NCERT Solutions for Class 12 Maths Chapter 9: Differential Equations - Exercise 9.4

Exercise 9.4

Refer to pages 18-40 for exercise 9.4 in the PDF
1. Solve the differential equation $\dfrac{{dy}}{{dx}} = \dfrac{{1 - \cos x}}{{1 + \cos x}}$

Ans: The differential equation is given by,

$\dfrac{{dy}}{{dx}} = \dfrac{{1 - \cos x}}{{1 + \cos x}}$…(i)

We have,  $1 - \cos x = 2{\sin ^2}\dfrac{x}{2}$

$1 + \cos x = 2co{\operatorname{s} ^2}\dfrac{x}{2}$

Applying both formula in the equation (i),

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

By simplifying,

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

$\dfrac{{dy}}{{dx}} = {\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}} \right)^2}$…(ii)

We have, $\dfrac{{\sin x}}{{\cos x}} = \tan x$

Now the equation (ii) becomes, 

$\dfrac{{dy}}{{dx}} = {\left( {\tan \dfrac{x}{2}} \right)^2}$

So,$\dfrac{{dy}}{{dx}} = {\tan ^2}\dfrac{x}{2}$

Since, ${\sec ^2}x - {\tan ^2}x = 1$

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

By separating the variables,

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

Taking integration on each side of the equation,

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

$\int {dy}  = \int {{{\sec }^2}\dfrac{x}{2}dx - \int {dx} } $

By integrating,

$y = 2\tan \dfrac{x}{2} - x + C$, $C$is an arbitrary constant.  

Hence, the required general solution is  $y = 2\tan \dfrac{x}{2} - x + C$  , $C$is an arbitrary constant.

2. Solve the differential equation $\dfrac{{dy}}{{dx}} = \sqrt {4 - {y^2}} \left( { - 2 < y < 2} \right)$

Ans: The differential equation is given by,

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

By separating variables,

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

Taking integration on each side of the equation,

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

By integrating,

${\sin ^{ - 1}}\dfrac{y}{2} = x + C$  , $C$is an arbitrary constant.  

This implies,

$\dfrac{y}{2} = \sin \left( {x + C} \right)$, $C$is an arbitrary constant.  

$y = 2\sin \left( {x + C} \right)$, $C$is an arbitrary constant.  

Hence, the required general solution is$y = 2\sin \left( {x + C} \right)$, $C$is an arbitrary constant.

3. Solve the differential equation $\dfrac{{dy}}{{dx}} + y = 1\left( {y \ne 1} \right)$

Ans: The differential equation is given by,

$\dfrac{{dy}}{{dx}} + y = 1$

Multiplying both sides with$dx$,

$dy + ydx = dx$

$dy = dx - ydx$

$dy = \left( {1 - y} \right)dx$

By separating variables,

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

Taking integration on each side of the equation,

$\int {\dfrac{{dy}}{{\left( {1 - y} \right)}}}  = \int {dx} $

By integrating,

$-\log \left( {1 - y} \right) = x + \log C$, $C$is an arbitrary constant. 

$\log \left( {1 - y} \right) +\log C = -x$, $C$ is an arbitrary constant.  

$\log C\left( {1 - y} \right) =  - x$

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

Dividing both sides with $C$,

$1 - y = frac{{{e^{ - x}}}}{C}$

Thus,$y = 1 - \frac{{{e^{ - x}}}}{C}$

Take  $A =  - \frac{1}{C}$,                   

$y = 1 + A{e^{ - x}}$ , $A$is an arbitrary constant.  

Hence, the required general solution is$y = 1 + A{e^{ - x}}$ , $A$is an arbitrary constant.

4. Solve the differential equation ${\sec ^2}x\tan ydx + {\sec ^2}y\tan xdy = 0$

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

5. Solve the differential equation \[\left( {{e^x} + {e^{ - x}}} \right)dy - \left( {{e^x} - {e^{ - x}}} \right)dx = 0\]

Ans: The differential equation is given by,

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

\[\left( {{e^x} + {e^{ - x}}} \right)dy = \left( {{e^x} - {e^{ - x}}} \right)dx\]

Divided by  \[\left( {{e^x} + {e^{ - x}}} \right)\],

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

Taking integration in each side of the equation,

\[\int {dy}  = \int {\left[ {\dfrac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}} \right]dx} \]

By integrating,

\[y = \int {\left[ {\dfrac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}} \right]dx}  + C\] , $C$ is an arbitrary constant.….(i)

Assume ${e^x} + {e^{ - x}} = t$

By differentiating with respect to $x$,

$\dfrac{d}{{dx}}\left( {{e^x} + {e^{ - x}}} \right) = \dfrac{{dt}}{{dx}}$

${e^x} - {e^{ - x}} = \dfrac{{dt}}{{dx}}$

This implies,

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

Substituting this value in the equation (i),

\[y = \int {\dfrac{1}{t}} dt + C\], $C$ is an arbitrary constant

By integrating,

\[y = \log t + C\], $C$ is an arbitrary constant

Putting value of $t$,

\[y = \log \left( {{e^x} + {e^{ - x}}} \right) + C\], $C$ is an arbitrary constant

Hence, the required general solution is\[y = \log \left( {{e^x} + {e^{ - x}}} \right) + C\], $C$ is an arbitrary constant.

6. Solve the differential equation $\dfrac{{dy}}{{dx}} = \left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)$

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. Solve the differential equation $y\log ydx - xdy = 0$

Ans: The differential equation is given by,

$y\log ydx - xdy = 0$

$y\log ydx = xdy$

By separating variables,

$\dfrac{{dy}}{{y\log y}} = \dfrac{{dx}}{x}$

Taking integration in each side of the equation,

$\int {\dfrac{{dy}}{{y\log y}}}  = \int {\dfrac{{dx}}{x}} $…(i)

Assume,$\log y = t$

By differentiating,

$\dfrac{1}{y}dy = dt$

Substituting this value in the equation (i),

$\int {\dfrac{{dt}}{t}}  = \int {\dfrac{{dx}}{x}} $

By integrating,

$\log t = \log x + \log C$ , $C$ is an arbitrary constant

$\log t = \log Cx$

So, $t = Cx$

By substituting value of $t$,

$\log y = Cx$ , $C$ is an arbitrary constant

So, $y = {e^{Cx}}$, $C$ is an arbitrary constant

Hence, the required general solution is$y = {e^{Cx}}$, $C$ is an arbitrary constant.

8. Solve the differential equation ${x^5}\dfrac{{dy}}{{dx}} =  - {y^5}$

Ans: The differential equation is given by,

${x^5}\dfrac{{dy}}{{dx}} =  - {y^5}$

By separating variables,

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

$\dfrac{{dy}}{{{y^5}}} + \dfrac{{dx}}{{{x^5}}} = 0$

Taking integration in each side of the equation,

$\int {\dfrac{{dy}}{{{y^5}}}}  + \int {\dfrac{{dx}}{{{x^5}}}}  = C$ , $C$ is an arbitrary constant.

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

By integrating,

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

This implies,

${y^{ - 4}} + {x^{ - 4}} =  - 4C$

Take $ - 4C = K$

${y^{ - 4}} + {x^{ - 4}} = K$ , $K$is an arbitrary constant.

Hence, the required general solution is ${y^{ - 4}} + {x^{ - 4}} = K$, $K$ is an arbitrary constant.

9. Solve the differential equation $\dfrac{{dy}}{{dx}} = {\sin ^{ - 1}}x$

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

$\Rightarrow -2xdx=2tdt$ 

$\Rightarrow -xdx=tdt$ 

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

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

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

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

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

Ans: The differential equation is given by,

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

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

By separating variables,

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

Taking integration in each side of the equation,

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

Let $\tan y = u$ ,

By differentiating,

${\sec ^2}ydy = du$

So, $\int {\dfrac{{{{\sec }^2}y}}{{\tan y}}} dy = \int {\dfrac{{du}}{u}} $

$\int {\dfrac{{{{\sec }^2}y}}{{\tan y}}} dy = \log u$ [Integrating]

Substituting,

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

Also let $1 - {e^x} = t$

By differentiating,

$ - {e^x}dx = dt$

So, $\int {\dfrac{{ - {e^x}}}{{\left( {1 - {e^x}} \right)}}} dx = \int {\dfrac{{dt}}{t}} $

$\int {\dfrac{{ - {e^x}}}{{\left( {1 - {e^x}} \right)}}} dx = \log t$ [Integrating]

Substituting,

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

Substituting all value of $\int {\dfrac{{{{\sec }^2}y}}{{\tan y}}} dy$ and $\int {\dfrac{{ - {e^x}}}{{\left( {1 - {e^x}} \right)}}} dx$ in the equation (i),

$\log \left( {\tan y} \right) = \log \left( {1 - {e^x}} \right) + \log C$ , $C$ is an arbitrary constant.

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

$\tan y = C\left( {1 - {e^x}} \right)$ , $C$ is an arbitrary constant

Hence, the required general solution is $\tan y = C\left( {1 - {e^x}} \right)$ , $C$ is an arbitrary constant.

11. Solve the differential equation $\left( {{x^3} + {x^2} + x + 1} \right)\dfrac{{dy}}{{dx}} = 2{x^2} + x;y = 1$ when $x = 0$

Ans: The differential equation is given by, 

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

We can write,

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

By separation variables,

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

Taking integration in each side of the equation,

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

By integrating,

$y = \int {\dfrac{{2{x^2} + x}}{{\left( {{x^3} + {x^2} + x + 1} \right)}}dx}  + K$ , $K$ is an arbitrary constant.

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

$y = \int {\dfrac{{2{x^2} + x}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}}dx}  + K$…(i)

Let $\dfrac{{2{x^2} + x}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}} = \dfrac{A}{{\left( {x + 1} \right)}} + \dfrac{{Bx + C}}{{\left( {{x^2} + 1} \right)}}$…(ii)

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

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

By simplifying,

$2{x^2} + x = A{x^2} + A + B{x^2} + Bx + Cx + C$

$2{x^2} + x = \left( {A + B} \right){x^2} + \left( {B + C} \right)x + \left( {A + C} \right)$

Comparing coefficient of $x$and ${x^2}$ ,

$A + B = 2$…(iii)

$B + C = 1$…(iv)

$A + C = 0$…(v)

From equation (v),

$A =  - C$

Substituting $A$ in the equation (iii),

$ - C + B = 2$

$B - C = 2$…(vi)

Adding equation (iv) and (vi),

$2B = 3$

So, $B = \dfrac{3}{2}$

Substituting value of $B$ in equation (iii),

$A + \dfrac{3}{2} = 2$

$A = 2 - \dfrac{3}{2}$

So, $A = \dfrac{1}{2}$

Then $C =  - \dfrac{1}{2}$

Substituting value of $A,B$ and $C$ in the equation (ii),

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

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

So, the equation (i) becomes,

$y = \int {\left[ {\dfrac{1}{2}\dfrac{1}{{\left( {x + 1} \right)}} + \dfrac{1}{2}\dfrac{{3x - 1}}{{\left( {{x^2} + 1} \right)}}} \right]dx}  + K$

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

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

$y = \dfrac{1}{2}\int {\dfrac{1}{{\left( {x + 1} \right)}}dx}  + \dfrac{3}{4}\int {\dfrac{{2x}}{{\left( {{x^2} + 1} \right)}}} dx - \dfrac{1}{2}\int {\dfrac{1}{{\left( {{x^2} + 1} \right)}}} dx + K$…(vii)

We have,

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

So, $d\left( {{x^2} + 1} \right) = 2xdx$

Then the equation (vii) becomes,

$y = \dfrac{1}{2}\int {\dfrac{1}{{\left( {x + 1} \right)}}dx}  + \dfrac{3}{4}\int {\dfrac{{d\left( {{x^2} + 1} \right)}}{{\left( {{x^2} + 1} \right)}}}  - \dfrac{1}{2}\int {\dfrac{1}{{\left( {{x^2} + 1} \right)}}} dx + K$

By integrating,

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

$y = \dfrac{1}{4}\left[ {2\log \left( {x + 1} \right) + 3\log \left( {{x^2} + 1} \right)} \right] - \dfrac{1}{2}{\tan ^{ - 1}}x + K$

$y = \dfrac{1}{4}\left[ {\log {{\left( {x + 1} \right)}^2} + \log {{\left( {{x^2} + 1} \right)}^3}} \right] - \dfrac{1}{2}{\tan ^{ - 1}}x + K$

$y = \dfrac{1}{4}\log \left[ {{{\left( {x + 1} \right)}^2}{{\left( {{x^2} + 1} \right)}^3}} \right] - \dfrac{1}{2}{\tan ^{ - 1}}x + K$…(viii)

When$x = 0$,

Put $y = 1$ ,

$1 = \dfrac{1}{4}\log \left( 1 \right) - \dfrac{1}{2}{\tan ^{ - 1}}\left( 0 \right) + K$

$1 = 0 - 0 + K$

$K = 1$

Substituting $K$ in the equation (viii),

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

Hence, the required general solution is$y = \dfrac{1}{4}\log \left[ {{{\left( {x + 1} \right)}^2}{{\left( {{x^2} + 1} \right)}^3}} \right] - \dfrac{1}{2}{\tan ^{ - 1}}x + 1$ .

12. Solve the differential equation $x\left( {{x^2} - 1} \right)\dfrac{{dy}}{{dx}} = 1;y = 0$ when $x = 2$

Ans: The differential equation is given by,

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

By separating variables,

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

We have, 

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

So, $dy = \dfrac{1}{{x\left( {x - 1} \right)\left( {x + 1} \right)}}dx$

Taking integration in each side of the equation,

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

Let $\dfrac{1}{{x\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{A}{x} + \dfrac{B}{{x - 1}} + \dfrac{C}{{x + 1}}$…(ii)

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

So, $A\left( {x - 1} \right)\left( {x + 1} \right) + Bx\left( {x + 1} \right) + Cx\left( {x - 1} \right) = 1$

$A{x^2} - A + B{x^2} + Bx + C{x^2} - Cx = 1$

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

Comparing coefficient of $x$and ${x^2}$ ,

$A + B + C = 0$…(iii)

$B - C = 0$…(iv)

$ - A = 1$

So, $A =  - 1$

From equation (iv),

$B = C$

Substituting value of $A$and $B$ ,

$ - 1 + C + C = 0$

$2C = 1$

$C = \dfrac{1}{2}$

So, $B = \dfrac{1}{2}$

Substituting the value of $A,B$ and $C$ in the equation (ii),

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

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

Substituting value of $\dfrac{1}{{x\left( {x - 1} \right)\left( {x + 1} \right)}}$ in the equation (i),

$\int {dy}  = \int {\left[ { - \dfrac{1}{x} + \dfrac{1}{2}\dfrac{1}{{x - 1}} + \dfrac{1}{2}\dfrac{1}{{x + 1}}} \right]} dx$

$\int {dy}  =  - \int {\dfrac{1}{x}dx}  + \dfrac{1}{2}\int {\dfrac{1}{{x - 1}}} dx + \dfrac{1}{2}\int {\dfrac{1}{{x + 1}}} dx$

By integrating,

$y =  - \log x + \dfrac{1}{2}\log \left( {x - 1} \right) + \dfrac{1}{2}\log \left( {x + 1} \right) + \log K$, $K$ is an arbitrary constant

$y = \dfrac{1}{2}\left[ { - 2\log x + \log \left( {x - 1} \right) + \log \left( {x + 1} \right) + 2\log K} \right]$

$y = \dfrac{1}{2}\left[ { - \log {x^2} + \log \left( {x - 1} \right) + \log \left( {x + 1} \right) + \log {K^2}} \right]$

$y = \dfrac{1}{2}\log \left[ {\dfrac{{{K^2}\left( {x - 1} \right)\left( {x + 1} \right)}}{{{x^2}}}} \right]$…(v)

When$x = 2$,

Put $y = 0$

$0 = \dfrac{1}{2}\log \left[ {\dfrac{{{K^2}\left( {2 - 1} \right)\left( {2 + 1} \right)}}{{{2^2}}}} \right]$

$\log \left[ {\dfrac{{3{K^2}}}{4}} \right] = 0$

$\dfrac{{3{K^2}}}{4} = {e^0}$

$\dfrac{{3{K^2}}}{4} = 1$

${K^2} = \dfrac{4}{3}$

Substituting value of ${K^2}$ in the equation (v),

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

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

Hence, the required general solution is$y = \dfrac{1}{2}\log \left[ {\dfrac{{4\left( {{x^2} - 1} \right)}}{{3{x^2}}}} \right]$ .

13. Solve the differential equation $\cos \left( {\dfrac{{dy}}{{dx}}} \right) = a$$\left( {a \in \mathbb{R}} \right);y = 1$ when $x = 0$

Ans: The differential equation is given by,

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

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

$dy = {\cos ^{ - 1}}adx$

Taking integration in each side of the equation,

$\int {dy}  = \int {{{\cos }^{ - 1}}adx} $

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

By integrating,

$y = {\cos ^{ - 1}}a \cdot x + K$, $K$ is an arbitrary constant

$y = x{\cos ^{ - 1}}a + K$…(i)

When $x = 0$

Put $y = 1$,

$1 = 0 \cdot {\cos ^{ - 1}}a + K$

$K = 1$

Substituting value of $K$ in the equation (i),

$y = x{\cos ^{ - 1}}a + 1$

Subtracting $1$ from both sides,

$y - 1 = x{\cos ^{ - 1}}a$

Dividing by $x$ ,

$\dfrac{{y - 1}}{x} = {\cos ^{ - 1}}a$

This implies,

$\cos \left( {\dfrac{{y - 1}}{x}} \right) = a$

Hence, the required general solution is $\cos \left( {\dfrac{{y - 1}}{x}} \right) = a$.

14.  Solve the differential equation $\dfrac{{dy}}{{dx}} = y\tan x,y = 1$ when $x = 0$

Ans: The differential equation is given by,

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

By separating variables,

$\dfrac{{dy}}{y} = \tan xdx$

Taking integration in each side of the equation,

$\int {\dfrac{{dy}}{y}}  = \int {\tan xd} x$

$\int {\dfrac{{dy}}{y}}  = \int {\dfrac{{\sin x}}{{\cos x}}d} x$...(i)

Let $\cos x = u$

By differentiating,

$ - \sin xdx = du$

$\sin xdx =  - du$

Substituting in the equation (i),

$\int {\dfrac{{dy}}{y}}  = \int {\dfrac{{ - 1}}{u}d} u$

$\int {\dfrac{{dy}}{y}}  =  - \int {\dfrac{1}{u}d} u$

$\log y =  - \log u + \log C$, $C$is an arbitrary constant

Putting value of $u$ ,

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

$\log y = \log \left( {\dfrac{C}{{\cos x}}} \right)$

$\log y = \log \left( {C\sec x} \right)$

$y = C\sec x$…(ii)

When$x = 0$

Put $y = 1$ ,

$1 = C\sec 0$

$C = 1$

Then the equation (ii) becomes,

$y = \sec x$

Hence, the required general solution is$y = \sec x$.

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

Ans: The differential equation of the curve is given by,

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

We have, 

$\dfrac{{dy}}{{dx}} \equiv y'$

So, $\dfrac{{dy}}{{dx}} = {e^x}\sin x$

By separating variable,

$dy = {e^x}\sin xdx$

Taking integration in each side of the equation,

$\int {dy}  = \int {{e^x}\sin xdx} $

By integrating,

$y = \int {{e^x}\sin xdx}  + C$…(i) ,$C$ is an arbitrary constant

Let $I = \int {{e^x}\sin xdx} $

By using by-parts formula $\int {u \cdot vdx = u\int {vdx - \int {\left[ {\dfrac{d}{{dx}}u\int {vdx} } \right]} } } dx$ ,

$I = \sin x\int {{e^x}dx}  - \int {\left[ {\dfrac{d}{{dx}}\left( {\sin x} \right)\int {{e^x}dx} } \right]} dx$

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

Again by using by-parts formula,

$I = \sin x{e^x} - \left[ {\cos x\int {{e^x}} dx - \int {\left( {\dfrac{d}{{dx}}\cos x\int {{e^x}dx} } \right)} dx} \right]$

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

$I = \sin x{e^x} - \left[ {\cos x{e^x} + \int {\sin x} {e^x}dx} \right]$

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

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

Adding $I$ to the both sides,

$2I = \sin x{e^x} - \cos x{e^x}$

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

Dividing by $2$ ,

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

Substituting value in the equation (i),

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

The curve is passing through the point $\left( {0,0} \right)$,

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

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

$0 = \dfrac{{ - 1}}{2} + C$

$C = \dfrac{1}{2}$

Now, substituting value of $C$in the equation (ii),

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

This implies,

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

Subtracting$\left( 1 \right)$ from both sides,

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

Hence, the required equation of the curve is$2y - 1 = {e^x}\left( {\sin x - \cos x} \right)$.

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

Ans: The differential equation of the curve is given by,

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

By separating the variables,

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

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

Taking integration in each side of the equation,

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

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

By integrating,

$y - 2\log \left( {y + 2} \right) = x + 2\log x + C$ , $C$is an arbitrary constant

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

This can be written as,

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

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

The curve passes through the point $\left( {1, - 1} \right)$,

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

$ - 2 - C = \log 1$

$ - 2 - C = 0$

$C =  - 2$

Substituting value of $C$ in the equation (i),

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

Hence, the required equation of the curve is$y - x + 2 = \log \left[ {{x^2}{{\left( {y + 2} \right)}^2}} \right]$.

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

Ans: Assume $x$ and $y$ be the $x$-coordinate and $y$-coordinate of the curve respectively.
We have, the slope of a tangent to the curve in the coordinate axis which is given by 

$\dfrac{{dy}}{{dx}}$ .

According to the question,

$y\dfrac{{dy}}{{dx}} = x$

By separating the variables,

$ydy = xdx$

Taking integration in each side of the equation,

$\int {ydy}  = \int {xdx} $

By integrating,

$\dfrac{{{y^2}}}{2} = \dfrac{{{x^2}}}{2} + C$ , $C$ is an arbitrary constant

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

${y^2} - {x^2} = 2C$…(i)

The curve passes through point$\left( {0, - 2} \right)$,

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

$4 = 2C$

$C = 2$

Substituting value of $C$ in the equation (i),

${y^2} - {x^2} = 2 \cdot 2$

${y^2} - {x^2} = 4$

Hence, the required equation of the curve is${y^2} - {x^2} = 4$.

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

Ans: From the question we have, $\left( {x,y} \right)$ is the point of contact of the curve and its tangent.

The slope $\left( {{m_1}} \right)$ of the line segment, joining $\left( {x,y} \right)$ and $\left( { - 4, - 3} \right)$ is,

${m_1} = \dfrac{{y + 3}}{{x + 4}}$

We have, the slope of a tangent to the curve is given by  $\dfrac{{dy}}{{dx}}$ .

Therefore, the slope $\left( {{m_2}} \right)$ of the tangent is,

${m_2} = \dfrac{{dy}}{{dx}}$

According to the given question,

${m_2} = 2{m_1}$

Substituting the values,

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

By separating variable,

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

Taking integration in each side of the equation,

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

By integrating,

$\log \left( {y + 3} \right) = 2\log \left( {x + 4} \right) + \log C$ , $C$ is an arbitrary constant

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

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

$\left( {y + 3} \right) = C{\left( {x + 4} \right)^2}$…(i)

This is the general equation of the curve.

The curve is passes through the point $\left( { - 2,1} \right)$ .

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

$4 = C{\left( 2 \right)^2}$

$4C = 4$

$C = 1$

Substituting value of $C$ in the equation (i),

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

Hence, the required equation of the curve is$\left( {y + 3} \right) = {\left( {x + 4} \right)^2}$.

19. The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

Ans: Let us assume the rate of change of the volume of the balloon is $k$.

Where $k$ is a constant.

Also Assume volume of spherical balloon is $v$

According to the given question,

$\dfrac{{dv}}{{dt}} = k$…(i)

We know that, volume of spherical balloon with radius $r$ is,

$v = \dfrac{4}{3}\pi {r^3}$

Substituting value of $v$ in the equation (i),

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

$\dfrac{4}{3}\pi  \cdot 3{r^2} \cdot \dfrac{{dr}}{{dt}} = k$

$4\pi {r^2}dr = kdt$

Taking integration in each side of the equation,

$4\pi \int {{r^2}d} r = k\int d t$

By integrating,

$4\pi \dfrac{{{r^3}}}{3} = kt + C$ , $C$ is an arbitrary constant

$4\pi {r^3} = 3\left( {kt + C} \right)$…(i)

At $t = 0$

$r = 3$,

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

$4\pi  \cdot 27 = 3C$

Divided by $3$ ,

$C = 36\pi $

At $t = 3$

$r = 6$,

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

Substituting value of $C$ ,

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

$864\pi  = 9k + 108\pi $

$864\pi  - 108\pi  = 9k$

$9k = 756\pi $

Dividing by $9$ ,

$k = 84\pi $

Substituting the values of $k$ and $C$ in equation (i),

$4\pi {r^3} = 3\left( {84\pi t + 36\pi } \right)$

$4\pi {r^3} = 3 \cdot 4\pi \left( {21t + 9} \right)$

Dividing by $4\pi $ ,

${r^3} = 3\left( {21t + 9} \right)$

${r^3} = 63t + 27$

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

Hence, the radius of the balloon after t seconds will be ${\left( {63t + 27} \right)^{\dfrac{1}{3}}}$.

20. In a bank, principal increases continuously at the rate of$r\% $ per year. Find the value of $r$ if Rs$100$ doubles itself in $10$years$\left( {{{\log }_e}2 = 0.6931} \right)$.

Ans: Let $p$ is the principle ,$t$ denotes time, and $r$ represents rate of interest.

Given that, the principal increases continuously at the rate of $r\% $per year.

We get,

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

By separating variables,

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

Taking integration in each side of the equation,

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

$\int {\dfrac{{dp}}{p}}  = \dfrac{r}{{100}}\int {\left( 1 \right)dt} $

By integrating,

$\log p = \dfrac{{rt}}{{100}} + k$ ; $k$ is an arbitrary constant

$p = {e^{\dfrac{{rt}}{{100}} + k}}$…(i)

When $t = 0$

$p = 100$

$100 = {e^{\dfrac{{r \cdot 0}}{{100}} + k}}$

$100 = {e^k}$

If $t = 10$

Then $p = 2 \times 100$

$p = 200$

Then the equation (i) becomes,

$200 = {e^{\dfrac{{r \cdot 10}}{{100}} + k}}$

$200 = {e^{\dfrac{{r \cdot 10}}{{100}}}} \cdot {e^k}$

Substituting value of ${e^k}$ ,

$200 = {e^{\dfrac{{r \cdot 10}}{{100}}}} \cdot 100$

Simplifying,

$\dfrac{{200}}{{100}} = {e^{\dfrac{r}{{10}}}}$

$2 = {e^{\dfrac{r}{{10}}}}$

Taking logarithm to both sides,

${\log _e}2 = {\log _e}{e^{\dfrac{r}{{10}}}}$

${\log _e}2 = \dfrac{r}{{10}}{\log _e}e$

Since, ${\log _e}e = 1$

${\log _e}2 = \dfrac{r}{{10}}$

Substituting $\left( {{{\log }_e}2 = 0.6931} \right)$ ,

$0.6931 = \dfrac{r}{{10}}$

$r = 0.6931 \times 10$

$r = 6.931$

Hence, the value of $r$is $6.93\% $.

21. In a bank, principal increases continuously at the rate of $5\% $ per year. An amount of ${\text{Rs }}1000$is deposited with this bank, how much will it worth after $10$years$\left( {{e^{0.5}} = 1.648} \right)$.

Ans: Let $p$ is the principle and $t$ denotes time.

Given that the principal increases continuously at the rate of $5\% $ per year.

We get,

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

Simplifying,

$\dfrac{{dp}}{{dt}} = \dfrac{p}{{20}}$

By separating variables,

$\dfrac{{dp}}{p} = \dfrac{{dt}}{{20}}$

Taking integration in each side of the equation,

$\int {\dfrac{{dp}}{p}}  = \int {\dfrac{{dt}}{{20}}} $

By integrating,

$\log p = \dfrac{t}{{20}} + k$

$p = {e^{\dfrac{t}{{20}} + k}}$…(i)

When $t = 0$

$p = 1000$,

$1000 = {e^{\dfrac{0}{{20}} + k}}$

${e^k} = 1000$

At $t = 10$ , the equation (i) becomes,

$p = {e^{\dfrac{{10}}{{20}} + k}}$

$p = {e^{\dfrac{{10}}{{20}}}} \cdot {e^k}$

Substituting value of ${e^k}$ ,

$p = {e^{\dfrac{{10}}{{20}}}} \cdot 1000$

Simplifying,

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

Substituting  ${e^{0.5}} = 1.648$ ,

$p = 1.648 \times 1000$

$p = 1648$

Hence, the amount will worth ${\text{Rs }}1648$after 10 years.

22. In a culture, the bacteria count is$1,00,000$. The number is increased by $10\% $ in $2$ hours. In howmany hours will the count reach$2,00,000$, if the rate of growth of bacteria is proportional to thenumber present?

Ans: Let us assume $x$ be the number of bacteria at any instant $t$.

Given that, the rate of growth of the bacteria is proportional to the number present.

According to the given information,

$\dfrac{{dx}}{{dy}} \propto x$

It can be written as,

$\dfrac{{dx}}{{dy}} = kx$

By separating the variables,

$\dfrac{{dx}}{x} = kdt$ ; $k$ is a constant

Taking integration in each side of the equation,

$\int {\dfrac{{dx}}{x}}  = k\int {dt} $

By integrating,

$\log x = kt + C$ ; $C$ is an arbitrary constant …(i)

Let us assume ${x_0}$ be the number of bacteria at$t = 0$ .

So, the equation (i) becomes,

$\log {x_0} = C$

Substituting the value of $C$ in the equation (i),

$\log x = kt + \log {x_0}$

$\log x - \log {x_0} = kt$

$\log \left( {\dfrac{x}{{{x_0}}}} \right) = kt$

$kt = \log \left( {\dfrac{x}{{{x_0}}}} \right)$…(ii)

Also, given that the number of bacteria increases by$10\% $ in $2$ hours.

This means,

$x = \dfrac{{110}}{{100}}{x_0}$

Simplifying,

$\dfrac{x}{{{x_0}}} = \dfrac{{11}}{{10}}$

Substituting this in the equation (ii),

When $t = 2$

$k \cdot 2 = \log \left( {\dfrac{{11}}{{10}}} \right)$

$k = \dfrac{1}{2}\log \left( {\dfrac{{11}}{{10}}} \right)$

Substituting the value of $k$ in the equation (ii),

$\dfrac{1}{2}\log \left( {\dfrac{{11}}{{10}}} \right)t = \log \left( {\dfrac{x}{{{x_0}}}} \right)$

$t = \dfrac{{2\log \left( {\dfrac{x}{{{x_0}}}} \right)}}{{\log \left( {\dfrac{{11}}{{10}}} \right)}}$…(iii)

When the number of bacteria increases from $1,00,000$to $2,00,000$, let us assume the time is ${t_1}$ .

This means, at $t = {t_1}$

$x = 2{x_0}$

So, the equation (iii) will be,

${t_1} = \dfrac{{2\log \left( {\dfrac{{2{x_0}}}{{{x_0}}}} \right)}}{{\log \left( {\dfrac{{11}}{{10}}} \right)}}$

${t_1} = \dfrac{{2\log 2}}{{\log \left( {\dfrac{{11}}{{10}}} \right)}}$

Hence, the number of bacteria increases from $1,00,000$ to $2,00,000$in $\dfrac{{2\log 2}}{{\log \left( {\dfrac{{11}}{{10}}} \right)}}$ hours.

23. The general solution of the differential equation $\dfrac{{dy}}{{dx}} = {e^{x + y}}$ is

a. ${e^x} + {e^{ - y}} = C$

b. ${e^x} + {e^y} = C$

c. ${e^{ - x}} + {e^y} = C$

d. ${e^{ - x}} + {e^{ - y}} = C$

Ans: The differential equation is given by,

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

This can be written as,

$\dfrac{{dy}}{{dx}} = {e^x} \cdot {e^y}$

By separating the variables,

$\dfrac{{dy}}{{{e^y}}} = {e^x}dx$

This implies,

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

Taking integration in each side of the equation,

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

By integrating,

$ - {e^{ - y}} = {e^x} + k$ ; $k$is an arbitrary constant

$ - {e^{ - y}} - {e^x} = k$

Multiplying both sides with $\left( { - 1} \right)$,

${e^{ - y}} + {e^x} =  - k$

Taking$ - k = C$ , $C$is an arbitrary constant

${e^{ - y}} + {e^x} = C$

${e^x} + {e^{ - y}} = C$

Hence, the general solution of the differential equation is${e^x} + {e^{ - y}} = C$ .

The correct answer is option (A).

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.4

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