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The general solution of the differential equation \(\left(1+e^{\frac{x}{y}}\right) d x+\left(1-\frac{x}{y}\right) e^{\frac{x}{y}} d y=0\) is (c is an arbitrary constant)
(A) \(x-y e^{\frac{x}{y}}=c\)
(B) \(y-x e^{y}=c\)
(C) \(x+y e^{\frac{x}{y}}=c\)
(D) \(y+x e^{\frac{x}{y}}=c\)

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Ans: (C)
Hint : Putting \(\frac{x}{y}=v \Rightarrow x=v y\)
$$
\Rightarrow \frac{d x}{d t}=v+y \frac{d v}{d y} \Rightarrow y \frac{d v}{d y}=-\left(\frac{v+e^{v}}{1+e^{v}}\right) \Rightarrow\left(\frac{1+e^{v}}{v+e^{v}}\right) d v+\frac{d y}{y}=0
$$
Integrating
$$
\begin{aligned}
&\Rightarrow \ln \left(\mathrm{v}+\mathrm{e}^{\mathrm{v}}\right)+\ln \mathrm{y}=\mathrm{c}_{1} \\
&\Rightarrow \mathrm{y}\left(\frac{\mathrm{x}}{\mathrm{y}}+\mathrm{e}^{\frac{x}{y}}\right)=\mathrm{c} \Rightarrow \mathrm{x}+\mathrm{y} \mathrm{e}^{\frac{x}{y}}=\mathrm{c}
\end{aligned}
$$
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