I tried solving this by bernoulli type , but coefficient didn't came appropriate.
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Maybe try for a solution of the form $y=\lambda x^n$? – lulu Oct 25 '16 at 11:56
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Divide both sides by $x^2$, and you'll find that this is indeed a Bernoulli differential equation – Ben Grossmann Oct 25 '16 at 11:57
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@Omnomnomnom i found $dv/dx -1/x = y^2/(-2(x^2))$ where $v=y^(1-n) = y^(-2)$ but can't go any further – codemonkey Oct 25 '16 at 12:06
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Divide by $y^2$ to get $\frac{x^2}{y^3}\frac{dy}{dx}+\frac{2x}{y^2}=1$.
Substitute $\frac{1}{y^2}=z$ which gives $\frac{1}{y^3}\frac{dy}{dx} =\frac{-1}{2}\frac{dz}{dx}$ to get Linear differential equation
$\frac{dz}{dx}-\frac{4}{x}z=\frac{-2}{x^2}$ which you can solve.
Nitin Uniyal
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