If y=y(x) is the solution of x frac{d y}{d x} | vedclass

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(B) Given differential equation: $x \frac{d y}{d x} = y + x e^{-\left(\frac{y}{x}\right)}$.Dividing by $x$,we get $\frac{d y}{d x} = \frac{y}{x} + e^{

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$e \log (1+e)$

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(B) Given differential equation: $x \frac{d y}{d x} = y + x e^{-\left(\frac{y}{x}\right)}$.Dividing by $x$,we get $\frac{d y}{d x} = \frac{y}{x} + e^{-\left(\frac{y}{x}\right)}$.This is a homogeneous differential equation. Let $v = \frac{y}{x}$,so $y = vx$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$.Substituting these into the equation: $v + x \frac{d v}{d x} = v + e^{-v}$.$x \frac{d v}{d x} = e^{-v} \Rightarrow e^v d v = \frac{d x}{x}$.Integrating both sides: $\int e^v d v = \int \frac{d x}{x} \Rightarrow e^v = \log |x| + C$.Substituting $v = \frac{y}{x}$: $e^{\frac{y}{x}} = \log |x| + C$.Given $y(1) = \log e = 1$,so at $x=1, y=1$: $e^{\frac{1}{1}} = \log(1) + C \Rightarrow e = 0 + C \Rightarrow C = e$.Thus,the solution is $e^{\frac{y}{x}} = \log x + e$.To find $y(e)$,put $x=e$: $e^{\frac{y(e)}{e}} = \log e + e = 1 + e$.Taking $\log$ on both sides: $\frac{y(e)}{e} = \log(1+e) \Rightarrow y(e) = e \log(1+e)$.

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

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