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So I have to solve $2t^2y''+(y')^3=2ty'$, I start by making $v=y'$, so then I have

$2t^2v'+v^3=2tv$, divide whole equation by $2t^2$, so

$v'+v^3/(2t^2)=v/t$, where this is Bernnoulli, so I let $u=v^{-2}$, and it's now

$u'-u/(t^2)=-2/t$, now this is linear, so I need an integrating factor, which is the integral of $-1/(t^2)$ to the power of e, doing that I get the integrating factor to be $e^{1/t}$, and skipping the step of multiplying it into the equation,

$d/dt[ue^{1/t}]=-2e^{1/t}/t^2$, integrating both sides I get

$ue^{1/t}=2e^{1/t}+C$, substituting v then y' back in, I get

$y'^{-2}=2+C/e^{1/t}$, and this is where I get stuck, as I can't solve the integral if I do variable separable, thanks in advance

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    $\begingroup$ Verify your equation for $u$. I have found $u^{\prime}+2u/t=1/t^2$. But of course I can be wrong... $\endgroup$
    – Kelenner
    Feb 2, 2015 at 16:22
  • $\begingroup$ I got that know, I must've made a stupid mistake, thanks! $\endgroup$
    – Jeff
    Feb 2, 2015 at 16:45

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