# I need some hints for solving yy''-(y')^2=xy^2 I noticed that the

I need some hints for solving $$\displaystyle{y}{y}{''}-{\left({y}'\right)}^{{2}}={x}{y}^{{2}}$$
I noticed that the left hand side is close to $$\displaystyle{\left({y}{y}'\right)}'$$:
$$\displaystyle{y}{y}{''}-{\left({y}'\right)}^{{2}}={x}{y}^{{2}}\Rightarrow{y}{y}{''}+{\left({y}'\right)}^{{2}}-{2}{\left({y}'\right)}^{{2}}={x}{y}^{{2}}\Rightarrow{\left({y}{y}'\right)}'-{2}{\left({y}'\right)}^{{2}}={x}{y}^{{2}}$$
But I don't know how to continue expressing the terms as derivatives of some functions.

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George Blue
Take the change of variable $$\displaystyle{z}={\ln{{y}}}$$. So you would end up with the following relations
$$\displaystyle{y}={e}^{{z}}$$
$$\displaystyle{y}'={z}'{e}^{{z}}$$
$$\displaystyle{y}{''}={e}^{{z}}{z}{''}+{\left({z}'\right)}^{{2}}{e}^{{z}}$$
Upon substitution (all $$\displaystyle{e}^{{z}}$$'s will cancel out), leaving you with a 2-nd order differential equation, which is easy to solve:
$$\displaystyle{z}{''}={x}$$
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Stephanie Mann
Consider $$\displaystyle{\frac{{{y}'}}{{{y}}}}$$ instead of $$\displaystyle{y}'{y}$$
$$\displaystyle{y}{y}{''}-{\left({y}'\right)}^{{2}}={x}{y}^{{2}}$$
$$\displaystyle{\left({\frac{{{y}'}}{{{y}}}}\right)}'={x}$$
Integrate
$$\displaystyle{\frac{{{y}'}}{{{y}}}}={\frac{{{x}^{{2}}}}{{{2}}}}+{k}$$
$$\displaystyle\int{\frac{{{\left.{d}{y}\right.}}}{{{y}}}}=\int{\frac{{{x}^{{2}}}}{{{2}}}}+{k}{\left.{d}{x}\right.}$$
$$\displaystyle{\ln{{y}}}={\frac{{{x}^{{3}}}}{{{6}}}}+{k}_{{1}}{x}+{k}_{{2}}$$
$$\displaystyle{y}{\left({x}\right)}={k}_{{2}}{e}^{{{\frac{{{x}^{{3}}}}{{{6}}}}+{k}_{{1}}{x}}}$$