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If $x \frac{d y}{d x}+y=x \frac{f(x y)}{f^{\prime}(x y)}$, then $|f(x y)|$ is equal to
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$\mathrm{Ce}^{\mathrm{x}^2 / 2}$
$x \frac{d y}{d x}+y=x \frac{f(x y)}{f^{\prime}(x y)} \Rightarrow \frac{d(x y)}{d x}=x \frac{f(x y)}{f^{\prime}(x y)} \Rightarrow \frac{f^{\prime}(x y)}{f(x y)} d(x y)=x d x \Rightarrow \ln |f(x y)|=\frac{x^2}{2}+k \Rightarrow f(x y)=C e^{\frac{x^2}{2}}$
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