# Is there any quick method to solve this second-order semilinear ODE

I want to solve this second-order semilinear ODE $$x''(s)-3x(s) x'(s)+x^3(s)=0, ~s\in\mathbb{R}.$$ I have tried this substitution $$p(x)=x'(s),$$ which implies $$x''(s)=p(x)\frac{d p(x)}{dx},$$ with which I reduced the original ODE to a first-order nonlinear non-homogeneous ODE. But this method is too clumsy. Is there any quick solver?

Maple has given the solution as $$x \left( s \right) ={\frac {-2\,{ C_1}\,s-2\,{C_2}}{{ C_1 }\,{s}^{2}+2\,{ C_2}\,s+2}},$$ and I am very curious of how do Maple solve?

• wolfram alpha has a different answer. wolframalpha.com/input/?i=y''-3yy'%2By%5E3+%3D+0 – Guy Mar 16 '14 at 8:35
• maybe that is what you meant and don't know $\TeX$. should I edit? – Guy Mar 16 '14 at 8:35
• Yeah it is okay now. – Guy Mar 16 '14 at 8:41
• Sorry, I have just convert Maple's result into LaTeX directly, without considering the convention. I have modified Maple's solution. – nuage Mar 16 '14 at 8:41
• yeah it is okay now. – Guy Mar 16 '14 at 8:41 • I get $t=1-\frac{c_1}{X}\pm\sqrt{\dots}$ – ccorn Mar 23 '14 at 2:10
• That sign error in the quadratic formula for $t$ gets carried through to $x$. Just flip the sign of $x$ and the solution is OK. – ccorn Mar 23 '14 at 3:22