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I need some hints for solving $yy''-(y')^2=xy^2$.

I noticed that the left hand side is close to $(yy')'$:

$yy''-(y')^2=xy^2\ \Leftrightarrow\ yy''+(y')^2-2(y')^2=xy^2\ \Leftrightarrow\ (yy')'-2(y')^2=xy^2$.

But I don't know how to continue expressing the terms as derivatives of some functions.

Thanks

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Consider $\frac {y'}{y}$ instead of $y'y$ $$yy''-(y')^2=xy^2$$ $$(\frac {y'}{y})'=x$$ Integrate $$\frac {y'}y=\frac {x^2}2+k$$ $$\int \frac {dy}y=\int \frac {x^2}2+kdx$$ $$\ln y=\frac {x^3}6+k_1x+k_2$$ $$y(x)=k_2e^{\frac {x^3}6+k_1x}$$

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    $\begingroup$ Thank you Isham for this and for the other one I posted. $\endgroup$ – El_Bastaix Aug 30 '18 at 12:16
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    $\begingroup$ @Victor you're welcome...you were on the right track... $\endgroup$ – Isham Aug 30 '18 at 12:16
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Take the change of variable $z = \ln y$. So you would end up with the following relations $$y = e^z$$ $$y' = z'e^z$$ $$y'' = e^z z'' + (z')^2 e^z$$ Upon substitution (all $e^z$'s will cancel out), leaving you with a $2^{nd}$ order differential equation, which is easy to solve: $$z'' = x$$

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  • $\begingroup$ Thanks Ahmad, great approach. $\endgroup$ – El_Bastaix Aug 30 '18 at 12:14

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