I need to solve the Cauchy problem for a differential equation: $$yy''x^2+3yy'x+y^2=(y'x+y)^2(1-y'x^2-yx),\ y(1)=2,\ y'(1)=0$$
I solved this equation: $$Let\ x=e^t,\ y=ze^{-t},\ y'=e^{-2t}(z'-z),\ y''=e^{-3t}(z''-3z'+2z)$$ $$=>zz''-(z')^2+(z')^3=0$$ $$Let\ z'=p, z''=pp'$$ $$=>zp'=p-p^2$$ $$\int \frac{dp}{p-p^2}=\int \frac{dz}{z}$$ $$\frac{p}{1-p}=zC$$ $$z'=p=\frac{zC}{zC+1}=\frac{z}{C+z}$$
What should I do next? How can I solve the Cauchy problem?
So you set $y(e^t)=z(t)e^{-t}$ or $z(\ln x)=xy(x)$. Then indeed \begin{align} z'(t)&=e^{2t}y'(e^t)+e^ty(e^t)=x(xy(x))'\\ z''(t)&=e^{3t}y''(e^t)+3e^{2t}y'(e^t)+e^ty(e^t)\\ &=x(x^2y''(x)+3xy'(x)+y(x))\\ &=\frac{x}{y}\left((xy(x))'\right)^2(1-x(xy(x))')\\ &=\frac{z'(t)^2(1-z'(t))}{z(t)} \end{align} This can then be separated into integrable terms, as you did indirectly, \begin{align} \frac{z''}{z'(1-z')}=\frac{z'}{z}\\ \frac{z'}{1-z'}=Cz\implies z'=\frac{Cz}{1+Cz} \end{align} Now apply again separation-of-variables to get $$\begin{align} \frac{z'}{Cz}=1-z'\implies \frac{\ln|z|}{C}+z=t+D. \end{align}$$ Apply the initial conditions to determine the constants, $z(0)=y(1)=2$, $z'(0)=y'(1)+y(1)=2$ which gives $C=-1$ and then $D=2-\ln(2)$. You can semi-formally solve this by applying the Lambert-W function, $$ ze^{-z}=e^{-t-2+\ln 2}\iff -ze^{-z}=-2e^{-t-2}\implies z=-W_{-1}(-2e^{-t-2}) $$