# Solve dy/dx−1/2(1+1/x)y+3/xy^3=0 ?

Solve $\frac{dy}{dx}-\frac{1}{2}\left(1+\frac{1}{x}\right)y+\frac{3}{x}{y}^{3}=0$?
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Kimberly Evans
Making the substitution $u\left(x\right)=\frac{1}{y\left(x\right)}$ and $u\prime \left(x\right)=-\frac{1}{{y}^{2}\left(x\right)}y\prime \left(x\right)$
we have
$\frac{\left(1+x\right){u}^{2}+2xuu\prime -6}{2x{u}^{3}}=0$ and now supposing that $2x{u}^{3}\ne 0$ we have
$\left(1+x\right){u}^{2}+2xuu\prime -6=0$ Now making $z={u}^{2}$ we have the new linear differential equation
$xz\prime +\left(1+x\right)z-6=0$
This differential equation can be easily solved giving
$z=\frac{1}{x}\left(6-C{e}^{-x}\right)$ where
and then
$u=±\sqrt{\frac{C{e}^{-x}+6}{x}}$ and finally
$y=\frac{1}{u}=±\frac{1}{\sqrt{\frac{C{e}^{-x}+6}{x}}}$
NOTE:
y(x)=0 is also solution.