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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.

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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.

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