Differential equation with initial conditions problem: how do we solve $yy'' - 2(y')^2 - y^2 = 0$?

Question

I have problem with solving following equation with initial conditions: $$y*y''-2(y')^2-y^2=0 $$ $$y(0)=1; y'(0)=0 $$ The problem is that i've tried substitution $ u(y)=y' $ and I end up with $$u'*u-2u^2/y =y $$ which is basically bernouli equation. I've done z sub so that: $z(y)=u^2$ and got equation $$z'-4z/y=2y $$ I solved that and got (with my initial condition ) $z=y^4-y^2 $ and that implies $ u^2=y^4-y^2 $ I have no idea what should be next step Any help appreciated !

Answer 1

HINT

To begin with, notice that \begin{align*} yy'' - 2(y')^{2} - y^{2} = 0 \Longleftrightarrow \frac{y''}{y} - \frac{(y')^{2}}{y^{2}} - \left(\frac{y'}{y}\right)^{2} - 1 = 0 \end{align*} Moreover, we do also have that \begin{align*} \frac{y''}{y} - \frac{(y')^{2}}{y^{2}} = \frac{y''y - (y')^{2}}{y^{2}} = \left(\frac{y'}{y}\right)' \end{align*}

Hence, if we make the substitution $y' = uy$, we obtain the following ODE \begin{align*} u' - u^{2} - 1 = 0 \Longleftrightarrow u' = u^{2} + 1 \Longleftrightarrow \frac{u'}{u^{2}+1} = 1 \end{align*}

Can you take it from here?