The general solution of left(1+e^{frac{x}{y}} | vedclass

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(D) Given equation: $\left(1+e^{x / y}\right) d x+e^{x / y}\left(1-\frac{x}{y}\right) d y=0$ Rearranging the terms: $\frac{d x}{d y}=-\frac{e^{x / y}(

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

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(D) Given equation: $\left(1+e^{x / y}\right) d x+e^{x / y}\left(1-\frac{x}{y}\right) d y=0$ Rearranging the terms: $\frac{d x}{d y}=-\frac{e^{x / y}(1-x / y)}{1+e^{x / y}}$ Let $x=v y$,then $\frac{d x}{d y}=v+y \frac{d v}{d y}$. Substituting these into the equation: $v+y \frac{d v}{d y}=-\frac{e^v(1-v)}{1+e^v}$ $y \frac{d v}{d y}=-\frac{e^v-v e^v}{1+e^v}-v = \frac{-e^v+v e^v-v-v e^v}{1+e^v} = -\frac{v+e^v}{1+e^v}$ Separating variables: $\frac{1+e^v}{v+e^v} d v=-\frac{d y}{y}$ Integrating both sides: $\int \frac{1+e^v}{v+e^v} d v=-\int \frac{d y}{y}$ Let $v+e^v=t$,then $(1+e^v) d v=d t$. So,$\int \frac{d t}{t}=-\int \frac{d y}{y} \Rightarrow \ln|t|=-\ln|y|+\ln|c|$ $\ln|t|+\ln|y|=\ln|c| \Rightarrow \ln|t y|=\ln|c| \Rightarrow t y=c$ Substituting $t=v+e^v$ and $v=x/y$: $(x/y+e^{x/y}) y=c$ $x+y e^{x/y}=c$.

The general solution of $\left(1+e^{\frac{x}{y}}\right) d x+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y=0$ is

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