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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?

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  • $\begingroup$ wolfram alpha has a different answer. wolframalpha.com/input/?i=y''-3yy'%2By%5E3+%3D+0 $\endgroup$ – Guy Mar 16 '14 at 8:35
  • $\begingroup$ maybe that is what you meant and don't know $\TeX$. should I edit? $\endgroup$ – Guy Mar 16 '14 at 8:35
  • $\begingroup$ Yeah it is okay now. $\endgroup$ – Guy Mar 16 '14 at 8:41
  • $\begingroup$ Sorry, I have just convert Maple's result into LaTeX directly, without considering the convention. I have modified Maple's solution. $\endgroup$ – nuage Mar 16 '14 at 8:41
  • $\begingroup$ yeah it is okay now. $\endgroup$ – Guy Mar 16 '14 at 8:41
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The homogeneous EDO is successively transformed to homogeneous then separable EDO as shown below :

enter image description here

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  • $\begingroup$ I get $t=1-\frac{c_1}{X}\pm\sqrt{\dots}$ $\endgroup$ – ccorn Mar 23 '14 at 2:10
  • $\begingroup$ 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. $\endgroup$ – ccorn Mar 23 '14 at 3:22

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