# Solution differential equation $\frac{d^2x}{dt^2}=ax+bx^3$

I'm trying to solve the following differential equation:
$$\frac{d^2x}{dt^2}=ax+bx^3$$

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?

• No you can not you missunderstand dx^2 which is a aymbol for the second derivative and not dx*dx – trula Sep 24 '20 at 17:56
• Wolfram alpha obtain this solution. – callculus Sep 24 '20 at 17:58
• @callculus nasty, with elliptic functions. Somehow I doubt this is nicely solvable, $x^3$ makes it wildly non-linear – gt6989b Sep 24 '20 at 18:01
• Duffing's equation – Narasimham Sep 24 '20 at 18:13
• 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? – 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.

Then

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

Analytical integration is possible, but difficult.