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How to solve? \(x"'+x'(x^2-2)=0\) Please, help
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Is that a third derivative? x''' ?
Yes,
Thanks anyway, I think I got it.
What did you get?
Actually, it comes from traveling wave equation, which is u(x,t) = f(x-ct) find the solution for \(u_t+u^2u_x+u_{xxx}=0\)
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This is what I have \(u_t= -cf'\\u_x=f'\\u_{xxx}=f"'\) Combine all, I have \(f"' =cf' -f^2f'\) Integral both sides I have
\(f" = cf -\dfrac{f^3}{3}\)
multiple both sides by f', then take integral both sides again, I have \(\dfrac{f'^2}{2}=\dfrac{cf^2}{2}-\dfrac{f^4}{12}\)
Simplify and factor f^2 out, transfer to the LHS \((\dfrac{f'}{f})^2=c-f^2/6\)
integral both sides again, \(ln f = cf -f^3/18\)
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so, from here, just solve cubic polynomial.
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