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The general solution of the differential equation $x^2 d y$ $-\left(x y-y^2\right) d x=0$ is
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$y \log x=x+c y$
$\begin{aligned} & \text { (c) } x^2 d y-\left(x y-y^2\right) d x=0 \\ & \Rightarrow x^2 d y=\left(x y-y^2\right) d x \\ & \Rightarrow \frac{d y}{d x}=\frac{x y-y^2}{x^2} \\ & \text { Let } y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \\ & \Rightarrow v+x \frac{d v}{d x}=\frac{x^2 v-v^2 x^2}{x^2}\end{aligned}$
$\begin{aligned} & \Rightarrow \frac{d x}{x}=\frac{d v}{-v^2} \Rightarrow \int \frac{d x}{x}=-\int \frac{d v}{v^2} \\ & \Rightarrow \ln x=-\left(-\frac{1}{v}\right)+C \Rightarrow \ln x=\frac{1}{v}+C \\ & \Rightarrow \ln x=\frac{x}{y}+C \Rightarrow y \log x=x+C y .\end{aligned}$
$\begin{aligned} & \Rightarrow \frac{d x}{x}=\frac{d v}{-v^2} \Rightarrow \int \frac{d x}{x}=-\int \frac{d v}{v^2} \\ & \Rightarrow \ln x=-\left(-\frac{1}{v}\right)+C \Rightarrow \ln x=\frac{1}{v}+C \\ & \Rightarrow \ln x=\frac{x}{y}+C \Rightarrow y \log x=x+C y .\end{aligned}$
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