Determine the solution of $x^4y''=(y')^2+3x^6$ Determine the solution of $x^4y''=(y')^2+3x^6$
My idea: 
suppose $(y')^2=t \to 2y'y''=\frac{dt}{dx}\to y''=\frac{1}{2\sqrt{t}}\frac{dt}{dx}$
then equation reduced into $x^4\frac{1}{2\sqrt{t}}\frac{dt}{dx}=t+3x^6$
how to solve from here ig ot no idea

First put $u = \dfrac{y'}{x^2}\implies y'' = 2xu+x^2u' = u^2+3x^2\implies x^2u'=u^2-2xu+x^2+2x^2 = (u-x)^2+2x^2\implies u' = \left(\dfrac{u-x}{x}\right)^2+2 $ . Next put $v = \dfrac{u-x}{x} \implies u = vx+x \implies u' = xv'+v + 1\implies xv'+v+1 = v^2+2\implies xv' = v^2-v+1\implies \dfrac{dv}{v^2-v+1} = \dfrac{dx}{x}$ . Can you take it from here to solve for $v$, then $u$, and finally then $y$ ? I hope it helps...
