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The solution of $\frac{d y}{d x}=\frac{y^2}{x y-x^2}$ is
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Verified Answer
The correct answer is:
$e^{y / x}=k y$
Given,
$\frac{d y}{d x}=\frac{y^2}{x y-x^2}$
Put
$y=v x$
$\begin{array}{rlrl}\Rightarrow & \frac{d y}{d x} & =v+x \frac{d v}{d x} \\ & \therefore & v+x \frac{d v}{d x} & =\frac{v^2 x^2}{v x^2-x^2} \\ \Rightarrow & x \frac{d v}{d x} & =\frac{v^2}{v-1}-v \\ \Rightarrow & & x \frac{d v}{d x} & =\frac{v}{v-1} \\ \Rightarrow & & \left(1-\frac{1}{v}\right) d v & =\frac{d x}{x} \\ \Rightarrow & v-\log v & =\log x+\log k \\ \Rightarrow & y / x & =\log x \cdot k \cdot \frac{y}{x}\end{array}$
$\begin{array}{llrl}\Rightarrow & \frac{y}{x} & =\log y k \\ \Rightarrow & e^{y / x} & =y k\end{array}$
$\frac{d y}{d x}=\frac{y^2}{x y-x^2}$
Put
$y=v x$
$\begin{array}{rlrl}\Rightarrow & \frac{d y}{d x} & =v+x \frac{d v}{d x} \\ & \therefore & v+x \frac{d v}{d x} & =\frac{v^2 x^2}{v x^2-x^2} \\ \Rightarrow & x \frac{d v}{d x} & =\frac{v^2}{v-1}-v \\ \Rightarrow & & x \frac{d v}{d x} & =\frac{v}{v-1} \\ \Rightarrow & & \left(1-\frac{1}{v}\right) d v & =\frac{d x}{x} \\ \Rightarrow & v-\log v & =\log x+\log k \\ \Rightarrow & y / x & =\log x \cdot k \cdot \frac{y}{x}\end{array}$
$\begin{array}{llrl}\Rightarrow & \frac{y}{x} & =\log y k \\ \Rightarrow & e^{y / x} & =y k\end{array}$
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