Solve differential equation $yy''=2x(y')^2$

Question

I can solve it when $y$ is not there on left-hand side. Not getting any approach how to do it. Please help.

Answer 1

Suppose $y\neq 0$. Let $$ p=\ln y$$ and then $$ p'=\frac{y'}{y}. $$ So $$p''=\frac{y''y-(y')^2}{y^2}=\frac{2x(y')^2-(y')^2}{y^2}=(2x-1)\bigg(\frac{y'}{y}\bigg)^2=(2x-1)(p')^2. $$ and hence $$ \frac{p''}{(p')^2}=2x-1. $$ Integrating both sides gives $$ -\frac{1}{p'}=x^2-x+C. $$ You can do the rest.

Answer 2

We can use $(\frac{y}{y'})'=\frac{y'^2-yy''}{y'^2}$.

$\frac{y'^2-yy''}{y'^2}=\frac{y'^2-2xy'^2}{y'^2}=1-2x$.

So $\frac{y}{y'}=x-x^2+c$ and $\frac{y'}{y}=\frac{1}{x-x^2+c}$.

Now, $lny=\int \frac{1}{x-x^2+c_1} +c_2$