{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}} \right)\frac{\text{dy}}{\text{dx}}\text{-xy=0}$ 

Ans: The given function is:
${{x}^{2}}=2{{y}^{2}}\log y$ 

Take derivative on both side:

\[\Rightarrow 2x=2\left( 2y\log y+{{y}^{2}}\left( \frac{1}{y} \right) \right)\frac{dy}{dx}\] 

\[\Rightarrow \frac{dy}{dx}=\frac{x}{\left( 2y\log y+y \right)}\]

Multiply numerator and denominator by $\text{y}$:

\[\Rightarrow \frac{dy}{dx}=\frac{xy}{\left( 2{{y}^{2}}\log y+{{y}^{2}} \right)}\] 

\[\Rightarrow \frac{dy}{dx}=\frac{xy}{\left( {{x}^{2}}+{{y}^{2}} \right)}\]  

The given differential equation is:
$\left( {{x}^{2}}+{{y}^{2}} \right)\frac{dy}{dx}-xy=0$ 

Solving LHS:

$\Rightarrow \left( {{x}^{2}}+{{y}^{2}} \right)\frac{dy}{dx}-xy$ 

$\Rightarrow \left( {{x}^{2}}+{{y}^{2}} \right)\left( \frac{xy}{{{x}^{2}}+{{y}^{2}}} \right)-xy$ 

$\Rightarrow xy-xy$ 

$\Rightarrow 0$ 

Thus LHS=RHS, the given function is the solution of the given differential equation. 


3. Form the differential equation representing the family of curves given by ${{\left( \text{x-a} \right)}^{\text{2}}}\text{+2}{{\text{y}}^{\text{2}}}\text{=}{{\text{a}}^{\text{2}}}$  , where $\text{a}$  is an arbitrary constant.

Ans: Given family of curve:

${{\left( x-a \right)}^{2}}+2{{y}^{2}}={{a}^{2}}$ 

$\Rightarrow {{x}^{2}}-2ax+{{a}^{2}}+2{{y}^{2}}={{a}^{2}}$ 

$\Rightarrow {{x}^{2}}+2{{y}^{2}}=2ax$ ……(1)

Differentiate both side with respect to $\text{x}$:

$\Rightarrow 2\left( x-a \right)+4y\frac{dy}{dx}=0$ 

$\Rightarrow \frac{dy}{dx}=-\frac{2\left( x-a \right)}{4y}$ 

$\Rightarrow \frac{dy}{dx}=\frac{2a-2x}{4y}$ 

Multiply numerator and denominator by $\text{x}$:

$\Rightarrow \frac{dy}{dx}=\frac{2ax-2{{x}^{2}}}{4xy}$ 

Using expression (1) back substitute $\text{2ax}$:

$\Rightarrow \frac{dy}{dx}=\frac{{{x}^{2}}+2{{y}^{2}}-2{{x}^{2}}}{4xy}$ 

$\Rightarrow \frac{dy}{dx}=\frac{2{{y}^{2}}-{{x}^{2}}}{4xy}$ 

Thus the differential equation for given family of curve is $\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{2}{{\text{y}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}}}{\text{4xy}}$.


4. Prove that ${{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}\text{=c}{{\left( {{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}} \right)}^{\text{2}}}$  is the general solution of differential equation $\left( {{\text{x}}^{\text{3}}}\text{-3x}{{\text{y}}^{\text{2}}} \right)\text{dx=}\left( {{\text{y}}^{\text{3}}}\text{-3}{{\text{x}}^{\text{2}}}\text{y} \right)\text{dy}$ , where $\text{c}$ is a parameter.

 Ans: Given differential equation:

$\left( {{x}^{3}}-3x{{y}^{2}} \right)dx=\left( {{y}^{3}}-3{{x}^{2}}y \right)dy$ 

$\Rightarrow \frac{dy}{dx}=\frac{{{x}^{3}}-3x{{y}^{2}}}{{{y}^{3}}-3{{x}^{2}}y}$ 

As it can be seen that this is an homogenous equation. Substitute $\text{y=vx}$:

$\Rightarrow \frac{d\left( vx \right)}{dx}=\frac{{{x}^{3}}-3x{{\left( vx \right)}^{2}}}{{{\left( vx \right)}^{3}}-3{{x}^{2}}\left( vx \right)}$ 

$\Rightarrow v+x\frac{dv}{dx}=\frac{{{x}^{3}}\left( 1-3{{v}^{2}} \right)}{{{x}^{3}}\left( {{v}^{3}}-3v \right)}$ 

$\Rightarrow v+x\frac{dv}{dx}=\frac{1-3{{v}^{2}}}{{{v}^{3}}-3v}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{1-3{{v}^{2}}}{{{v}^{3}}-3v}-v$ 

$\Rightarrow x\frac{dv}{dx}=\frac{1-3{{v}^{2}}-{{v}^{4}}+3{{v}^{2}}}{{{v}^{3}}-3v}$ 

$\Rightarrow x\frac{dv}{dx}=\frac{1-{{v}^{4}}}{{{v}^{3}}-3v}$ 

Separate the differentials:

$\frac{{{v}^{3}}-3v}{1-{{v}^{4}}}dv=\frac{dx}{x}$ 

Integrate both side:

$\int \frac{{{v}^{3}}-3v}{1-{{v}^{4}}}dv=\int \frac{dx}{x}$ 

$\Rightarrow \int \frac{{{v}^{3}}-3v}{1-{{v}^{4}}}dv=\log x+\log C$ 

\[\Rightarrow I=\log x+\log C\,\,\left( I=\int \frac{{{v}^{3}}-3v}{1-{{v}^{4}}}dv \right)\] ……(1)

Solving integral $\text{I}$:

\[\Rightarrow I=\int \frac{{{v}^{3}}-3v}{1-{{v}^{4}}}dv\] 

\[\Rightarrow \frac{{{v}^{3}}-3v}{1-{{v}^{4}}}=\frac{{{v}^{3}}-3v}{\left( 1-{{v}^{2}} \right)\left( 1+{{v}^{2}} \right)}\] 

\[\Rightarrow \frac{{{v}^{3}}-3v}{1-{{v}^{4}}}=\frac{{{v}^{3}}-3v}{\left( 1-v \right)\left( 1+v \right)\left( 1+{{v}^{2}} \right)}\] 

Using partial fraction:

\[\Rightarrow \frac{{{v}^{3}}-3v}{1-{{v}^{4}}}=\frac{A}{1-v}+\frac{B}{1+v}+\frac{Cv+D}{1+{{v}^{2}}}\] 

Solving for $\text{A,B,C}\,\text{and}\,\text{D}$:

$A=-\frac{1}{2}$ 

$B=\frac{1}{2}$ 

$C=-2$ 

$D=0$ 

\[\Rightarrow \frac{{{v}^{3}}-3v}{1-{{v}^{4}}}=\frac{-\frac{1}{2}}{1-v}+\frac{\frac{1}{2}}{1+v}+\frac{-2v+0}{1+{{v}^{2}}}\] 

$I=-\frac{1}{2}\int \frac{1}{1-v}dv+\frac{1}{2}\int \frac{1}{1+v}dv-\int \frac{2v}{1+{{v}^{2}}}dv$

$I=-\frac{1}{2}\left( -\log \left( 1-v \right) \right)+\frac{1}{2}\left( \log \left( 1+v \right) \right)-\log \left( 1+{{v}^{2}} \right)$ 

\[I=\frac{1}{2}\left( \log \left( 1-{{v}^{2}} \right) \right)-\frac{2}{2}\log \left( 1+{{v}^{2}} \right)\] 

\[I=\frac{1}{2}\left( \log \frac{\left( 1-{{v}^{2}} \right)}{{{\left( 1+{{v}^{2}} \right)}^{2}}} \right)\] 

\[\Rightarrow I=\frac{1}{2}\left( \log \frac{\left( 1-\frac{{{y}^{2}}}{{{x}^{2}}} \right)}{{{\left( 1+\frac{{{y}^{2}}}{{{x}^{2}}} \right)}^{2}}} \right)\] 

\[\Rightarrow I=\frac{1}{2}\left( \log \frac{{{x}^{2}}\left( {{x}^{2}}-{{y}^{2}} \right)}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}} \right)\]

\[\Rightarrow I=\frac{1}{2}\left( \log \frac{\left( {{x}^{2}}-{{y}^{2}} \right)}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}} \right)+\frac{1}{2}\log {{x}^{2}}\] 

\[\Rightarrow I=\frac{1}{2}\left( \log \frac{\left( {{x}^{2}}-{{y}^{2}} \right)}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}} \right)+\log x\] 

 

Back substitute $\text{I}$ in expression (1):

\[\Rightarrow I=\log x+\log C\,\] 

\[\Rightarrow \frac{1}{2}\left( \log \frac{\left( {{x}^{2}}-{{y}^{2}} \right)}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}} \right)+\log x=\log x+\log C\]

\[\Rightarrow \log \frac{\left( {{x}^{2}}-{{y}^{2}} \right)}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}}=2\log C\] 

\[\Rightarrow \frac{{{x}^{2}}-{{y}^{2}}}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}}={{C}^{2}}\] 

\[\Rightarrow {{x}^{2}}-{{y}^{2}}=c{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}\,\,\left( c={{C}^{2}} \right)\] 

Thus for given differential equation, its general solution is \[{{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}\text{=c}{{\left( {{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}} \right)}^{\text{2}}}\].


5. Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

Ans: Draw a circle according to :


(Image will be uploaded soon)


The equation of the given circle will be:

${{\left( x-a \right)}^{2}}+{{\left( y-a \right)}^{2}}={{a}^{2}}$ 

Differentiate both side with respect to $\text{x}$:

$\Rightarrow 2\left( x-a \right)+2\left( y-a \right)\frac{dy}{dx}=0$ 

$\Rightarrow x-a+yy'-ay'=0$ 

$\Rightarrow a=\frac{x+yy'}{1+y'}$ 

Using equation of circle:

$\Rightarrow {{\left( x-a \right)}^{2}}+{{\left( y-a \right)}^{2}}={{a}^{2}}$

\[\Rightarrow {{\left( x-\left( \frac{x+yy'}{1+y'} \right) \right)}^{2}}+{{\left( y-\left( \frac{x+yy'}{1+y'} \right) \right)}^{2}}={{\left( \frac{x+yy'}{1+y'} \right)}^{2}}\]  

\[\Rightarrow {{\left( \frac{x+xy'-x-yy'}{1+y'} \right)}^{2}}+{{\left( \frac{y+yy'-x-yy'}{1+y'} \right)}^{2}}={{\left( \frac{x+yy'}{1+y'} \right)}^{2}}\]

\[\Rightarrow {{\left( \frac{y'\left( x-y \right)}{1+y'} \right)}^{2}}+{{\left( \frac{y-x}{1+y'} \right)}^{2}}={{\left( \frac{x+yy'}{1+y'} \right)}^{2}}\]

\[\Rightarrow {{\left( x-y \right)}^{2}}\left( 1+y{{'}^{2}} \right)={{\left( x+yy' \right)}^{2}}\]   

Thus the differential equation for given family of curve is \[{{\left( \text{x-y} \right)}^{\text{2}}}\left( \text{1+y}{{\text{ }\!\!'\!\!\text{ }}^{\text{2}}} \right)\text{=}{{\left( \text{x+yy }\!\!'\!\!\text{ } \right)}^{\text{2}}}\].


6. Find the general solution of the differential equation $\frac{\text{dy}}{\text{dx}}\text{+}\sqrt{\frac{\text{1-}{{\text{y}}^{\text{2}}}}{\text{1-}{{\text{x}}^{\text{2}}}}}\text{=0}$.

Ans: The given differential equation is:

 $\frac{dy}{dx}+\sqrt{\frac{1-{{y}^{2}}}{1-{{x}^{2}}}}=0$ 

$\frac{dy}{dx}=-\sqrt{\frac{1-{{y}^{2}}}{1-{{x}^{2}}}}$ 

$\Rightarrow \frac{dy}{\sqrt{1-{{y}^{2}}}}=-\frac{dx}{\sqrt{1-{{x}^{2}}}}$ 

Integrate both side:

$\Rightarrow \int \frac{dy}{\sqrt{1-{{y}^{2}}}}=-\int \frac{dx}{\sqrt{1-{{x}^{2}}}}$ 

$\Rightarrow {{\sin }^{-1}}y=-{{\sin }^{-1}}x+C$

$\Rightarrow {{\sin }^{-1}}y+{{\sin }^{-1}}x=C$ 

Thus the general solution of given differential equation is $\text{si}{{\text{n}}^{\text{-1}}}\text{y+si}{{\text{n}}^{\text{-1}}}\text{x=C}$.


7. Show that the general solution of the differential equation $\frac{\text{dy}}{\text{dx}}\text{+}\frac{{{\text{y}}^{\text{2}}}\text{+y+1}}{{{\text{x}}^{\text{2}}}\text{+x+1}}\text{=0}$ is given by $\left( \text{x+y+1} \right)\text{=A}\left( \text{1-x-y-2xy} \right)$  where $\text{A}$ is a parameter. 

Ans: The given differential equation is:

$\frac{dy}{dx}+\frac{{{y}^{2}}+y+1}{{{x}^{2}}+x+1}=0$

$\Rightarrow \frac{dy}{dx}=-\frac{{{y}^{2}}+y+1}{{{x}^{2}}+x+1}$

$\Rightarrow \frac{dy}{dx}=-\frac{{{y}^{2}}+2\left( \frac{1}{2} \right)y+\frac{1}{4}-\frac{1}{4}+1}{{{x}^{2}}+2\left( \frac{1}{2} \right)x+\frac{1}{4}-\frac{1}{4}+1}$ 

$\Rightarrow \frac{dy}{dx}=-\frac{{{\left( y+\frac{1}{2} \right)}^{2}}+\frac{3}{4}}{{{\left( x+\frac{1}{2} \right)}^{2}}+\frac{3}{4}}$   

$\Rightarrow \frac{dy}{{{\left( y+\frac{1}{2} \right)}^{2}}+\frac{3}{4}}=-\frac{dx}{{{\left( x+\frac{1}{2} \right)}^{2}}+\frac{3}{4}}$ 

Integrate both side:

$\Rightarrow \int \frac{dy}{{{\left( y+\frac{1}{2} \right)}^{2}}+{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}}=-\int \frac{dx}{{{\left( x+\frac{1}{2} \right)}^{2}}+{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}}$

$\Rightarrow \frac{1}{\left( \frac{\sqrt{3}}{2} \right)}{{\tan }^{-1}}\left[ \frac{y+\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right]=-\frac{1}{\left( \frac{\sqrt{3}}{2} \right)}{{\tan }^{-1}}\left[ \frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right]+C$ 

$\Rightarrow \frac{2}{\sqrt{3}}\left( {{\tan }^{-1}}\left[ \frac{2y+1}{\sqrt{3}} \right]+{{\tan }^{-1}}\left[ \frac{2x+1}{\sqrt{3}} \right] \right)=C$ 

$\Rightarrow {{\tan }^{-1}}\left[ \frac{2y+1}{\sqrt{3}} \right]+{{\tan }^{-1}}\left[ \frac{2x+1}{\sqrt{3}} \right]=\frac{\sqrt{3}}{2}C$ 

Thus the general solution for given differential equation is $\text{ta}{{\text{n}}^{\text{-1}}}\left[ \frac{\text{2y+1}}{\sqrt{\text{3}}} \right]\text{+ta}{{\text{n}}^{\text{-1}}}\left[ \frac{\text{2x+1}}{\sqrt{\text{3}}} \right]\text{=}\frac{\sqrt{\text{3}}}{\text{2}}\text{C}$.


8. Find the equation of the curve passing through the point $\left( 0,\frac{\pi }{4} \right)$ whose differential equation is $\text{sin}\,\text{x}\,\text{cos}\,\text{ydx+cos}\,\text{x}\,\text{sin}\,\text{ydy=0}$.

Ans: Given differential equation is:

$\sin \,x\,\cos \,ydx+\cos \,x\,\sin \,ydy=0$

$\Rightarrow \sin \,x\,\cos \,ydx+\cos \,x\,\sin \,ydy=0$

Divide both side by $\text{cos}\,\text{x}\,\text{cos}\,\text{y}$:

$\Rightarrow \frac{\sin \,x\,\cos \,ydx+\cos \,x\,\sin \,ydy}{\cos x\cos y}=0$ 

$\Rightarrow \tan xdx+\tan ydy=0$ 

$\Rightarrow \tan ydy=-\tan xdx$ 

Integrate both side:

$\Rightarrow \int \tan ydy=-\int \tan xdx$ 

$\Rightarrow \log \left( \sec y \right)=-\log \left( \sec x \right)+C$ 

$\Rightarrow \log \left( \sec y \right)+\log \left( \sec x \right)=C$ 

$\Rightarrow \log \left( \sec x\sec y \right)=C$ 

$\Rightarrow \sec x\sec y=k\,\,\left( k={{e}^{C}} \right)$ 

As curve passes through $\left( \text{0,}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right)$:

$\sec 0\sec \left( \frac{\pi }{4} \right)=k$ 

$\Rightarrow k=\sqrt{2}$ 

$\Rightarrow \sec x\sec y=\sqrt{2}$ 

Thus the equation of required curve is $\text{sec}\,\text{x}\,\text{sec}\,\text{y=}\sqrt{\text{2}}$.


9. Find the particular solution of the differential equation $\left( \text{1+}{{\text{e}}^{\text{2x}}} \right)\text{dy+}\left( \text{1+}{{\text{y}}^{\text{2}}} \right){{\text{e}}^{\text{x}}}\text{dx=0}$ given that $\text{y=1}$ when $\text{x=0}$.

Ans: The given differential equation is:

$\left( 1+{{e}^{2x}} \right)dy+\left( 1+{{y}^{2}} \right){{e}^{x}}dx=0$ 

Divide both side $\left( \text{1+}{{\text{e}}^{\text{2x}}} \right)\left( \text{1+}{{\text{y}}^{\text{2}}} \right)$:

\[\frac{dy}{\left( 1+{{y}^{2}} \right)}+\frac{{{e}^{x}}}{\left( 1+{{e}^{2x}} \right)}dx=0\] 

\[\int \frac{dy}{\left( 1+{{y}^{2}} \right)}=-\int \frac{{{e}^{x}}}{\left( 1+{{e}^{2x}} \right)}dx\] 

\[{{\tan }^{-1}}y=-\int \frac{{{e}^{x}}}{\left( 1+{{\left( {{e}^{x}} \right)}^{2}} \right)}dx\]

Substitute $t={{e}^{x}}$:

$dt={{e}^{x}}dx$ 

\[\Rightarrow {{\tan }^{-1}}y=-\int \frac{1}{\left( 1+{{t}^{2}} \right)}dt\] 

\[\Rightarrow {{\tan }^{-1}}y=-{{\tan }^{-1}}t+C\] 

\[\Rightarrow {{\tan }^{-1}}y=-{{\tan }^{-1}}{{e}^{x}}+C\] 

$\Rightarrow {{\tan }^{-1}}y+{{\tan }^{-1}}{{e}^{x}}=C$ 

As $\text{y=1}$ when $\text{x=0}$:

${{\tan }^{-1}}\left( 1 \right)+{{\tan }^{-1}}\left( {{e}^{0}} \right)=C$ 

\[\Rightarrow \frac{\pi }{4}+\frac{\pi }{4}=C\] 

\[\Rightarrow C=\frac{\pi }{2}\] 

$\Rightarrow {{\tan }^{-1}}y+{{\tan }^{-1}}{{e}^{x}}=\frac{\pi }{2}$ 

Thus the required particular solution is $\text{ta}{{\text{n}}^{\text{-1}}}\text{y+ta}{{\text{n}}^{\text{-1}}}{{\text{e}}^{\text{x}}}\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$.


10. Solve the differential equation $\text{y}{{\text{e}}^{\frac{\text{x}}{\text{y}}}}\text{dx=}\left( \text{x}{{\text{e}}^{\frac{\text{x}}{\text{y}}}}\text{+}{{\text{y}}^{\text{2}}} \right)\text{dy}\,\,\left( \text{y}\ne \text{0} \right)$.

Ans: The given differential equation is:

$y{{e}^{\frac{x}{y}}}dx=\left( x{{e}^{\frac{x}{y}}}+{{y}^{2}} \right)dy$ 

$y{{e}^{\frac{x}{y}}}\frac{dx}{dy}=x{{e}^{\frac{x}{y}}}+{{y}^{2}}$ 

$\Rightarrow y{{e}^{\frac{x}{y}}}\frac{dx}{dy}-x{{e}^{\frac{x}{y}}}={{y}^{2}}$ 

$\Rightarrow {{e}^{\frac{x}{y}}}\frac{\left[ y\frac{dx}{dy}-x \right]}{{{y}^{2}}}=1$ 

Substitute $\text{z=}{{\text{e}}^{\frac{\text{x}}{\text{y}}}}$:

$z={{e}^{\frac{x}{y}}}$ 

$\frac{d}{dy}z=\frac{d}{dy}{{e}^{\frac{x}{y}}}$ 

$\Rightarrow \frac{dz}{dy}=\frac{d}{dy}\left( {{e}^{\frac{x}{y}}} \right)$ 

$\Rightarrow \frac{dz}{dy}={{e}^{\frac{x}{y}}}\frac{d}{dy}\left( \frac{x}{y} \right)$ 

$\Rightarrow \frac{dz}{dy}={{e}^{\frac{x}{y}}}\left[ \left( \frac{1}{y} \right)\frac{dx}{dy}-\frac{x}{{{y}^{2}}} \right]$ 

$\Rightarrow \frac{dz}{dy}={{e}^{\frac{x}{y}}}\left[ \frac{y\frac{dx}{dy}-x}{{{y}^{2}}} \right]$ 

$\Rightarrow \frac{dz}{dy}=1$ 

\[\Rightarrow dz=dy\] 

\[\Rightarrow \int dz=\int dy\]

\[\Rightarrow z=y+C\] 

\[\Rightarrow {{e}^{\frac{x}{y}}}=y+C\] 

Thus the required general solution is \[{{\text{e}}^{\frac{\text{x}}{\text{y}}}}\text{=y+C}\].


11. Find a particular solution of the differential equation $\left( \text{x-y} \right)\left( \text{dx+dy} \right)\text{=dx-dy}$ given that $\text{y=-1}$ when $\text{x=0}$. Hint (put $\text{x-y=t}$).

Ans: Given differential equation is:

$\left( x-y \right)\left( dx+dy \right)=dx-dy$ 

$\Rightarrow \left( x-y \right)dx-dx=\left( y-x \right)dy-dy$

$\Rightarrow \left( x-y+1 \right)dy=\left( 1-x+y \right)dx$ 

$\Rightarrow \frac{dy}{dx}=\frac{1-x+y}{x-y+1}$ 

Put $\text{x-y=t}$:

$x-y=t$ 

$\Rightarrow 1-\frac{dy}{dx}=\frac{dt}{dx}$ 

$\Rightarrow 1-\frac{dt}{dx}=\frac{dy}{dx}$ 

$\Rightarrow 1-\frac{dt}{dx}=\frac{1-t}{1+t}$ 

$\Rightarrow \frac{dt}{dx}=1-\frac{1-t}{1+t}$

$\Rightarrow \frac{dt}{dx}=\frac{1+t-1+t}{1+t}$ 

$\Rightarrow \frac{dt}{dx}=\frac{2t}{1+t}$ 

$\Rightarrow \frac{1+t}{t}dt=2dx$ 

Integrate both side:

$\Rightarrow \int \frac{1+t}{t}dt=2\int dx$

$\Rightarrow \int \frac{1}{t}dt+\int dt=2x+C$ 

$\Rightarrow \log \left| t \right|+t=2x+C$ 

$\Rightarrow \log \left| x-y \right|+x-y=2x+C$ 

$\Rightarrow \log \left| x-y \right|-y=x+C$ 

As $\text{y=-1}$ when $\text{x=0}$:

$\Rightarrow \log \left| 0-\left( -1 \right) \right|-\left( -1 \right)=0+C$ 

$\Rightarrow \log 1+1=C$ 

$\Rightarrow C=1$ 

Thus the required particular solution is:

$\log \left| x-y \right|-y=x+1$.


12. Solve the differential equation $\left[ \frac{{{\text{e}}^{\text{-2}\sqrt{\text{x}}}}}{\sqrt{\text{x}}}\text{-}\frac{\text{y}}{\sqrt{\text{x}}} \right]\frac{\text{dx}}{\text{dy}}\text{=1}\,\left( \text{x}\ne \text{0} \right)$.

Ans: Given differential equation is:
$\left[ \frac{{{e}^{-2\sqrt{x}}}}{\sqrt{x}}-\frac{y}{\sqrt{x}} \right]\frac{dx}{dy}=1$ 

$\Rightarrow \frac{dy}{dx}=\frac{{{e}^{-2\sqrt{x}}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}$ 

$\Rightarrow \frac{dy}{dx}+\frac{y}{\sqrt{x}}=\frac{{{e}^{-2\sqrt{x}}}}{\sqrt{x}}$ 

It is linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$:

$p=\frac{1}{\sqrt{x}}$ 

$Q=\frac{{{e}^{-2\sqrt{x}}}}{\sqrt{x}}$ 

Calculating integrating factor:

$I.F={{e}^{\int pdx}}$

$I.F={{e}^{\int \frac{1}{\sqrt{x}}dx}}$ 

$I.F={{e}^{2\sqrt{x}}}$  

The general solution is given by:

$y\times I.F=\int \left( Q\times I.F \right)dx+C$ 

$y\times \left( {{e}^{2\sqrt{x}}} \right)=\int \left( \frac{{{e}^{-2\sqrt{x}}}}{\sqrt{x}}\times {{e}^{2\sqrt{x}}} \right)dx+C$ 

$\Rightarrow y{{e}^{2\sqrt{x}}}=\int \left( \frac{{{e}^{-2\sqrt{x}+2\sqrt{x}}}}{\sqrt{x}} \right)dx+C$ 

$\Rightarrow y{{e}^{2\sqrt{x}}}=\int \frac{1}{\sqrt{x}}dx+C$ 

$\Rightarrow y{{e}^{2\sqrt{x}}}=2\sqrt{x}+C$ 

Thus the general solution for the given differential equation is

$y{{e}^{2\sqrt{x}}}=2\sqrt{x}+C$.


13. Find a particular solution of the differential equation $\frac{\text{dy}}{\text{dx}}\text{+ycotx=4x}\,\text{cosec}\,\text{x}\left( \text{x}\ne \text{0} \right)$ given that $\text{y=0}$ when $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$.

Ans:
The given differential equation is:

 $\frac{dy}{dx}+y\cot x=4x\,\text{cosec}x$ 

It is linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$:

$p=\cot x$ 

$Q=4x\text{cosec}\,x$ 

Calculating integrating factor:

$I.F={{e}^{\int pdx}}$

$I.F={{e}^{\int \cot xdx}}$ 

$I.F={{e}^{\log \left| \sin x \right|}}$ 

$I.F=\sin x$ 

The general solution is given by:

$y\times I.F=\int \left( Q\times I.F \right)dx+C$

$\Rightarrow y\times \sin x=\int \left( 4x\cos \text{ec}x \right)\sin xdx+C$ 

$\Rightarrow y\sin x=4\int xdx+C$ 

$\Rightarrow y\sin x=4\left( \frac{{{x}^{2}}}{2} \right)+C$ 

$\Rightarrow y\sin x=2{{x}^{2}}+C$ 

As $\text{y=0}$ when $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$:

$0\times \sin \left( \frac{\pi }{2} \right)=2{{\left( \frac{\pi }{2} \right)}^{2}}+C$ 

$C=-2\left( \frac{{{\pi }^{2}}}{4} \right)$ 

$C=-\frac{{{\pi }^{2}}}{2}$ 

Thus the required particular solution is:

$y\sin x=2{{x}^{2}}-\frac{{{\pi }^{2}}}{2}$


14. Find a particular solution of the differential equation $\left( \text{x+1} \right)\frac{\text{dy}}{\text{dx}}\text{=2}{{\text{e}}^{\text{-y}}}\text{-1}$ given that $\text{y=0}$ when $\text{x=0}$.

Ans:
The given differential equation is:

 $\left( x+1 \right)\frac{dy}{dx}=2{{e}^{-y}}-1$

$\Rightarrow \frac{dy}{2{{e}^{-y}}-1}=\frac{dx}{x+1}$ 

Integrate both side:

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=\int \frac{dx}{x+1}$ 

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=\log \left( x+1 \right)+\log C$ ……(1) 

Evaluating LHS integral:

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=\int \frac{{{e}^{y}}dy}{2-{{e}^{y}}}$ 

Put $\text{t=2-}{{\text{e}}^{\text{y}}}$:

$t=2-{{e}^{y}}$

$dt=-{{e}^{y}}dy$  

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=-\int \frac{dt}{t}$

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=-\log \left( t \right)$ 

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=\log \frac{1}{t}$ 

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=\log \frac{1}{2-{{e}^{y}}}$ 

Back substituting in expression (1):

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=\log \left( x+1 \right)+\log C$ 

$\Rightarrow \log \left( \frac{1}{2-{{e}^{y}}} \right)=\log C\left( x+1 \right)$ 

$\Rightarrow 2-{{e}^{y}}=\frac{1}{C\left( x+1 \right)}$ 

As $\text{y=0}$ when $\text{x=0}$:

$\Rightarrow 2-{{e}^{0}}=\frac{1}{C\left( 0+1 \right)}$ 

$\Rightarrow 2-1=\frac{1}{C}$ 

$\Rightarrow C=1$ 

Thus the required particular solution is:

$\Rightarrow 2-{{e}^{y}}=\frac{1}{\left( x+1 \right)}$

$\Rightarrow {{e}^{y}}=2-\frac{1}{\left( x+1 \right)}$ 

$\Rightarrow {{e}^{y}}=\frac{2x+2-1}{\left( x+1 \right)}$ 

$\Rightarrow {{e}^{y}}=\frac{2x+1}{x+1}$ 

$\Rightarrow y=\log \left( \frac{2x+1}{x+1} \right)$  

Thus for given conditions the particular solution is $\text{y=log}\left( \frac{\text{2x+1}}{\text{x+1}} \right)$ .


15. The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was $\text{20000}$  in $\text{1999}$ and $\text{25000}$ in the year$\text{2004}$ , what will be the population of the village in $\text{2009}$?

Ans: Let the population at time $\text{t}$ be $\text{y}$. Then according to :

$\frac{dy}{dt}\propto y$ 

$\Rightarrow \frac{dy}{dt}=ky$ 

$\Rightarrow \frac{dy}{y}=kdt$ 

Integrate both side:

$\Rightarrow \int \frac{dy}{y}=k\int dt$ 

$\Rightarrow \log y=kt+C$ 

In $1999$ (taking reference year) $\text{t=0}$ population was $\text{y=20000}$.

$\Rightarrow \log \left( 20000 \right)=k\left( 0 \right)+C$ 

$\Rightarrow C=\log \left( 20000 \right)$ 

In $2004$, $\text{t=5}$ population was $\text{y=25000}$.

$\Rightarrow \log \left( 25000 \right)=k\left( 5 \right)+\log \left( 20000 \right)$ 

$\Rightarrow 5k=\log \left( 25000 \right)-\log \left( 20000 \right)$ 

$\Rightarrow 5k=\log \left( \frac{25000}{20000} \right)$ 

$\Rightarrow k=\frac{1}{5}\log \left( \frac{5}{4} \right)$ 

Thus the population relation will be:

$\log y=\frac{1}{5}\log \left( \frac{5}{4} \right)t+\log \left( 20000 \right)$ 

For $2009$, $t=10$ 

$\Rightarrow \log y=\frac{1}{5}\log \left( \frac{5}{4} \right)\times 10+\log \left( 20000 \right)$ 

$\Rightarrow \log y=2\log \left( \frac{5}{4} \right)+\log \left( 20000 \right)$ 

$\Rightarrow \log y=\log {{\left( \frac{5}{4} \right)}^{2}}+\log \left( 20000 \right)$ 

$\Rightarrow \log y=\log \left( \frac{25}{16}\times 20000 \right)$ 

$\Rightarrow y=31250$ 

Thus the population in the year $\text{2009}$, is $31250$.


16. The general solution of the differential equation $\frac{\text{ydx-xdy}}{\text{y}}\text{=0}$.

A. $xy=C$ 

B. $x=C{{y}^{2}}$ 

C. $y=Cx$ 

D. $y=C{{x}^{2}}$ 


Ans: Given differential equation:

$\frac{ydx-xdy}{y}=0$

Divide both side by $\text{x}$ :

$\Rightarrow \frac{ydx-xdy}{xy}=0$

$\Rightarrow \frac{dx}{x}-\frac{dy}{y}=0$ 

Integrate both side:

$\Rightarrow \int \frac{dx}{x}-\int \frac{dy}{y}=0$ 

$\Rightarrow \log \left| x \right|-\log \left| y \right|=\log k$ 

$\Rightarrow \log \left| \frac{x}{y} \right|=\log k$ 

$\Rightarrow \frac{x}{y}=k$ 

$\Rightarrow y=Cx\,\,\,\left( C=\frac{1}{k} \right)$

Thus the correct option is (C)


17. Find the general solution of a differential equation of the type $\frac{\text{dx}}{\text{dy}}\text{+}{{\text{P}}_{\text{1}}}\text{x=}{{\text{Q}}_{\text{1}}}$.

A. \[y{{e}^{\int {{P}_{1}}dy}}=\int \left( {{Q}_{1}}{{e}^{\int {{P}_{1}}dy}} \right)dy+C\]   

B. \[y{{e}^{\int {{P}_{1}}dx}}=\int \left( {{Q}_{1}}{{e}^{\int {{P}_{1}}dx}} \right)dy+C\] 

C. \[x{{e}^{\int {{P}_{1}}dy}}=\int \left( {{Q}_{1}}{{e}^{\int {{P}_{1}}dy}} \right)dy+C\]   

D. \[x{{e}^{\int {{P}_{1}}dy}}=\int \left( {{Q}_{1}}{{e}^{\int {{P}_{1}}dx}} \right)dy+C\]   

Ans: The given differential equation is:

$\frac{dx}{dy}+{{P}_{1}}x={{Q}_{1}}$ 

It is a linear differential equation and its general solution is:

$x{{e}^{\int {{P}_{1}}dy}}=\int \left( {{Q}_{1}}{{e}^{\int {{P}_{1}}dy}} \right)dy+C$ 

With integrating factor $I.F={{e}^{\int {{P}_{1}}dy}}$.

Thus the correct option is (C).


18. Find the general solution of the differential equation ${{\text{e}}^{\text{x}}}\text{dy+}\left( \text{y}{{\text{e}}^{\text{x}}}\text{+2x} \right)\text{dx=0}$.

A. $\text{x}{{\text{e}}^{\text{y}}}\text{+}{{\text{x}}^{\text{2}}}\text{=C}$ 

B. $\text{x}{{\text{e}}^{\text{y}}}\text{+}{{\text{y}}^{\text{2}}}\text{=C}$ 

C. $\text{y}{{\text{e}}^{\text{x}}}\text{+}{{\text{x}}^{\text{2}}}\text{=C}$ 

D. $\text{y}{{\text{e}}^{\text{y}}}\text{+}{{\text{x}}^{\text{2}}}\text{=C}$ 

Ans: The given differential equation is:
${{e}^{x}}dy+\left( y{{e}^{x}}+2x \right)dx=0$ 

$\Rightarrow {{e}^{x}}\frac{dy}{dx}+y{{e}^{x}}=-2x$ 

$\Rightarrow \frac{dy}{dx}+y=-2x{{e}^{-x}}$   

The given differential equation is of the form:

$\frac{dy}{dx}+Py=Q$ 

$\Rightarrow P=1$ 

$\Rightarrow Q=-2x{{e}^{-x}}$ 

Calculating integrating factor:
$I.F={{e}^{\int Pdx}}$

$\Rightarrow I.F={{e}^{\int dx}}$ 

$\Rightarrow I.F={{e}^{x}}$  

It is a linear differential equation and its general solution is:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$

 $y\left( {{e}^{x}} \right)=\int \left( -2x{{e}^{-x}}\times {{e}^{x}} \right)dy+C$ 

$\Rightarrow y{{e}^{x}}=-2\int xdx+C$ 

$\Rightarrow y{{e}^{x}}=-2\left( \frac{{{x}^{2}}}{2} \right)+C$ 

$\Rightarrow y{{e}^{x}}+{{x}^{2}}=C$ 

Thus the correct answer is option (C).


NCERT Solutions for Class 12 Maths Chapter 9 - Differential Equations

NCERT Solutions for Class 12 Maths – Free PDF Download

Differential Equations Class 12 Solutions PDF is an ideal respite from the busy lifestyle of students. Having readymade solutions not only gives you the required help with complex problems but they also relieve all the stress that comes from managing many academic things and other extracurricular activities in your life. Get 100 per cent accurate NCERT Books for Class 12 Maths Chapter 9 (Differential Equations) solved by expert Maths teachers.


Chapter 9 – Differential Equations 

NCERT Solutions for Class 12 Maths Chapter 9

9.1 Introduction

The introduction part of Chapter 9 Class 12 Maths will remind you of what you learned in the previous class of differentiation. You would recall how you could differentiate a function f by varying an independent variable. You would also remember that in the last chapter on integration, how you could find a function when its derivative is given. 

In this chapter, you would learn about differential equations which are based on the concepts described above. You would learn how a differential equation is represented employing a dependent variable, independent variable and differential coefficients of the dependent variable concerning the independent variable. You would get to know how a differential equation is denoted like (a2 + b2)da = (a2 - b2)db.

Ch 9 Maths Class 12 has many solved examples which are explained in detail to make you understand the concepts well. Some of the other main topics discussed in this chapter are:

  • Differential equation order

  • Differential equation degree

  • How to form a differential equation that can represent a family of curves

  • Homogeneous differential equations 

  • Linear differential equations 

  • How to solve a first-degree differential equation


9.2 Basic Concepts

In this topic of Class 12 Maths Ch 9 NCERT Solutions, you will be introduced to the fundamentals of the differential equation and their general representation. You would learn that an equation that has, apart from the dependent and independent variable, derivatives of the dependent variable regarding the independent variable is called a differential equation. An example of such an equation is: x * dy/dx + y = 0.

You would further expand this knowledge of differential equations to define ordinary and partial differential equations. A differential equation which involves the derivative of the dependent variable regarding only one independent variable is referred to as an ordinary differential equation like expressed above. There could be differential equations that have derivatives of the dependent variable concerning more than one independent variable and are called partial differential equations.

You would also understand what is meant by order of a differential equation which is the order of the highest order derivative of the dependent variable occurring in the equation. For example, dy/dx = ex is a first-order differential equation whereas d2y/dx2 +y = 0 is a 2nd order equation. 

You would get acquainted with the degree of a differential equation which is possible only if the differential equation is a polynomial equation. It is determined by the highest power (must be a positive integral index) of the highest order derivative involved in the differential equation. So, one can find out the degree of a differential equation for below-mentioned equations like:

  • (dy/dx)2 + dy/dx – sin2y = 0 -> This has 1st degree equation

  • dy/dx + sin(dy/dx) = 0 – Here it cannot be determined since it is not a polynomial equation.


Exercise 9.1

Solutions: 12 Questions (10 Short Questions, 2 MCQs)

9.3 General and Particular Solutions of a Differential Equation - H2

Here you will get to know what is meant by general and particular solutions of a differential equation. A general solution is the one where the independent arbitrary constants of the equation are equal to the order of the equation.  So for an equation d2y/dx2 + y = 0, its general solution would be given as y = K Cos x + C sin x,  since it has 2 arbitrary constants K and C which is equal to the order of the equation that is 2.

To find the particular solution of a differential equation, the arbitrary constants need to be given particular values. So, in the example, above if we replace K = C = 1, we get the solution y = cos x + sin x which is termed as the particular solution of the differential equation.


Exercise 9.2

Solutions: 12 Questions (10 Short Questions, 2 MCQs)

9.4 Formation of a Differential Equation whose General Solution is Given

In this, you would learn how to formulate differential equations given “n” arbitrary constants and differentiating the equation n times over to get the n + 1 equations. When you eliminate the arbitrary constants from these n + 1 equations, you will get the required differential equation. You would use this method to derive a differential equation that represents a family of curves. 


Exercise 9.3

Solutions: 12 Questions (5 Short Questions, 5 Long Questions, 2 MCQs)

9.5 Methods of Solving First Order, First Degree Differential Equations

In this section, you would learn 3 methods of solving a differential equation of first order and first degree. The 3 methods described are:

  • Differential Equation with Variable Separable – If we can reduce a differential equation to the form f(x) dx + g(s) ds = 0, it means the variables have been separated. Hence integrating this reduced form of the equation, you would get ∫ f(x) dx +  ∫ g(s) ds = K, where K is an arbitrary constant.

  • Homogeneous Differential Equation – Any first-order differential equation is called a homogeneous equation if it can be represented in the form: dy/dx = F(y/x). To solve such an equation, you would learn that you need to create a new variable v =y/x and follow the steps below:

  1. Y = vx

  2. dy/dx = d(vx)/dx = v dx/dx + x dv/dx –> as per product rule

  3. The above can be simplified as dy/dx = v + xdv/dx

  • Linear Differential Equations – A differential equation of the form dy/dx + Ky = C where K and C are constants or functions of x only, is a linear differential equation of first order. In a linear differential equation, the dependent variable and its derivatives occur only in the first power, and there is no product of these involved in the equation. You would learn the steps on how to solve the first-order linear differential equations.


Exercise 9.4

Solutions: 23 Questions (10 Short Questions, 12 Long Questions, 1 MCQs)


Exercise 9.5

Solutions: 17 Questions (15 Short Questions, 2 MCQs)


Exercise 9.6

Solutions: 19 Questions (2 Long Questions, 15 Short Questions, 2 MCQs)


Key Features of NCERT Solutions for Class 12 Maths Chapter 9

All the students want to score well in their CBSE exams and also get into any of the well-known engineering or medical colleges. By finding the right guidance and high-quality solutions, they can easily achieve it. Some of the key benefits of accessing our online solutions for Class 12th Maths Chapter 9 are:

  • The solutions are made by experts in education domain and meet the needs of CBSE curriculum

  • You are provided with a detailed stepwise explanation of every question

  • You will also get tips and tricks for remembering all the important formulas and concepts

  • In case you get stuck anywhere, our team of experts is always there to help you.

FAQs (Frequently Asked Questions)

1. What is a differential equation as stated in Class 12 books?

Differential equations assist you in differentiating any function w.r.t an independent variable. It describes a relationship between the functions and their derivatives. The functions refer to the physical quantities whereas their derivatives refer to the rate at which the function is changing and this relationship is differential equations. A D.E. takes the form dy/dx = g(x). Here, y signifies the function. And the function cited here is f(x).

2. How do you solve differential equations?

Differential equations can be of different orders. There are various approaches to solve these equations. For solving first-order linear differential equations, we need to perform the substitution. Then, after part factorization, one can perform the separation of variables. Once this step is successfully done, we can substitute the original value. Perform these steps until the solution is obtained for the original equation.

3. Is Class 12 Maths Chapter 9 tough?

Practice makes Maths easy for any student. Once the student is able to clearly grasp the basic concepts and learns the tricks to solve each question, only then will Maths be easy for that student. Hence, it is important to learn the basics well and practice thoroughly for all types of questions with sincerity. The Class 12 Maths syllabus is a mix of different questions in terms of difficulty. Visit the page NCERT Solutions Class 12 Maths Chapter 9 for the solutions

4. What is the order and degree of the differential equation? Give an example.

Order simply helps to find out the order of the highest term in any D.E., which is basically the term with the highest exponent value. For instance, in this equation, dy/dx + 4 = 2, the order is 1. The degree of any D.E. is associated with its order. The power raised to which the highest term is characterized by is the degree of the D.E. In the above-cited example of the 1st order D.E., the degree of the D.E. is 1. Modules relating to this topic or other topics covered in this chapter can be found on the Vedantu website or on the Vedantu app at free of cost.

5. What is the best Solution book for NCERT Class 12 Maths Chapter 9?

You may acquire NCERT Class 12 Math Solutions by going to the Vedantu website and searching for Class 12 Math solutions. Aside from that, you may access a variety of modules that will assist you in achieving high marks in math examinations. The exercise solutions are provided on the page NCERT Solutions Class 12 Maths Chapter 9. Click on it to download a PDF of the solutions.

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