The solution of x frac{dy}{dx} = y + x e^{y/x | vedclass

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(D) The given differential equation is $x \frac{dy}{dx} = y + x e^{y/x}$.Dividing by $x$,we get $\frac{dy}{dx} = \frac{y}{x} + e^{y/x}$.This is a homo

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$e^{-y/x} + \log x = 1$

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(D) The given differential equation is $x \frac{dy}{dx} = y + x e^{y/x}$.Dividing by $x$,we get $\frac{dy}{dx} = \frac{y}{x} + e^{y/x}$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation: $v + x \frac{dv}{dx} = v + e^v$.This simplifies to $x \frac{dv}{dx} = e^v$,or $e^{-v} dv = \frac{1}{x} dx$.Integrating both sides: $\int e^{-v} dv = \int \frac{1}{x} dx$,which gives $-e^{-v} = \log x + c$.Substituting $v = y/x$,we get $-e^{-y/x} = \log x + c$.Given the condition $y(1) = 0$,we substitute $x = 1$ and $y = 0$: $-e^0 = \log 1 + c$,which implies $-1 = 0 + c$,so $c = -1$.Thus,$-e^{-y/x} = \log x - 1$,which rearranges to $e^{-y/x} + \log x = 1$.

The solution of $x \frac{dy}{dx} = y + x e^{y/x}$ with $y(1) = 0$ is

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