If $y=y(x)$ is the solution of $x \frac{d y}{d x}=y+x e^{-\left(\frac{y}{x}\right)}, y(1)=$ $\log…

1. $\log \left(\frac{1}{\mathrm{e}}+1\right)$
2. e $\log (1+\mathrm{e})$
3. $\mathrm{e} \log \left(\frac{1}{\mathrm{e}}+1\right)$
4. $\operatorname{elog}\left(1-\frac{1}{\mathrm{e}}\right)$

Solution

Which is a homogenous differential equation So let $\frac{y}{x}=v \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$ ...(ii) From $\mathrm{eq}^{\mathrm{n}}$ (i) \& (ii), we get $\begin{aligned} & v+x \frac{d v}{d x}=v+e^{-v} \\ & \Rightarrow x \frac{d v}{d x}=e^{-v} \Rightarrow e^v d \mu=\frac{d x}{x}\end{aligned}$ Integrating both sides, we get
$\begin{aligned} & y(1)=\log e=1 \\ & \therefore \log (1)=e^1+c \Rightarrow c=-e\end{aligned}$ Putting the value of $c$ in equation (1), we get $\Rightarrow \log x=e^{\frac{y}{x}}-e$

Asked in: AP EAMCET 2023 (17 May Shift 2)

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