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

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(D) 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 homogen

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

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(D) 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 this into the equation: $v + x \frac{dv}{dx} = v + e^v$.This simplifies to $x \frac{dv}{dx} = e^v$.Separating variables: $e^{-v} dv = \frac{1}{x} dx$.Integrating both sides: $\int e^{-v} dv = \int \frac{1}{x} dx$, which gives $-e^{-v} = \log x + c$.Substituting $v = y/x$: $-e^{-y/x} = \log x + c$.Using the condition $y(1) = 0$: $-e^{-0/1} = \log 1 + c \Rightarrow -1 = 0 + c \Rightarrow c = -1$.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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