# Nonlinear ODE of second order

How to solve the ODE:

$$yy'' + (y')^2 = x.$$

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What is $(y y')'$ ? –  Raskolnikov Aug 31 '11 at 9:46
The LHS is the derivative of $yy'$, and we get $yy'=\frac{x^2}2+C$ where $C$ is a constant. Now, note that $2 yy'$ is the derivative of $y^2$. –  Davide Giraudo Aug 31 '11 at 10:06

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