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Answer: $x^{2}y-xy^{2}=C$

Given: $\left ( y^{2}-2xy \right )dx=\left ( x^{2}-2xy \right )dy$

To find: we have to solve the given differential equation

Hint: In homogeneous differential equation put   $y=vx$ and$\frac{dy}{dx}=v+x\frac{dv}{dx}$

Solution: we have

$\left ( y^{2}-2xy \right )dx=\left ( x^{2}-2xy \right )dy$

$\Rightarrow \frac{dy}{dx}=\frac{y^{2}-2xy}{x^{2}-2xy}$

It is homogeneous differential equation.

Put $y=vx$ and $\frac{dy}{dx}=v+x\frac{dv}{dx}$

So,$v+x\frac{dv}{dx}=\frac{v^{2}x^{2}-2vx^{2}}{x^{2}-2vx^{2}}$

\begin{aligned} &\Rightarrow v+x \frac{d v}{d x}=\frac{v^{2}-2 v}{1-2 v} \\ &\Rightarrow x \frac{d v}{d x}=\frac{v^{2}-2 v}{1-2 v}-v \\ &\Rightarrow x \frac{d v}{d x}=\frac{v^{2}-2 v-v+2 v^{2}}{1-2 v} \\ &\Rightarrow x \frac{d v}{d x}=\frac{3 v^{2}-3 v}{1-2 v} \end{aligned}

$\Rightarrow \frac{1-2 v}{3\left(v^{2}-v\right)} d v=\frac{d v}{x} \\$

$\Rightarrow \frac{-(2 v-1)}{3\left(v^{2}-v\right)} d v=\frac{d x}{d x} \\$

$\Rightarrow \int \frac{2 v-1)}{\left.v^{2}-v\right)} d v=-3 \int \frac{d x}{x} \\$                $[Integrating \; \; on \: \: both\: \: side]$

$\Rightarrow \log \left|v^{2}-v\right|=-3 \log |x|+\log c . \\$

$\Rightarrow v^{2}-v=\frac{c}{x^{3}} \\$

$\Rightarrow \frac{y^{2}}{x^{2}}-\frac{y}{x}=\frac{c}{x^{3}} \\$

$\Rightarrow \frac{y^{2}-x y}{x^{2}}=\frac{c}{x^{3}} \\$

$\Rightarrow y^{2}-x y=\frac{c}{x} \\$

$\Rightarrow x y^{2}-x^{2} y=c \\$

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