# Solving the ODE $\,\,x^4yy''+x^4(y')^2+3x^3yy'-1=0$

I'm currently trying to solve the following ODE:

$$x^4yy''+x^4(y')^2+3x^3yy'-1=0$$

I've tried the substitution $$\upsilon=\frac{y}{x}$$which didn't simplify the whole lot. Then I tried rewriting it by division by $y^2$ which yields:

$$x^4\frac{y''}{y}+x^4\left(\frac{y'}{y}\right)^2+3x^3\frac{y'}{y}-\frac{1}{y^2}=0.$$

Now if it wasn't for the $1/y^2$-term the substitution $v=y'/y$ might be a very logical choice. Unfortinately the last term makes this substitution cumbersome.

Now I was wondering if there was any logical substitution that I could make to simplify/solve this differential equation. I'm pretty stuck on this one.

Edit. The substitution $\,\upsilon=yy'\,$ is starting to look promising ...

$$x^4yy''+x^4(y')^2+3x^3yy'-1=0,$$ or $$x^4\big(yy''+(y')^2\big)+3x^3yy'-1=0,$$ or $$x^4\big(yy'\big)'+3x^3yy'-1=0,$$ and setting $z=yy'$ we obtain $$x^3z'+3x^2z-\frac 1x=0,$$ or $$(x^3z)'=(\log x)'$$ or $$x^3z-\log x=c,$$ for some constant $c$. Hence $$yy'=z=cx^{-3}+x^{-3}\log x$$ or $$\frac{1}{2}y^2=-\frac{c}{2x^2}-\frac{\log x+1}{2x^2}=-\frac{\log x}{2x^2}+\frac{c'}{x^2}+c''$$ and thus for suitable constants $c_1,c_2$ we have the expression $$y=\pm\left(c_1+\frac{c_2}{x^2}-\frac{\log x}{x^2}\right)^{1/2}\,.$$