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How to solve the ODE:

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

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    What is $(y y')'$ ?2011-08-31
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    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 $y^2$.2011-08-31
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    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!\\2011-08-31
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    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?2011-08-31

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