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The solution of $\left(x^2+y^2\right) d x=2 x y d y$ is :
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Verified Answer
The correct answer is:
$c\left(x^2-y^2\right)=x$
$\because \quad \frac{x^2+y^2}{2 x y}=\frac{d y}{d x}$
Put $y=v x$ and $\frac{d y}{d x}=v+x \frac{d v}{d x}$
$\therefore \quad v+x \frac{d v}{d x}=\frac{x^2+v^2 x^2}{2 x^2 v}$
$\Rightarrow \quad v+x \frac{d v}{d x}=\frac{1+v^2}{2 v}$
$\Rightarrow \quad \frac{x d v}{d x}=\frac{1+v^2-2 v^2}{2 v}$
$\Rightarrow \quad \frac{2 v}{1-v^2} d v=\frac{1}{x} d x$
$\Rightarrow \quad-\log \left(1-v^2\right)=\log x+\log c$
$\Rightarrow \quad-\log \left(\frac{x^2-y^2}{x^2}\right)=\log x+\log c$
$\Rightarrow \quad-\log \left(x^2-y^2\right)+2 \log x=\log x+\log c$
$\Rightarrow \quad \log x=\log c+\log \left(x^2-y^2\right)$
$\Rightarrow \quad x=c\left(x^2-y^2\right)$
Put $y=v x$ and $\frac{d y}{d x}=v+x \frac{d v}{d x}$
$\therefore \quad v+x \frac{d v}{d x}=\frac{x^2+v^2 x^2}{2 x^2 v}$
$\Rightarrow \quad v+x \frac{d v}{d x}=\frac{1+v^2}{2 v}$
$\Rightarrow \quad \frac{x d v}{d x}=\frac{1+v^2-2 v^2}{2 v}$
$\Rightarrow \quad \frac{2 v}{1-v^2} d v=\frac{1}{x} d x$
$\Rightarrow \quad-\log \left(1-v^2\right)=\log x+\log c$
$\Rightarrow \quad-\log \left(\frac{x^2-y^2}{x^2}\right)=\log x+\log c$
$\Rightarrow \quad-\log \left(x^2-y^2\right)+2 \log x=\log x+\log c$
$\Rightarrow \quad \log x=\log c+\log \left(x^2-y^2\right)$
$\Rightarrow \quad x=c\left(x^2-y^2\right)$
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