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

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$e^{- w x} + 2 \log x = 1$

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

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

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

answer is D.

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Detailed Solution

Put $y = v x . Then \frac{d y}{d x} = v + x \frac{d v}{d x}$

Given equation is $x \frac{d y}{d x} = y + x e^{y / x} \Rightarrow \frac{d y}{d x} = \frac{y}{x} + e^{y / x} \Rightarrow v + x \frac{d v}{d x} = v + e^{v}$

$\begin{array}{l} \Rightarrow x \frac{d v}{d x} = e^{v} \Rightarrow e^{- v} d v = \frac{d x}{x} \Rightarrow \int e^{- v} d v = \int \frac{d x}{x} \Rightarrow - e^{- v} = \log x - c \Rightarrow \log x + e^{- v} = c \\ \Rightarrow \log x + e^{y / x} = c \Rightarrow y (1) = 0 \Rightarrow \log 1 + e^{0} = c \Rightarrow c = 1 \end{array}$