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

Halkadvalseln 2021-11-19 Answered
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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Expert Answer

George Blue
Answered 2021-11-20 Author has 1039 answers
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
Answered 2021-11-21 Author has 1607 answers
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
\(\displaystyle{\frac{{{y}'}}{{{y}}}}={\frac{{{x}^{{2}}}}{{{2}}}}+{k}\)
\(\displaystyle\int{\frac{{{\left.{d}{y}\right.}}}{{{y}}}}=\int{\frac{{{x}^{{2}}}}{{{2}}}}+{k}{\left.{d}{x}\right.}\)
\(\displaystyle{\ln{{y}}}={\frac{{{x}^{{3}}}}{{{6}}}}+{k}_{{1}}{x}+{k}_{{2}}\)
\(\displaystyle{y}{\left({x}\right)}={k}_{{2}}{e}^{{{\frac{{{x}^{{3}}}}{{{6}}}}+{k}_{{1}}{x}}}\)
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