Differential Equation Class 12 Notes CBSE Maths Chapter 9 [Free PDF Download]

Section–A (1 Mark Questions)

1. The degree of the differential equation $\left ( \frac{d^{2}y}{dx^{2}} \right )^{2}+\left ( \frac{dy}{dx} \right )^{2}=xsin\left ( \frac{dy}{dx} \right )$  is______.

Ans. The degree of above differential equation is not defined because on solving $sin\left ( \frac{dy}{dx} \right )$  we will get an infinite series in the increasing powers of $\frac{dy}{dx}$ . Therefore, its degree is not defined.

2. The integrating factor of differential equation $cos\;x\frac{dy}{dx}+ysinx=1$  is_______.

Ans. Given, $cos\;x\frac{dy}{dx}+ysinx=1$ 

$\Rightarrow \frac{dy}{dx}+y\;tan\;x=sec\;x$

Here, P=tan x and Q=sec x 

$I.F.\;=e^{Pdx}=e^{\int tan\;x\;dx}=e^{log\;sec\;x}=sec\;x $

3. Integrating factor of the differential of the form $\frac{dy}{dx}+P_{1}x=Q_{1}$

is given by _____.

Ans. Given differential equation

$\frac{dy}{dx}+P_{1}x=Q_{1}$ 

$I.F.=e^{\int P_{1}dy}$.

4. State True or False. Correct substitution for the solution of differential equation of the type $\frac{dy}{dx}=f(x,y)$ where $f(x,y)$ is a homogeneous function of zero degree is v=xy.

Ans. False, The correct substitution is y=vx.

5. Find the degree of the differential equation $\left [ 1+\left ( \frac{dy}{dx} \right )^{2} \right ]^{\frac{3}{2}}=\frac{d^{2}y}{dx^{2}}$ .

Ans. Given is, $\left [ 1+\left ( \frac{dy}{dx} \right )^{2} \right ]^{\frac{3}{2}}=\frac{d^{2}y}{dx^{2}}$ 

On squaring both sides, we get

$\left [ 1+\left ( \frac{dy}{dx} \right )^{2} \right ]^{3}=\frac{d^{2}y}{dx^{2}}$

So, the degree of differential equation is 2.

Section–B (2 Marks Questions)

6. Find the solution of the differential equation $x \frac{d y}{d x}+2 y=x^2$.

Ans. Given differential equation is $x \frac{d y}{d x}+2 y=x^2$

$$\Rightarrow \frac{d y}{d x}+\frac{2 y}{x}=x$$

This equation of the form $\frac{d y}{d x}+P y=Q$

$$\therefore \text { I.F. }=e^{\int_{\frac{1}{x}}^2 d x}=e^{2 \log x}=x^2$$

The general solution is

$$\begin{aligned}& y x^2=\int x \cdot x^2 d x+C \\& \Rightarrow y x^2=\frac{x^4}{4}+C \\& \Rightarrow y=\frac{x^2}{4}+C x^{-2} .\end{aligned}$$

7. Find the order and degree of the differential equation $\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^{\frac{1}{4}}+x^{\frac{1}{5}}=0$

Ans. Given that, $\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^{\frac{1}{4}}+x^{\frac{1}{5}}=0$

$$\Rightarrow\left(\frac{d y}{d x}\right)^{\frac{1}{4}}=-\left(x^{\frac{1}{5}}+\frac{d^2y}{d x^2}\right)$$

On squaring both sides, we get

$$\left(\frac{d y}{d x}\right)^{\frac{1}{2}}=\left(x^{\frac{1}{5}}+\frac{d^2 y}{d x^2}\right)^2$$

Again, squaring both sides, we get

$$\frac{d y}{d x}=\left(x^{\frac{1}{5}}+\frac{d^2 y}{d x^2}\right)^4$$

Thus, order $=2$, degree $=4$.

8. Form the differential equation for $y=A \cos \alpha x+B \sin \alpha x$, where $\mathrm{A}$ and $\mathrm{B}$ are arbitrary constants.

Ans. Given, $y=A \cos \alpha x+B \sin \alpha x$

On differentiating both sides w.r.t., $x$ we get

$$\frac{d y}{d x}=-\alpha A \sin \alpha x+\alpha B \cos \alpha x$$

Again, differentiating both sides w.r.t. $x$, we get

$$\begin{aligned}& \frac{d^2 y}{d x^2}=-A \alpha^2 \cos \alpha x-\alpha^2 B \sin \alpha x \\& \Rightarrow \frac{d^2 y}{d x^2}=-\alpha^2(A \cos \alpha x+B \sin \alpha x) \\& \Rightarrow \frac{d^2 y}{d x^2}=-\alpha^2 y \\& \Rightarrow \frac{d^2 y}{d x^2}+\alpha^2 y=0 .\end{aligned}$$

9. Show that the solution of differential equation $x d y-y d x=0$ represents a straight line passing through origin.

Ans. Given is, $x d y-y d x=0$

$$\begin{aligned}& \Rightarrow x d y=y d x \\& \Rightarrow \frac{d y}{y}=\frac{d x}{x}\end{aligned}$$

On integrating both sides, we get

$$\begin{aligned}& \Rightarrow \log y=\log x+\log C \\& \Rightarrow \log y=\log C x \\& \Rightarrow y=C x\end{aligned}$$ which represents is a straight line passing through origin.

10. What is/are the number of solutions of $\frac{d y}{d x}=\frac{y+1}{x-1}$, when $y(1)=2$.

Ans. Given is, $\frac{d y}{d x}=\frac{y+1}{x-1}$

$$\Rightarrow \frac{d y}{y+1}=\frac{d x}{x-1}$$

On integrating both sides, we get

$$\begin{aligned}& \log (y+1)=\log (x-1)-\log C \\& \Rightarrow C(y+1)=(x-1) \\& \Rightarrow C=\frac{x-1}{y+1}\end{aligned}$$

When $x=1$ and $y=2$, then $C=0$

So, the required solution is $x-1=0$

Hence, only one solution exists.

11. Show that the General solution of the differential equation of the type $\frac{d x}{d y}+P_1 x=Q_1$ is given by $$x e^{\int A d y}=\int Q_1 e^{\int A d y} d y+C .$$

Ans. Given differential equation is $\frac{d x}{d y}+P_1 x=Q_1$

To get the general solution of this equation, we multiply both sides by integrating factor,

$$\begin{aligned}& \text { I.F. }=e^{\int P_1 d y} \\& \Rightarrow e^{\int P_1 d y}\left(\frac{d x}{d y}+P_1 x\right)=Q_1 e^{\int P_1 d y} \\& \Rightarrow \frac{d x}{d y} e^{\int P_{d y}}+P_1 x e^{\int P_1 d y}=Q_1 e^{\int P_d d y} \\& \Rightarrow \frac{d}{d y}\left(x e^{\int P_1 d y}\right)=Q_1 e^{\int P_1 d y} \\& \Rightarrow \int \frac{d}{d y}\left(x e^{\int P_1 d y}\right) d y=\int Q_1 e^{\int P_1 d y} d y \\& \Rightarrow x e^{\int P_1 d y}=\int Q_1 e^{\int P_1 d y} d y+C\end{aligned}$$

This is the required solution of the given differential equation.

12. Find the differential equation of all non-vertical lines in a plane.

Ans. The family of all non-vertical line is represented as:

$y=m x+c$, where $m \neq \tan \frac{\pi}{2}$

On differentiating above equation w.r.t. $x$, we get

$$\frac{d y}{d x}=m$$

Again, differentiating equation (i) w.r.t. $x$, we get

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

13. Solve: $\frac{d y}{d x}-y=1, y(0)=1$.

Ans. Given is, $\frac{d y}{d x}-y=1$

$$\begin{aligned}& \Rightarrow \frac{d y}{d x}=1+y \\

& \Rightarrow \frac{d y}{1+y}=d x\end{aligned}$$

On integrating both sides, we get

$$\log (1+y)=x+C$$

When $x=0$ and $y=1$, then

$$\begin{aligned}& \log 0+C \\& \Rightarrow C=\log 2\end{aligned}$$

The required solution is

$$\begin{aligned}& \log (1+y)=x+\log 2 \\& \Rightarrow \log \left(\frac{1+y}{2}\right)=x \\& \Rightarrow \frac{1+y}{2}=e^x \\& \Rightarrow 1+y=2 e^x \\& \Rightarrow y=2 e^x-1 .\end{aligned}$$

PDF Summary - Class 12 Maths Differential Equations Notes (Chapter 9)

Definition:

An equation involving the dependent variable and independent variable and also the derivatives of the dependable variable is known as differential equation. This can be mathematically written as $x\dfrac{dy}{dx}+y=0$.

The derivative $\dfrac{dy}{dx}$ can also be written as $f'(x)\text{ }or\text{ }y'(x)$. Similarly,

$ \dfrac{{{d}^{2}}y}{d{{x}^{2}}}\Rightarrow f''(x)\text{ or }y''(x) $ 

$ \dfrac{{{d}^{3}}y}{d{{x}^{3}}}\Rightarrow f'''(x)\text{ or }y'''(x) $ 

Some examples can be \[\dfrac{dy}{dx}=\dfrac{x}{{{y}^{\dfrac{1}{3}}}\left( 1+{{x}^{\dfrac{1}{3}}} \right)}\] , $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-{{p}^{2}}y$  or \[{{x}^{2}}{{\left( \dfrac{dy}{dx} \right)}^{2}}={{y}^{2}}+1\] . 

Differential equations which involve only one independent variable are called ordinary differential equations.

Order of Differential Equations:

The order of a differential equation is the order of the highest derivative involved in the differential equation. This can be understood clearly by looking at a few examples. 

1. First order differential equation - ${{\left( \dfrac{dy}{dx} \right)}^{4}}+{{\left( \dfrac{dy}{dx} \right)}^{2}}-5x=0$. The maximum derivative of $y$ with respect to $x$ is $\dfrac{dy}{dx}$.

2. Second order differential equation - $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+7y=0$. The maximum derivative of $y$ with respect to $x$ is $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$.

3. Third order differential equation - \[{{\left( \dfrac{{{d}^{3}}y}{d{{x}^{3}}} \right)}^{2}}-3\left( \dfrac{dy}{dx} \right)+2=0\] . The maximum derivative of $y$ with respect to $x$ is \[\dfrac{{{d}^{3}}y}{d{{x}^{3}}}\].
   
   

Degree of Differential Equations:

The degree of a differential equation is the degree of the highest differential coefficient when the equation has been made rational and integral as far as the differential coefficients are concerned. This can be understood clearly by looking at a few examples.

1. First degree differential equation - \[\dfrac{dy}{dx}=\dfrac{5x}{{{y}^{\dfrac{1}{3}}}\left( 1-{{x}^{\dfrac{1}{3}}} \right)}\]. The power of the highest order derivative \[\dfrac{dy}{dx}\] is $1$.

2. Second degree differential equation - \[{{\left( \dfrac{{{d}^{3}}y}{d{{x}^{3}}} \right)}^{2}}+6\left( \dfrac{dy}{dx} \right)=-2\] . The power of the highest order derivative \[\dfrac{{{d}^{3}}y}{d{{x}^{3}}}\] is $2$.

3. Third degree differential equation - ${{\left[ 1+{{\left( \dfrac{dy}{dx} \right)}^{2}} \right]}^{{}^{1}/{}_{3}}}=3\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ . First, making it rational, $\left[ 1+{{\left( \dfrac{dy}{dx} \right)}^{2}} \right]=27{{\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)}^{3}}$. The power of highest order derivative $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ is $3$.
   
   

Illustration 1: Find the order and degree of the following differential equations.

i. $\sqrt{\dfrac{{{d}^{2}}y}{d{{x}^{2}}}}=\sqrt[3]{\dfrac{dy}{dx}+3}$ 

Ans: Rewriting it as ${{\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)}^{3}}={{\left( \dfrac{dy}{dx}+3 \right)}^{2}}$

So, the order $=2$ and the degree $=3$.

ii. \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}={{\left\{ 1+{{\left( \dfrac{dy}{dx} \right)}^{4}} \right\}}^{{}^{5}/{}_{3}}}\] 

Ans: Rewriting it as \[{{\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)}^{3}}={{\left[ 1+{{\left( \dfrac{dy}{dx} \right)}^{4}} \right]}^{5}}\].

So, the order $=2$ and the degree $=3$.

iii. $y=px+\sqrt{{{a}^{2}}{{p}^{2}}+{{b}^{2}}}\text{ where }p=\dfrac{dy}{dx}$ 

Ans: Substituting p and then rewriting it as ${{\left( y-x\dfrac{dy}{dx} \right)}^{2}}={{a}^{2}}{{\left( \dfrac{dy}{dx} \right)}^{2}}+{{b}^{2}}$.  

So, the order $=1$ and the degree $=2$.

Formation of Ordinary Differential Equation:

There may be some arbitrary constants in an equation containing variables and constants. An ordinary differential equation is formed as a result of elimination of these arbitrary constants. 

Consider an equation containing $n$ arbitrary constants. Differentiating this equation n times we get $n$ additional equations containing n arbitrary constants and derivatives. Eliminating $n$ arbitrary constants from the above $\left( n+1 \right)$ equations, differential equations involving nth derivative are obtained. After this is complete, the resulting equation will be of the form $\phi \left( x,y,\dfrac{dy}{dx},\dfrac{{{d}^{2}}y}{d{{x}^{2}}},.....,\dfrac{{{d}^{n}}y}{d{{x}^{n}}} \right)=0$

Illustration 2: Find the differential equation of the family of all circles which pass through the origin and whose centre lie on $y-$axis.

Ans: Let the equation of the circle be 

${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ 

If it passes through $(0,0)$, then $c=0$

$\therefore $ The equation of circle is ${{x}^{2}}+{{y}^{2}}+2gx+2fy=0$

Since the centre of the circle lies on $y-$axis then $g=0$.

$\therefore $ The equation of the circle is 

${{x}^{2}}+{{y}^{2}}+2fy=0......(i)$

This represents a family of circles. 

Differentiating gives, 

$2x+2y\dfrac{dy}{dx}+2f\dfrac{dy}{dx}=0...........(ii)$ 

From $(i)\text{ and }(ii)$, 

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

Hence, this is the required differential equation.

Solution of a Differential Equation:

The solution of the differential equation is a relation is a relation between the independent and dependent variable free from derivatives satisfying the given differential equation.

So, the solution of an equation given by $\dfrac{dy}{dx}=m$ can be obtained by integrating both the sides to remove the derivatives and obtain $y=mx+c$, where $c$ is an arbitrary constant.

a) General solution or primitive

The general solution of a differential equation is the relation between the variables (not involving the derivatives) which contain the same number of the arbitrary constants as the order of the differential equation. 

Thus the general solution of the differential equation 

$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=4y$ is $y=A\sin 2x+B\cos 2x$, where $A\text{ and }B$ are the constants.

b) Particular solution or integral 

A solution which is obtained by giving particular values to the arbitrary constants in the general solution is called a particular solution.

Illustration 3: Show that \[v=\dfrac{A}{r}+B\] is the general solution of the second order differential equation $\dfrac{{{d}^{2}}v}{d{{r}^{2}}}+\dfrac{2}{r}\dfrac{dv}{dr}=0$, where $A\text{ and }B$ are arbitrary constant. 

Ans: Given \[v=\dfrac{A}{r}+B\] 

Differentiating, $\dfrac{dv}{dr}=-\dfrac{A}{{{r}^{2}}}$.

Differentiating again, $\dfrac{{{d}^{2}}v}{d{{r}^{2}}}=\dfrac{2A}{{{r}^{3}}}.......(i)$ 

Rearranging the second term and substituting first derivative, 

$\dfrac{{{d}^{2}}v}{d{{r}^{2}}}=-\dfrac{2}{r}\dfrac{dv}{dr}$

$\dfrac{2A}{{{r}^{3}}}=-\dfrac{2}{r}\left( -\dfrac{A}{{{r}^{2}}} \right)$

$\dfrac{2A}{{{r}^{3}}}-\dfrac{2A}{{{r}^{3}}}=0$

Putting $A=4,B=5$ in \[v=\dfrac{A}{r}+B\] we get a particular solution of the differential equation 

$\dfrac{{{d}^{2}}v}{d{{r}^{2}}}+\dfrac{2}{r}\dfrac{dv}{dr}=0$ is $v=\dfrac{4}{r}+5$.

Illustration 4: Show that $y=a{{e}^{x}}+b{{e}^{2x}}+c{{e}^{-3x}}$ is a solution of the equation $\dfrac{{{d}^{3}}y}{d{{x}^{3}}}-7\dfrac{dy}{dx}+6y=0$.

Ans: Given that

\[y=a{{e}^{x}}+\text{ }b{{e}^{2x}}+c{{e}^{-3x}}\text{ }...\left( i \right)\] 

Differentiating, 

\[y'=a{{e}^{x}}+2b{{e}^{2x}}-3c{{e}^{-3x}}\text{ }...\left( ii \right)\] 

Differentiating \[\left( ii \right)\],

\[y''=a{{e}^{x}}+\text{ }4b{{e}^{2x}}+9c{{e}^{-3x}}\] 

Differentiating again,

$y'''=a{{e}^{x}}+8b{{e}^{2x}}-27c{{e}^{-3x}}$ 

The given differential equation is \[\dfrac{{{d}^{3}}y}{d{{x}^{3}}}-7\dfrac{dy}{dx}+6y=0\].

Considering the LHS and substituting the terms,

$\left[ a{{e}^{x}}+8b{{e}^{2x}}-27c{{e}^{-3x}} \right]-7\left[ a{{e}^{x}}+2b{{e}^{2x}}-3c{{e}^{-3x}} \right]+6\left[ a{{e}^{x}}+b{{e}^{2x}}+c{{e}^{-3x}} \right]$  

\[\Rightarrow a{{e}^{x}}+8b{{e}^{2x}}-27c{{e}^{-3x}}-7a{{e}^{x}}-14b{{e}^{2x}}+21c{{e}^{-3x}}+6a{{e}^{x}}+6b{{e}^{2x}}+6c{{e}^{-3x}}\]

\[\Rightarrow 0\] 

This is equal to RHS.

Since it satisfies the equation, $y=a{{e}^{x}}+b{{e}^{2x}}+c{{e}^{-3x}}$ is the solution for \[\dfrac{{{d}^{3}}y}{d{{x}^{3}}}-7\dfrac{dy}{dx}+6y=0\].

Method of solving an equation of the first order and first degree:

A differential equation of the first order and first degree can be written in the form $\dfrac{dy}{dx}=f\left( x,y \right)$ or, $Mdx+Ndy=0$, where $M\text{ and }N$  are functions of $x\text{ and }y$.

1. Method – 1

i. Variable Separation: 

The general form of such an equation is 

$f(x)dx+f(y)dy=0$ ...(i) 

Integrating it gives the solution as 

$\int{f(x)dx}+\int{f(y)dy}=c$

ii. Solution of differential equation of the type $\dfrac{dy}{dx}=f\left( ax+by+c \right)$:

Consider the differential equation $\dfrac{dy}{dx}=f\left( ax+by+c \right)$ ...(i) where $f\left( ax+by+c \right)$ is some function of $ax+by+c$. 

Let $z=ax+by+c$ 

$\therefore \dfrac{dz}{dx}=a+b\dfrac{dy}{dx}$ 

or, $\dfrac{dy}{dx}=\dfrac{\dfrac{dz}{dx}-a}{b}$ 

From (i), $\dfrac{\dfrac{dz}{dx}-a}{b}=f(z)$ 

or, $\dfrac{dz}{dx}=bf(z)+a$ 

or, $\dfrac{dz}{bf(z)+a}=dx$ ...(ii) 

In the differential equation (ii), the variables $x\text{ and }z$ are separated.

Integrating, we get 

 \[\int{\dfrac{dx}{bf(z)+a}=\int{dx}+c}\] 

or, \[\int{\dfrac{dx}{bf(z)+a}=x+c}\] , where $z=ax+by+c$

This represents the general solution of the differential equation (i)

Illustration 5: Solve ${{\left( x-y \right)}^{2}}\dfrac{dy}{dx}={{a}^{2}}$.

Ans: Let \[xy=v\] and differentiate it to get

$\Rightarrow \dfrac{dy}{dx}=1-\dfrac{dv}{dx}$

Substituting these in ${{\left( x-y \right)}^{2}}\dfrac{dy}{dx}={{a}^{2}}$ and rearranging terms in variable separable form,

$\Rightarrow dx=\dfrac{{{v}^{2}}}{{{v}^{2}}-{{a}^{2}}}dv$

Integrating

\[\int{dx}=\int{\dfrac{{{v}^{2}}}{{{v}^{2}}-{{a}^{2}}}dv}\] 

$x+c=\int{\dfrac{{{v}^{2}}-{{a}^{2}}+{{a}^{2}}}{{{v}^{2}}-{{a}^{2}}}}dv$ 

$x+c=\int{dv}+\int{\dfrac{{{a}^{2}}}{{{v}^{2}}-{{a}^{2}}}}dv$ 

\[x+c=v+\dfrac{{{a}^{2}}}{2a}\log \left| \dfrac{v-a}{v+a} \right|\]

$x+c=x-y+\dfrac{a}{2}\log \left| \dfrac{x-y-a}{x-y+a} \right|$

$c=-y+\dfrac{a}{2}\log \left| \dfrac{x-y-a}{x-y+a} \right|$

$y-\dfrac{a}{2}\log \left| \dfrac{x-y-a}{x-y+a} \right|=C$ 

Illustration 6: Solve, $\dfrac{dy}{dx}=\sin \left( x+y \right)+\cos \left( x+y \right)$ 

Ans: Let $z=x+y$ and differentiate it to get the variable separable form as

$\therefore \dfrac{dz}{dx}=1+\dfrac{dy}{dx}\Rightarrow \dfrac{dy}{dx}=\dfrac{dz}{dx}-1$ 

$\dfrac{dz}{dx}-1=\sin z+\cos z$  

$\dfrac{dz}{dx}=\sin z+\cos z+1$

Using identities $\sin x=2\sin \dfrac{x}{2}\text{cos}\dfrac{x}{2}\text{ }and\text{ }\cos x=2{{\cos }^{2}}\dfrac{x}{2}-1$,

$\Rightarrow \dfrac{dz}{dx}=2\sin \dfrac{z}{2}\text{cos}\dfrac{z}{2}+2{{\cos }^{2}}\dfrac{z}{2}$

Taking out $2{{\cos }^{2}}\dfrac{z}{2}$,

$\Rightarrow \dfrac{dz}{dx}=2{{\cos }^{2}}\dfrac{z}{2}\left( \tan \dfrac{z}{2}+1 \right)$

$\Rightarrow \dfrac{dz}{2{{\cos }^{2}}\dfrac{z}{2}\left( \tan \dfrac{z}{2}+1 \right)}=dx$ 

Integrating, 

$\Rightarrow \int{\dfrac{dz}{2{{\cos }^{2}}\dfrac{z}{2}\left( \tan \dfrac{z}{2}+1 \right)}}=\int{dx}$

Take $u=\tan \dfrac{z}{2}+1$.

So, $du=\dfrac{1}{2}{{\sec }^{2}}\dfrac{z}{2}dz$ 

Using identities $\sec x=\dfrac{1}{\cos x}$,

$du=\dfrac{1}{2{{\cos }^{2}}\dfrac{z}{2}}dz$

Substituting in the integral,

$\Rightarrow \int{\dfrac{du}{u}}=\int{dx}$

$\Rightarrow \log u=x+c$ 

Resubstituting back $u=\tan \dfrac{z}{2}+1$, 

$\Rightarrow \log \left( \tan \dfrac{z}{2}+1 \right)=x+c$

$\therefore \log \left( \tan \dfrac{x+y}{2}+1 \right)=x+c$  

This is the required general solution. 

i. Solution of differential equation of the type \[\dfrac{dy}{dx}=\dfrac{{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}}{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}},\text{ }\mathbf{where}\text{ }\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}\] 

Here \[\dfrac{dy}{dx}=\dfrac{{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}}{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}},\text{ }\mathbf{where}\text{ }\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}\] ...(i) 

Let \[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\lambda \] (say) 

$\therefore {{a}_{1}}=\lambda {{a}_{2}},{{b}_{1}}=\lambda {{b}_{2}}$   

From (i), \[\dfrac{dy}{dx}=\dfrac{\lambda {{a}_{2}}x+\lambda {{b}_{2}}y+{{c}_{1}}}{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}}\] 

$=\dfrac{dy}{dx}=\dfrac{\lambda \left( {{a}_{2}}x+{{b}_{2}}y \right)+{{c}_{1}}}{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}}$ ...(ii) 

Let $z={{a}_{2}}x+{{b}_{2}}y$ 

$\therefore \dfrac{dz}{dx}={{a}_{2}}+{{b}_{2}}\dfrac{dy}{dx}\Rightarrow \dfrac{dy}{dx}=\dfrac{\dfrac{dz}{dx}-{{a}_{2}}}{{{b}_{2}}}$ ...(iii) 

From (ii) and (iii), we get 

$\dfrac{\dfrac{dz}{dx}-{{a}_{2}}}{{{b}_{2}}}=\dfrac{\lambda z+{{c}_{1}}}{z+{{c}_{2}}}$

or, $\dfrac{dz}{dx}=\dfrac{{{b}_{2}}\lambda z+{{c}_{1}}}{z+{{c}_{2}}}+{{a}_{2}}=\dfrac{\lambda {{b}_{2}}z+{{b}_{2}}{{c}_{1}}+{{a}_{2}}z+{{a}_{2}}{{c}_{2}}}{z+{{c}_{2}}}$ 

or $dx=\dfrac{z+{{c}_{2}}}{\lambda {{b}_{2}}+{{a}_{2}}z+{{b}_{2}}{{c}_{1}}+{{a}_{2}}{{c}_{2}}}dz$, where $x\text{ and }z$ are separated.

Integrating, we get 

$x+c=\int{\dfrac{z+{{c}_{2}}}{\lambda {{b}_{2}}+{{a}_{2}}z+{{b}_{2}}{{c}_{1}}+{{a}_{2}}{{c}_{2}}}dz\text{ where }z={{a}_{2}}x+{{b}_{2}}y}$

2. Method – 2

i. Homogeneous differential equation: 

A function $f\left( x,y \right)$ is called homogeneous function of degree n if 

$f\left( \lambda x,\lambda y \right)={{\lambda }^{n}}f\left( x,y \right)$ 

For example: 

a) $f\left( x,y \right)={{x}^{2}}{{y}^{2}}-x{{y}^{3}}$ is a homogeneous function of degree four, since 

$f\left( \lambda x,\lambda y \right)=\left( {{\lambda }^{2}}{{x}^{2}} \right)\left( {{\lambda }^{2}}{{y}^{2}} \right)-\left( \lambda x \right)\left( {{\lambda }^{3}}{{y}^{3}} \right)$  

$={{\lambda }^{4}}\left( {{x}^{2}}{{y}^{2}}-x{{y}^{3}} \right)$ 

$=\lambda f\left( x,y \right)$ 

b) $f\left( x,y \right)={{x}^{2}}{{e}^{\dfrac{x}{y}}}+\dfrac{{{x}^{3}}}{y}+{{y}^{2}}\log \left( \dfrac{y}{x} \right)$ is a homogeneous function of degree two, since

$f\left( \lambda x,\lambda y \right)=\left( {{\lambda }^{2}}{{x}^{2}} \right){{e}^{\dfrac{\lambda x}{\lambda y}}}+\dfrac{{{\lambda }^{3}}{{x}^{3}}}{\lambda y}+\left( {{\lambda }^{2}}{{y}^{2}} \right)\log \left( \dfrac{\lambda y}{\lambda x} \right)$ 

\[={{\lambda }^{2}}\left[ {{x}^{2}}{{e}^{\dfrac{x}{y}}}+\dfrac{{{x}^{3}}}{y}+{{y}^{2}}\log \left( \dfrac{y}{x} \right) \right]\] 

$={{\lambda }^{2}}f\left( x,y \right)$  

A differential equation of the form $\dfrac{dy}{dx}=f(x,y)$ , where $f\left( x,y \right)$ is a homogeneous polynomial of degree zero is called a homogeneous differential equation. Such equations are solved by substituting $v=\dfrac{y}{x}\text{ or }\dfrac{x}{y}$ and then separating the variables.

Illustration 7: Solve $\dfrac{dy}{dx}=\dfrac{y\left( 2y-x \right)}{x\left( 2y+x \right)}$ 

Ans: Each of the given functions, i.e. $y\left( 2y-x \right)$ and $x\left( 2y+x \right)$ is a homogeneous function of degree $2$. Hence, the given equation is a homogeneous differential equation.

Putting $y=vx$ and differentiating w.r.t $x$,

$\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$ 

Substituting in given equation,

$v+x\dfrac{dv}{dx}=\dfrac{vx\left( 2vx-x \right)}{x\left( 2vx+x \right)}=\dfrac{v\left( 2v-1 \right)}{2v+1}$

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

After simplifying the RHS,

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

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

Integrating,

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

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

\[\Rightarrow v+\dfrac{1}{2}\log v=-\log x+\log c\] 

Resubstituting $v=\dfrac{y}{x}$,

\[\Rightarrow \dfrac{y}{x}+\dfrac{1}{2}\log \dfrac{y}{x}=-\log x+\log c\]

\[\therefore y+\dfrac{x}{2}\log \dfrac{y}{x}=-x\log x+C\]

ii. Differential equation reducible to homogeneous forms:

Equation of the form $\dfrac{dy}{dx}=\dfrac{ax+by+c}{{a}'x+{b}'y+{c}'}$, where $\dfrac{a}{{{a}'}}\ne \dfrac{b}{{{b}'}}$ can be reduced to homogeneous form by changing the variables $x,y\text{ to }{x}',{y}'$ by equations $x={x}'+h\text{ and }y={y}'+k$ where $h\text{ and }k$ are constants to be chosen so as to make the given equation homogeneous.

$dx=d{x}'\text{ and }dy=d{y}'$ 

The given equation becomes 

\[\dfrac{d{y}'}{d{x}'}=\dfrac{a\left( {x}'+h \right)+b\left( {y}'+k \right)+c}{{a}'\left( {x}'+h \right)+{b}'\left( {y}'+k \right)+{c}'}\] 

$=\dfrac{a{x}'+b{y}'+\left( ah+bk+c \right)}{{a}'{x}'+{b}'{y}'+\left( {a}'h+{b}'k+{c}' \right)}$ 

Now, choose $h\text{ and }k$ so that 

$ah+bk+c=0$ 

and ${a}'h+{b}'k+{c}'=0$

From these equations, the values of $h\text{ and }k$ in terms of the coefficients are obtained. 

Then the given equation reduces to 

$\dfrac{dy}{dx}=\dfrac{a{x}'+b{y}'}{{a}'{x}'+{b}'{y}'}$   

Which is the homogeneous form.

3. Method – 3

i. Linear differential equation: 

A differential equation is said to be linear if the dependent variable y and its derivative occur in the first degree. 

An equation of the form \[\dfrac{dy}{dx}+Py=Q\] ...(i) 

where $P\text{ and }Q$ are functions of x only or constant is called a linear equation of the first order.

Similarly $\dfrac{dx}{dy}+Px=Q$ is a linear differential equation where $P\text{ and }Q$ are functions of $y$ only. 

To get the general solution of the above equations, determination of a function $R\text{ of }x$ called Integrating function (I.F) is required. So, multiply both sides of the given equation by $R$

where, $R={{e}^{\int{Pdx}}}=\text{I}\text{.F}\text{.}$ ...(iii) 

From (i) and (iii), 

${{e}^{\int{Pdx}}}\dfrac{dy}{dx}+Py{{e}^{\int{Pdx}}}=Q{{e}^{\int{Pdx}}}$ 

\[\dfrac{d}{dx}\left( y{{e}^{\int{Pdx}}} \right)=Q.{{e}^{\int{Pdx}}}\]  

Integrating, 

$y{{e}^{\int{Pdx}}}=\int{Q{{e}^{\int{Pdx}}}dx}+c$ is the required solution. 

This can also be written and memorized as 

$y\left( \text{I}\text{.F}\text{.} \right)=\int{Q\left( \text{I}\text{.F}\text{.} \right)dx}+c$

Illustration 8: Solve $2x\dfrac{dy}{dx}=y+6{{x}^{\dfrac{5}{2}}}-2\sqrt{x}$
Ans: The given equation can be written as 

$\dfrac{dy}{dx}+\left( \dfrac{-1}{2x} \right)y=3{{x}^{\dfrac{3}{2}}}-\dfrac{1}{\sqrt{x}}$ 

This is the form of \[\dfrac{dy}{dx}+Py=Q\] 

Hence $\text{I}\text{.F}\text{.}={{e}^{\int{\dfrac{-1}{2x}dx}}}={{e}^{-\dfrac{1}{2}\log x}}=\dfrac{1}{\sqrt{x}}$ 

Now using $y\left( \text{I}\text{.F}\text{.} \right)=\int{Q\left( \text{I}\text{.F}\text{.} \right)dx}+c$,

$\Rightarrow \dfrac{y}{\sqrt{x}}=\int{\left( 3{{x}^{\dfrac{3}{2}}}-\dfrac{1}{\sqrt{x}} \right)\dfrac{1}{\sqrt{x}}}dx+c$ 

$\Rightarrow \dfrac{y}{\sqrt{x}}=\int{\left( 3x-\dfrac{1}{x} \right)}dx+c$

Integrating,

$\Rightarrow \dfrac{y}{\sqrt{x}}=3\dfrac{{{x}^{2}}}{2}-\log x+c$

$\Rightarrow y=\dfrac{3}{2}{{x}^{2}}\sqrt{x}-\sqrt{x}\log x+c\sqrt{x}$

Therefore, $y=\dfrac{3}{2}{{x}^{\dfrac{5}{2}}}-\sqrt{x}\log x+c\sqrt{x}$.

i. Differential equation reducible to the linear form: 

Sometimes equations which are not linear can be reduced to the linear form by suitable transformation. 

Here, ${f}'\left( y \right)\dfrac{dy}{dx}+f(y)P(x)=Q(x)$ ...(i) 

Let, $f\left( y \right)=u\Rightarrow {f}'\left( y \right)dy=du$ 

Then (i) reduces to 

$\dfrac{du}{dx}+uP(x)=Q(x)$ Which is of the linear differential equation form.

Illustration 9: Solve \[{{\sec }^{2}}\theta d\theta +\tan \theta \left( 1-r\tan \theta  \right)dr=0\]

Ans: The given equation can be written as 

$\dfrac{d\theta }{dr}+\dfrac{\tan \theta }{{{\sec }^{2}}\theta }=\dfrac{r{{\tan }^{2}}\theta }{{{\sec }^{2}}\theta }$ 

$\left( \dfrac{{{\sec }^{2}}\theta }{{{\tan }^{2}}\theta } \right)\dfrac{d\theta }{dr}+\dfrac{1}{\tan \theta }=r$ 

${{\csc }^{2}}\theta \dfrac{d\theta }{dr}+\cot \theta =r$ ...(i) 

Let $\cot \theta =u$ 

$\Rightarrow -{{\csc }^{2}}\theta d\theta =du$ 

Then (i) reduces to 

$-\dfrac{du}{dr}+u=r\text{ or }\dfrac{du}{dr}-u=-r$ ...(ii) 

Which is a linear differential equation. 

So, $\text{I}\text{.F}\text{.}={{e}^{\int{-1dr}}}={{e}^{-r}}$

Now using $y\left( \text{I}\text{.F}\text{.} \right)=\int{Q\left( \text{I}\text{.F}\text{.} \right)dx}+c$,

$\Rightarrow u{{e}^{-r}}=\int{-r}{{e}^{-r}}dr+c$ 

$\Rightarrow u{{e}^{-r}}=-\int{r{{e}^{-r}}dr}+c$

Using integration by parts with first function as $r$ and second function as ${{e}^{-r}}$,

$\Rightarrow u{{e}^{-r}}=-\left[ r\int{{{e}^{-r}}}dr-\int{\dfrac{d}{dr}(r).\int{{{e}^{-r}}}dr} \right]$

$\Rightarrow u{{e}^{-r}}=-\left[ -r{{e}^{-r}}-\int{{{e}^{-r}}}dr \right]$

\[\Rightarrow u{{e}^{-r}}=-\left[ -r{{e}^{-r}}+{{e}^{-r}} \right]+c\]

\[\Rightarrow u{{e}^{-r}}=r{{e}^{-r}}-{{e}^{-r}}+c\]

\[\Rightarrow u=r-1+C\]

Resubstituting,

$\therefore \cot \theta =r-1+C$
 

i. Extended form of linear equations: 

Bernoulli’s equation: 

An equation of the form $\dfrac{dy}{dx}+Py=Q{{y}^{n}}$, where $P\text{ and }Q$  are function of $x$ alone or constants and $n$ is constant, other than $0\text{ and }1$, is called a Bernoulli’s equation. 

Here $\dfrac{dy}{dx}+Py=Q{{y}^{n}}$

Dividing by ${{y}^{n}}$,

$\dfrac{1}{{{y}^{n}}}\dfrac{dy}{dx}+P.\dfrac{1}{{{y}^{n-1}}}=Q$

Putting $\dfrac{1}{{{y}^{n-1}}}=v$and differentiating w.r.t $x$, 

$-\dfrac{(n-1)}{{{y}^{n}}}\dfrac{dy}{dx}=\dfrac{dv}{dx}$ 

$\dfrac{1}{{{y}^{n}}}\dfrac{dy}{dx}=\dfrac{-1}{n-1}\dfrac{dv}{dx}$ 

$\dfrac{dv}{dx}=\left( 1-n \right){{y}^{-n}}\dfrac{dy}{dx}$ 

The equation becomes 

$\dfrac{dv}{dx}+\left( 1-n \right)Pv=Q\left( 1-n \right)$ 

Which is a linear equation with $v$ as an independent variable.

Illustration 10: Solve ${{\cos }^{2}}x\dfrac{dy}{dx}-y\tan 2x={{\cos }^{4}}x$, where $\left| x \right|=\dfrac{\pi }{4}$ and $y\left( \dfrac{\pi }{4} \right)=\dfrac{3\sqrt{3}}{8}$.

Ans: The given equation can be written as 

$\dfrac{dy}{dx}-y\tan 2x{{\sec }^{2}}x={{\cos }^{2}}x$

This is the form of $\dfrac{dy}{dx}+Py=Q$ 

Here $P=-\tan 2x{{\sec }^{2}}x,Q={{\cos }^{2}}x$ 

$\int{Pdx}=-\int{\tan 2x{{\sec }^{2}}xdx}$ 

$=-\int{\dfrac{2\tan x}{1-{{\tan }^{2}}x}{{\sec }^{2}}xdx}$

$=\int{\dfrac{dt}{t}}$ 

Putting $1-{{\tan }^{2}}x=t$ 

$\therefore -2\tan x{{\sec }^{2}}xdx=dt$ 

$=\log t=\log \left( 1-{{\tan }^{2}}x \right)$ 

$\therefore \text{I}\text{.F}\text{.}={{e}^{\int{P.dx}}}={{e}^{\log \left( 1-{{\tan }^{2}}x \right)}}=1-{{\tan }^{2}}x$ 

Now using $y\left( \text{I}\text{.F}\text{.} \right)=\int{Q\left( \text{I}\text{.F}\text{.} \right)dx}+c$,

$\Rightarrow y\left( 1-{{\tan }^{2}}x \right)=\int{{{\cos }^{2}}x\left( 1-{{\tan }^{2}}x \right)dx}+c$

$\Rightarrow y\left( 1-{{\tan }^{2}}x \right)=\int{\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right)dx}+c$

Using identity $\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$,

$\Rightarrow y\left( 1-{{\tan }^{2}}x \right)=\int{\cos 2xdx}+c$

$\Rightarrow y\left( 1-{{\tan }^{2}}x \right)=\dfrac{\sin 2x}{2}+C$ …..(i)

Given that \[x=\dfrac{\pi }{6},y=\dfrac{3\sqrt{3}}{8}\]

Substituting in (i),

$\Rightarrow \dfrac{3\sqrt{3}}{8}\left( 1-{{\tan }^{2}}\dfrac{\pi }{6} \right)=\dfrac{\sin \dfrac{\pi }{3}}{2}+C$

\[\Rightarrow \dfrac{3\sqrt{3}}{8}\left( 1-\dfrac{1}{3} \right)=\dfrac{\sqrt{3}}{4}+C\]

\[\Rightarrow \dfrac{3\sqrt{3}}{8}\left( \dfrac{2}{3} \right)=\dfrac{\sqrt{3}}{4}+C\]

\[\Rightarrow \dfrac{\sqrt{3}}{4}=\dfrac{\sqrt{3}}{4}+C\]

$\therefore C=0$

Hence from (i),

$\Rightarrow y\left( 1-{{\tan }^{2}}x \right)=\dfrac{\sin 2x}{2}$

$\therefore y=\dfrac{\sin 2x}{2\left( 1-{{\tan }^{2}}x \right)}$

4. Method – 4

Exact Differential Equation: 

A differential equation is said to be exact if it can be derived from its solution (primitive) directly by differentiation, without any elimination, multiplication etc. 

For example, the differential equation $xdy+ydx=0$ is an exact differential equation as it is derived by direct differentiation for its solution, the function $xy=c$.

Illustration 11: Solve $\left( 1+xy \right)ydx+\left( 1-xy \right)xdy=0$ 

Ans: The given equation can be written as 

$ydx+x{{y}^{2}}dx+xdy-{{x}^{2}}ydy=0$ 

$\left( ydx+xdy \right)+xy\left( ydx-xdy \right)=0$ 

$d\left( xy \right)+xy\left( ydx-xdy \right)=0$  

Dividing by ${{x}^{2}}{{y}^{2}}$,

$\dfrac{d\left( xy \right)}{{{x}^{2}}{{y}^{2}}}+\dfrac{ydx-xdy}{xy}=0$ 

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

Integrating,

$-\dfrac{1}{xy}+\log x-\log y=c$ 

Which is the required solution.

Application of Differential Equations: 

The below results are helpful when solving geometrical problems.

Consider the below diagram,

(Image Will Be Updated Soon)

Let PT and PN be the tangent and the normal at $P(x,y)$ respectively. Let the tangent at P make an angle $\theta $ with the x-axis.

Then the slope of the tangent at P $=\tan \theta ={{\left( \dfrac{dy}{dx} \right)}_{P}}$ 

The slope of the normal at P $=-\dfrac{1}{{{\left( \dfrac{dy}{dx} \right)}_{P}}}$  

Equation of the tangent at $P(x,y)$ is

$Y-y={{\left( \dfrac{dy}{dx} \right)}_{P}}\left( X-x \right)$

Equation of the normal at $P(x,y)$ is 

$Y-y=-\dfrac{1}{{{\left( \dfrac{dy}{dx} \right)}_{P}}}\left( X-x \right)$

From $\Delta PGT,\sin \theta =\dfrac{PG}{PT}=\dfrac{y}{PT}$ 

\[\therefore \] 

And $\tan \theta =\dfrac{PG}{TG}=\dfrac{y}{TG}$ 

$\Rightarrow $ 

From $\Delta PGN,\cos \theta =\dfrac{PG}{PN}=\dfrac{y}{PN}$ 

$\Rightarrow $ 

$\tan \theta =\dfrac{GN}{y}$ 

$\Rightarrow $

Illustration 12: If the length of the sub-normal at any point P on the curve is directly proportional to $\text{O}{{\text{P}}^{2}}$, where O is the origin, then form the differential equation of the family of curves and hence find the family of curves. 

Ans: Here $\text{AB}=y\tan \theta =y\dfrac{dy}{dx}$ 

Drawing the diagram,

(Image Will Be Updated Soon)

Also $\text{O}{{\text{P}}^{2}}={{x}^{2}}+{{y}^{2}}$ 

Given, length of the subnormal $=k.\text{O}{{\text{P}}^{2}}$ 

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

$2y\dfrac{dy}{dx}-2k{{y}^{2}}=2k{{x}^{2}}$ ….(i) 

Let ${{y}^{2}}=t\Rightarrow 2y\dfrac{dy}{dt}=\dfrac{dt}{dx}$ …(ii)

From (i) and (ii),

$\dfrac{dt}{dx}-2kt=2k{{x}^{2}}$ 

Which is a linear differential equation.

$\therefore \text{I}\text{.F}\text{.}={{e}^{\int{-2kdx}}}={{e}^{-2kx}}$  

The solution is 

$t.{{e}^{-2kx}}=\int{2k{{x}^{2}}{{e}^{-2kx}}dx}+c$ 

$=2k\left[ {{x}^{2}}\dfrac{{{e}^{-2kx}}}{-2k}+\dfrac{2}{2k}\int{x{{e}^{-2kx}}dx} \right]$ 

$=2k\left[ {{x}^{2}}\dfrac{{{e}^{-2kx}}}{-2k}+\dfrac{1}{k}\left\{ x\dfrac{{{e}^{-2kx}}}{-2k}+\dfrac{1}{2k}\int{{{e}^{-2kx}}dx} \right\} \right]$ 

$=-{{x}^{2}}{{e}^{-2kx}}-\dfrac{x{{e}^{-2kx}}}{k}+\dfrac{1}{k}\dfrac{{{e}^{-2kx}}}{2k}+c$ 

$\therefore {{y}^{2}}=-{{x}^{2}}-\dfrac{x}{k}+\dfrac{1}{2{{k}^{2}}}+c{{e}^{2kx}}$

Differential Equations Revision Notes Class 12

Introduction & Definition

Many important problems in Physical Science and Engineering lead to equations involving derivations on differentials when they are expressed in mathematical terms. Such equations are called differential equations. The following are some examples:

\[{\text{1}}{\text{. }}\frac{{dy}}{{dx}} = \sin x\]

\[{\text{2}}{\text{. }} {\left( {\frac{{dy}}{{dx}}} \right)^2} - y = {x^2}\]

\[{\text{3}}{\text{. }} \frac{{{d^2}y}}{{d{x^2}}} - 4\frac{{dy}}{{dx}} + 3 = 0\]

\[{\text{4}}{\text{. }} \left({\frac{{{d^2}y}}{{d{x^2}}}} \right) - 4{x^2} = 0\]

\[{\text{5}}{\text{. }} \frac{{{d^3}y}}{{d{x^3}}} + {\left( {\frac{{dy}}{{dx}}} \right)^3} - 4y = 0\]

\[{\text{ 6}}{\text{. }}{\left. {x = \sqrt[a]{{1 + \left( {\frac{{dy}}{{dx}}} \right.}}} \right)^2}\]

\[= {x^2} = {a^2}\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]\]

The order of a differential equation is the order of the highest order derivative in the given equation. Thus, equations 1, 2, & 6 are of the first order. Equations 3 & 4 are of 2nd order and equation 5 is of 3rd order.

The degree of a differential equation is the degree of the highest order derivative when the derivatives are free of radicals and negative indices. Thus, equations 1, 3, & 5 are of 1st degree while equations 2, 4, & 6 are of 2nd degree.

A differential equation is said to be linear of the unknown function and all the derivatives involved occur only in the first degree. The equations 1, 3 & 5 are linear while 2, 4, 6 are non-linear. A linear differential equation is always of 1st degree but the converse is not always true.

For ex: the equation \[\dfrac{{{d^2}y}}{{d{x^2}}} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2} + {\text{y}} = 0\]

Even though it has degree 1, it is not linear.

Formation of Differential Equations

Let \[f(x,y,c) = 0.....(i)\]

be an equation containing x, y and one arbitrary constant c. Differentiating with respect to x, we get the equation \[\varphi \left( {x,y,\dfrac{{dy}}{{dx}},c} \right) = 0\].....(ii)

Equation (ii) will in general contain c. If c is eliminated between (i) and (ii), we shall get a relation involving x, y and \[\dfrac{{dy}}{{dx}}\] which will be a differential equation of the 1st order.

For Ex: consider an equation y = ax + 1.....(iii)

where a is an arbitrary constant.

Differentiating both side of equation 3 with respect to x.

\[\dfrac{{dy}}{{dx}} = a\]  .....(iv)

Eliminating a between (iii) & (iv), we get \[y = x\dfrac{{dy}}{{dx}} + 1\].....(v)

which is a differential equation of 1st order & 1st degree.

Equation (iii) is the solution of the differential equation (v)

Similarly, if we have an equation \[f\left( {x,y,{c_1},{c_2}} \right) = 0\] containing two arbitrary constants \[{c_1}\] and \[{c_2}\], then by differentiating it twice we shall get two equations which along with the given equation can be used to eliminate \[{c_1}\] and \[{c_2}\]. We shall then obtain a differential equation of 2nd order.

Thus, we find that when there is only one arbitrary not only constant we get a first order differential, while with 2 arbitrary constants, a 2nd order differential equation is obtained.

The solution of a differential equation in which the number of independent arbitrary constants is equal to the order of the equation, i.e. called the general solution of the given differential equation.

The solution obtained by giving particular values to the arbitrary constants of the general solution is called a particular solution of the given differential equation.

Illustration

Find the differential equation from the relation \[x = a\cos t + b\sin t\] .....(1)

Differentiating equation (1) twice with respect to t, we get \[\dfrac{{dy}}{{dx}} + x = 0\] and \[\dfrac{{{d^2}y}}{{d{x^2}}} + x = 0\] is the required differential equation.

Differential Equations of the 1st order & 1st degree

A differential equation of the 1st order & 1st degree is represented in the form $Mdx + Ndy = 0$

where both $M$ & $N$ are functions of $x$ & $y$ or contents. The general solution of such type of equation involves only one independent arbitrary constant.

These are different methods for solving such equations.

A. Separation of Variables

Such a differential equation is solvable by this method if it can be reduced to the form:

${f_1}(x){\text{d}}x + {{\text{f}}_2}({\text{y}}){\text{dy}} = 0$

Where ${f_1}$  is a function of $x$ & ${f_2}$ is a function of $y.$

Integrating each term of the above equation separately we can write its general solution as

$\int {{f_1}} (x)dx + \int {{f_2}} (y)dy = c$

Where c is an arbitrary constant.

Illustration

Solve $\tan xdy - \tan ydx = 0.....(1)$

Solution:

$ \Rightarrow \tan x\;dy - \tan ydx = 0$

$ \Rightarrow \dfrac{{dy}}{{\tan y}} = \dfrac{{dx}}{{\tan x}}\quad  \Rightarrow \cot y\;{\text{d}}y = \cot x\;{\text{d}}x$

Integrating both sides we get

\[\log |\sin y| = \log |\sin x| + c\;\;\;\;\;\;\;\because \int {\cot xdx = \log |\sin x|} \]

\[\log \left| {\dfrac{{\sin y}}{{\sin x}}} \right| = \log {\text{k}} \Rightarrow \left| {\dfrac{{\sin y}}{{\sin x}}} \right| = {\text{k}}\]

\[\Rightarrow |\sin y| = k|\sin x|\].....(2)

Putting $x = \dfrac{\pi }{4}$ and ${\text{y}} = \dfrac{\pi }{2}$ in equation $(2),$ we get

\[\sin \dfrac{\pi }{2} = k\sin \dfrac{\pi }{4}\]

\[\Rightarrow 1 = k \cdot \dfrac{1}{{\sqrt 2 }} =  > k = \sqrt 2 \]

Putting value of ${\text{k}}$ in $(2),$ we get the particular solution of (1) \[|\sin {\text{y}}| = \sqrt 2 |\sin|\]

Note: equation (2) is the general solution of equation (1)

B. By Method of Substitution

In certain cases when the variables in the differential equation cannot be separated directly, suitable substitutions can facilitate the separation of variables as can be seen in the following example.

Illustration

Solve $\dfrac{{dy}}{{dx}} = {I^{x - y}} + 1$

Solution

We put $x - y = z$

Differentiating both sides with respect to $x$ we get

$1 - \dfrac{{dy}}{{dx}}{{\text{e}}^2} + 1$

$ \Rightarrow \dfrac{{dy}}{{dx}} - {{\text{e}}^2} \Rightarrow {{\text{e}}^2}\;{{\text{d}}_z} =  - {\text{d}}x$

Integrating $\int {{{\text{e}}^{ - 2}}} \;{{\text{d}}_2} =  - \int {\text{d}} x$

$ \Rightarrow \dfrac{{{e^{ - z}}}}{{ - 1}} =  - x - {\text{c}}$ (we have taken the arbitrary constant as c)

\[\Rightarrow  - {e^2} = x + c\;[\because z = x - y]\]

\[\Rightarrow {{\text{e}}^{ - x - x}} = x + c\quad\]

which is the required general solution.

C. When the Equation is Homogeneous

In the equation $Mdx + Ndy = 0$

If M & N are both homogeneous functions of $x$ & $y$. Such an equation is said to be homogeneous and can be expressed in the form $\dfrac{{dy}}{{dx}} - f\left( {\dfrac{y}{x}} \right)$

We generally take the substitution $y = vx,$ as can be seen in the following example.

Illustration

Solve $x\dfrac{{dy}}{{dx}} = y + x\tan \dfrac{y}{x}.....(1)$

Solution:

Clearly, the given expression equation (1) is homogeneous $x$ and $y.$

We put $y = vx \Rightarrow \dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}$

The equation (1) becomes

$x\left( {v + x\dfrac{{dv}}{{dx}}} \right) = vx + x\tan \left( {\dfrac{{vx}}{x}} \right)$

$ \Rightarrow v + x\dfrac{{dv}}{{dx}} \Rightarrow v + \tan v$ where $x \ne 0$

\[\Rightarrow x\dfrac{{dv}}{{dx}}{\text{ from }}v\]

\[\Rightarrow \dfrac{{dv}}{{\tan v}}\dfrac{{dx}}{x}\]

Integrating both sides, we get $\int {\cot } vdv = \int {\dfrac{{dx}}{x}} $

\[\begin{array}{*{20}{l}} { \Rightarrow \sin v = cx} \\ { \Rightarrow \sin \dfrac{y}{x} = cx} \end{array}\]

which is the general solution of equation (1)

D. Non-Homogeneous Equations Reducible to Homogeneous Equations

Of the form $\dfrac{{dy}}{{dx}} = \dfrac{{{a_1}x + {b_1}y + {c_1}}}{{{a_2}x + {b_2}y + {c_2}}}.....(1)$

The method of reducing ex: (1) to a homogeneous equation is equivalent to transferring the origin of coordinates to the point of intersection $({\text{h}},x)$ of the lines ${a_1}x + {b_i}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$

Is substituting $x = x + h$ 

$y = y + h.....(2)$

$x,y$ bring the axis referred to $({\text{h}},{\text{k}})$ as the origin $h$ and $k$ are to be chosen that equation (1)becomes homogeneous. From (2) $dx - dX = dy = dY$

Example (1) becomes $\dfrac{{dy}}{{dx}} = \dfrac{{{a_1}(x + h) + {b_1}(y + h) + {c_1}}}{{{a_2}(x + h) + {b_2}(y + h) + {c_2}}}.....(3)$

Now, $h$ and $k$ are so chosen that ${a_2}h + {b_2}h + {c_1} = 0$ and ${a_2}h + {b_2}h + {c_2} = 0$

Then equation (3) becomes $\dfrac{{dy}}{{dx}} = \dfrac{{{a_1}x + {b_1}y}}{{{a_2}x + {b_2}y}}$ which is homogeneous in $X$ and $Y$ and can be solved in the usual way by putting $Y = vx$ Finally we revert to the current coordinates by using equation (2) values of $h$ and $k$ being already determined.

Linear Equation of First Order

A linear Equation of first order is of the form $\dfrac{{dy}}{{dx}} + Py = Q$

where $P$ and $Q$ are functions of $x$ (or constants).

To find the general solution, we multiply both sides of (1) by ${{\text{e}}^{\int {Pdx} }}$ and obtain

Integrating both sides, we get $y \cdot {e^{\int p dx}} = \int {\left( {Q \cdot {e^{\int {Pdx} }}} \right)} dx + c$

Or, $y \cdot {e^{\int p dx}} \cdot \left\{ {\int {\left( {Q \cdot {e^{\int p dx}}} \right)} dx + c} \right\}$

which is the required general solution of the differential equation (1).

Remarks:

(i) If ${\text{P}}$ is constant, say ${\text{k}},$ then I.F. becomes ${{\text{e}}^{\text{k}}}$

(ii) if $Q = 0,$ then the general solution becomes $y = c \cdot {e^{ - \int {Pdx} }}$

(iii) Here the factor ${{\text{e}}^{\int {{\text{Pdx}}} }}$ is called an Integrating factor (I.F) of (i) I.F. of a differential equation is a factor multiplying by which the L.H.S. and the R.H.S. of the equation can be easily integrated

(iv) A linear differential equation contains neither products nor powers involving y or $\dfrac{{dy}}{{dx}}$.

Illustration

Solve $\dfrac{{dy}}{{dx}} + 3y = {e^{2x}}$

Solution: We have $\dfrac{{dy}}{{dx}} + 3y = {{\text{e}}^{2x}}.....(1)$

Here, $P = 3$ and $Q = {e^{2x}}$ Integrating Factor of (1) is ${e^{\int {pdx} }} = {e^{\int 3 dx}} = {e^{3x}}$

Multiplying both side of (1) by ${{\text{e}}^{3x}},$ we gat

${{\text{e}}^{3x}} \cdot \dfrac{{dy}}{{dx}} + 3y \cdot {e^{3x}} = {e^{2x}} \cdot {e^{3x}}$

or, $\dfrac{d}{{dx}}\left( {y \cdot {e^{3x}}} \right) = {e^{5x}}$

Integrating both side, we get

$y \cdot {e^{3x}} = \int {{e^{5x}}} dx + c$

$y \cdot {e^{3x}} = \dfrac{1}{5}{e^{5x}} + c,$ or $y = \dfrac{1}{5}{e^{2x}} + c{e^{ - 3x}},$ which is the required solution of (1)

Applications of Differential Equations

In solving some mathematical problems concerning physical science and engineering application of differential equations is useful.

Example: The role of increase of bacteria in a culture is proportional to the number of bacteria present and it is found that number doubles in 5 hours. Express this mathematically using the rate of increase of bacteria & time. Hence, calculate how many times the bacteria may be expected to grow at the end of 15 hours.

Solution

Let the original no. of bacteria present $ = {\text{b}}$

At the end of the hours, no. of bacteria = x

Then according to the problem $\dfrac{{dy}}{{dt}}.x \Rightarrow \dfrac{{dx}}{{dt}}kx,$ where ${\text{k}}$ is a constant

$ \Rightarrow \dfrac{{dy}}{{dt}}{\text{kt}}$

Integrating we get $\log x = kt + c.....(1)$

where $c$ is an arbitrary constant

Initially at $k = 0,x = b$

$\therefore $ Equation (1) gives 

$\therefore $ Equation (1) becomes $\log x = kt + \log b$

Now, when $t = 5,x = 2b$ (given) $\therefore $ Equation (2) given $\log 2\;b = 5k + \log b$

\[= 5k = \log 2b - \log b = \log \dfrac{{2b}}{b}\]

\[= k = \dfrac{1}{5}\log 2\]

$\therefore $ equation (2) becomes $\log x = \dfrac{1}{5}(\log 2) \cdot t + \log b.....(3)$

which the required mathematical expression

When $t = 15,$ equation (3) given

\[\log x = \dfrac{1}{5}(\log 2) \cdot 15 + \log b\]

\[\Rightarrow \log x - \log b = 3\log 2\]

\[\Rightarrow \log \dfrac{x}{6} = \log {2^3} = \log 8\]

\[\Rightarrow \dfrac{x}{6} = 8\]

\[\Rightarrow x = 8\;b\]

Hence, the no. of bacteria will be 8 times at the end of 15 hours.

Why are Differential Equations Class 12 Notes Important?

Though you have studied hard throughout the year, revision is considered a must before the exam. It's a human tendency that the things which we have not gone through for a long time will deteriorate with time, hence it is mandatory to revise all the concepts before the exam. If the topics that you studied months back are not revised, it is as good as not studied.

Differential Equations Class 12 Notes will help you in the following ways:

• While revising the topics covered in the notes, you can point out your weaknesses and strengths and then can make your preparation according to your need.

• You can refer to the notes to have a quick revision of the entire chapter before the exams.

• Differential Equation notes are a must before the exam, as this helps you to recall the concepts of each topic to answer effectively in the examination.
  
  

Download Revision Notes Class 12 Chapter 9 Differential Equations PDF

While preparing for the exams, it is important to have revision notes. To make CBSE Class 12 students Differential Equation preparation ace up, we at Vedantu offer you the most effective Class 12 Maths Notes Chapter 9 Differential Equations. Referring to these notes will not only help you in school exam preparation but also help you to prepare for competitive exams like the Olympiad. Differential Equations Class 12 Notes are designed by the subject experts in a simple language considering the latest exam pattern and CBSE guidelines. These notes are a great way to revise the Differential Equation topics quickly before the exams. Students can download Revision Notes Class 12 Chapter 9 Differential Equations free pdf through the link given.

Significance of Class 12 Maths Chapter 9 Differential Equations Notes

The syllabus of Class 12 Maths has been set to develop the conceptual foundation of the students. This foundation will be utilised to choose a professional course after the completion of the Higher Secondary level education. In fact, this foundation of conceptual knowledge will also allow the students to crack competitive exams.

One of the crucial parts of this syllabus is Class 12 Maths Chapter 9 Differential Equations. This chapter is a part of calculus, a prime topic of advanced Mathematics that students study at this level. This is where the notes designed by the experts will come in very handy. Apart from the theoretical explanation given in the chapter, these notes will act as the perfect guide to follow and focus on the concepts taught.

The simpler version of the concepts and principles of differential equations will enable students to focus on them better. They will also find out the context of the mathematical derivations and explanations of these concepts in a better way. Hence, adding these notes to the study material for the preparation of this chapter is quite beneficial for the students.

Benefits of Class 12 Maths Chapter 9 Differential Equations Notes

• These notes have been formulated by following the latest CBSE Class 12 Maths syllabus to include all the topics in a single file. This file can be accessed online at your convenience.

• You can also download and add this file to your study material so that you can refer to it offline.

• Resolve the doubts related to the fundamental concepts of differential equations added to this chapter without any hassle. Refer to the notes to get a simpler version of the topics and find the answer to all your queries in no time.

• Follow these notes for revising this chapter before an exam and recall what you have studied while solving questions in a better way. In this way, you can solve all the questions precisely and score more.

Download Class 12 Maths Chapter 9 Differential Equations Notes PDF

Get the free PDF version of these notes and focus on the concepts only. Check how the formulas have been derived to equate and operate on differential equations. Develop your fundamental concepts related to this chapter and learn how to score more in the exams.

Topics Covered in Class 12 Differential Equations

A Mathematical equation that involves an independent variable, dependent variable, and differential coefficients of the dependent variable with respect to the independent variable is called a differential equation. Revise more about these in Differential Equations Class 12 revision notes.

The topics and sub-topics covered in Differential Equations Class 12 revision notes are:

1. Introduction

2. Basic Concepts

3. Order of a differential equation

4. Degree of a differential equation

5. General and Particular Solutions of a Differential Equation

6. Formation of a Differential Equation 

7. Procedure to form a differential equation 

8. Solutions for First Order, First Degree Differential Equations

9. Differential equations with variables separable

10. Homogeneous differential equations

11. Linear differential equations.
    
    

Why Choose Vedantu’s Class 12 Differential Equations Notes?

Here are some of the benefits of referring to Vedantu’s Class 12 Differential Equations Notes:

• Students can easily understand and grasp the topics covered in the chapter.

• Studying from the notes will help you to attempt the complicated questions asked in the exam more easily.

• The notes are prepared by the subject experts who have vast years of experience in teaching Mathematics.

• All the relevant diagrams are provided to make the students understand the concepts more easily.

• The notes will help you to attempt the exams more confidently.
  
  

Conclusion

Class 12 Maths Differential Equations revision notes are very simple and explained in a very easy language so that students find them easy to understand and can retain the topics easily without any problem. Experts at Vedantu designed Class 12 Maths Differential Equations revision notes in such a way that they will greatly benefit you during exam time.

Other Chapters Revision Notes

Find here CBSE Class 12 Maths revision notes for other chapters:

• CBSE Class 12 Maths Chapter 1 Revision Notes

• CBSE Class 12 Maths Chapter 2 Revision Notes

• CBSE Class 12 Maths Chapter 3 Revision Notes

• CBSE Class 12 Maths Chapter 4 Revision Notes

• CBSE Class 12 Maths Chapter 5 Revision Notes

• CBSE Class 12 Maths Chapter 6 Revision Notes

• CBSE Class 12 Maths Chapter 7 Revision Notes

• CBSE Class 12 Maths Chapter 8 Revision Notes

• CBSE Class 12 Maths Chapter 9 Revision Notes

• CBSE Class 12 Maths Chapter 10 Revision Notes

• CBSE Class 12 Maths Chapter 11 Revision Notes

• CBSE Class 12 Maths Chapter 12 Revision Notes

• CBSE Class 12 Maths Chapter 13 Revision Notes

Other Related Links

• CBSE Class 12 Maths Revision Notes 

• CBSE Class 12 NCERT Maths Books PDF

• CBSE Class 12 Maths Important Questions

• CBSE Class 12 Maths Important Formulas

• Differential Equations

• Order and Degree of Differential Equation

• First Order Differential Equations

• Homogeneous Differential Equations