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Need Solution for R.D. Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 25 Maths Textbook Solution.

Answers (1)

Answer: $x+ye^{\frac{x}{y}}=c$

Given: $1+e^{\frac{x}{y}}dx+e^{\frac{x}{y}}\left ( 1-\frac{x}{y} \right )dy=0$

To solve: We have to solve the given differential equation.

Hint: We have to put $x=vy$ and $\frac{dx}{dy}=v+x\frac{dv}{dx}$

Solution: We have,

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

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

It is homogeneous equation.

Put $x=vy$ and $\frac{dx}{dy}=v+x\frac{dv}{dx}$

So, $v+x\frac{dv}{dx}=\frac{-e^{v}\left ( 1-v \right )}{1+e^{v}}-v$

$\begin{aligned} \Rightarrow y \frac{d v}{d y}=\frac{-e^{v}(1-v)}{1+e^{v}}-v \\ \end{aligned}$

$\begin{aligned} \Rightarrow y \frac{d v}{d y}=\frac{-e^{v}+v e^{v}-v-v e^{v}}{1+e^{v}} \\ \end{aligned}$

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

$\Rightarrow \frac{1+e^{v}}{v+e^{v}}=-\frac{d y}{y} \text { (on seperating the variables) }$

Integrating both sides

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

Put $v+e^{v}=t$

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

So equation becomes

$\Rightarrow \int \frac{d t}{t}=-\log y+\log c \\$

$\Rightarrow \log t=-\log y+\log c \\$

$\Rightarrow \log \left|v+e^{v}\right|+\log y=\log c\left[\therefore t=v+e^{v}\right] \\$

$\Rightarrow \log \left|y\left(v+e^{v}\right)\right|=\log c \\$

$\Rightarrow y\left(\frac{x}{y}+e^{\frac{x}{y}}\right)=c\left[\therefore v=\frac{x}{y}\right] \\$

$\Rightarrow x y e^{\frac{x}{y}}=c$

Hence this is required solution.

Posted by

infoexpert21

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