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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
That's correct Davide! So that the general solution is \\ $$y^2=\frac{1}{3}x^3+Ax+B$$ an elliptic curve. Indeed I put this ODE which I solved by\\ using $(yy')'=(y')^2+yy''. I put this ODE as I want to know people with dominion about\\ nonlinear ODE's. But the ODE I haven't been able to solve is\\ $$yy''+y'=x$$, maybe you have an interesting idea at respect. Thanks so much!\\ – perucho07 Aug 31 '11 at 11:20
Are you saying that even though your question is about $yy''+(y')^2=x$, the equation you really want to know about is $yy''+y'=x$? If so, why didn't you ask about the equation you really wanted to ask about? – Gerry Myerson Aug 31 '11 at 13:58

1 Answer 1

Integrating both sides with respect to $x$ gives \begin{equation*} -\frac{x^2}{2}+y'y=C_1 \end{equation*} where we have used the reverse product rule and $C_1$ is a constant. Solve for $y'$ and multiply both sides by $2y$ to get \begin{equation*} 2y'y=x^2+2C_1. \end{equation*} Integrate both sides with respect to $x$ to get \begin{equation*} y^2=\frac{x^3}{3}+2C_1x+C_2 \end{equation*} where $C_2$ is another constant. Take the square root. $~_{\square}$

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