I'm trying to solve the following differential equation:

I tried the following: $$\frac{d^2x}{ax+bx^3}=dt^2$$

But I'm not sure how to continue. Can I use $d^2x$ the same as $dx^2$ and use partial fraction decomposition to work out the left side of the equation? Like this: $$ \begin{split} \iint{\frac{1}{ax+bx^3}d^2x}&=\iint{dt^2}\\ \frac{1}{b}\iint{\frac{A}{x}+\frac{B}{x+\sqrt{\frac{a}{b}}}+\frac{C}{x-\sqrt{\frac{a}{b}}}d^2x} &= \frac{t^2}{2}+c_1t+c_2 \end{split} $$
But am I using $d^2x$ correctly here? If not, how can I solve this equation?

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    $\begingroup$ No you can not you missunderstand dx^2 which is a aymbol for the second derivative and not dx*dx $\endgroup$ – trula Sep 24 '20 at 17:56
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    $\begingroup$ Wolfram alpha obtain this solution. $\endgroup$ – callculus Sep 24 '20 at 17:58
  • $\begingroup$ @callculus nasty, with elliptic functions. Somehow I doubt this is nicely solvable, $x^3$ makes it wildly non-linear $\endgroup$ – gt6989b Sep 24 '20 at 18:01
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    $\begingroup$ Duffing's equation $\endgroup$ – Narasimham Sep 24 '20 at 18:13
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    $\begingroup$ I think it will be good to know the context in which you have this equation. May be it can be solved easier knowing the context. Is it a given equation of acceleration? $\endgroup$ – Math Lover Sep 24 '20 at 18:21

$$\frac{d^2x}{dt^2}=ax+bx^3$$ You can reduce the order. Multiply by $2x'$ both sides: $$2x'x''=2axx'+2bx^3x'$$ After integration, it gives: $$x'^2=ax^2+\dfrac 12bx^4+C_1$$ It's separable ( maybe not easy to integrate again).

Note that: $$dx^2=2xdx \ne d^2x$$ I am not sure that $d^2x$ alone has any meaning. So no you can't equate $dx^2$ and $d^2x$.You can write this: $$\frac{d^2x}{dt^2}=\frac{dx'}{dt}$$


This is the undamped, unforced Duffing equation. Wikipedia gives an explanation of the method of solving it

You can multiply the whole thing by $dx/dt$ to get an expression that can be integrated to a first order equation and then solved.


WLOG, $b=\pm2$, as you can achieve by rescaling $x$.

The equation is autonomous, so with $p:=\dot x$,

$$pp'=ax\pm 2x^3$$ and by integration

$$p^2=ax^2\pm x^4+c,$$

which is separable.


$$\frac{dx}{\sqrt{\pm x^4+ax^2+c}}=\pm dt.$$

Analytical integration is possible, but difficult.


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