# Solution of a nonlinear ODE

How can I solve this ODE$$-vU'=2(UU''+(U')^2),$$ where $v$ is a constant. I can see that the right hand side is $(U^2)''$ but is this useful.

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What is $v$? The independent variable? – Babak S. Jan 20 '14 at 11:02
$v$ is the variable or some constant? – Martín-Blas Pérez Pinilla Jan 20 '14 at 11:02
see the edit it's a constant. – Vaolter Jan 20 '14 at 11:04

You have $$-vU' = (U^2)'',$$ integrate this once and get $$-vU + C = (U^2)',$$ this is a regular first order ODE which you can solve $$-vU + C = 2UU'\\ U' = \frac{1}{2}(-v + \frac{C}{U}).$$ by integrating $$\int \frac{2dU}{-v + \frac{C}{U}} = \int 1 dt$$ if $t$ is your variable. You get $$t = -\frac{1}{v^2}(C\log(C - vU) + vU)+D$$ for constants $C$ and $D$. The inverse is of course a pain, but still, $U$ is defined in the equation above.