Reduce to first order and solve $yy'' = 3y'^2$
Dividing both sides by $y$ and re-arranging $y'' – 3y' = 0$. This is clearly an homogenous equation, a solution might be $e^{3x}$to test this:
$y' = 3e^{3x}, y'' = 9e^{3x} \implies 9e^x-9e^x = 0$
Substituting $y = ue^{3x}, y' = u'e^{3x}+3ue^{3x}, y'' = u''e^{3x}+3u'e^{3x}+3u'e^{3x}+9ue^{3x}$
Plugging these in $$(u''e^{3x}+3u'e^{3x}+3u'e^{3x}+9ue^{3x})-3\cdot (u'e^{3x}+3ue^{3x})=0$$
$$\implies u''e^{3x}+3u'e^{3x}=0$$
Substituting $v = u'$
$$\implies v'e^{3x}+3ve^{3x}=0$$
$$=\int\frac{dv}{v} = 3\int dx$$
However the solution to this exercise is $(c_1x+c_2)^{-\frac{1}{2}}$ , how did they get this?
Best Answer
The algebra is wrong, it would be $y'' - 3{(y')^2 \over y} = 0$.
It's better to write it as ${y'' \over y'} = 3{y' \over y}$ and view ${y'' \over y'}$ as the derivative of $\ln |y'|$, and similarly ${y' \over y}$ as the derivative of $\ln |y|$. Then you can reduce the order by integrating both sides.