I want to find the solution of the following differential equation.

$$ \frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}} = −\frac{1}{ax + \mathrm{d}x/\mathrm{d}t} $$

where a is a positive constant and x is a position of a free object that is not bounded or tied to a fixed object (e.g., a wall).

I think that the solution may behave like an oscillator, but it seem to be slightly different from that of the oscillator. However, due to the terms in the denominator, it has the singularity. How to solve this type of equation?

** As I studied differential equations for a while 20 years ago, I do not know how to solve this problem. Since the solution of the above differential equation is an important issue for me, I would appreciate it if you help.


$$\frac{d^2x}{dt^2}=−\frac{1}{a x + \frac{dx}{dt}} \tag 1$$ The usual change of function to reduce the order of this autonomous ODE is

$\frac{dx}{dt}=y(x) \quad;\quad \frac{d^2x}{dt^2}=\frac{dy}{dx}\frac{dx}{dt}=y\frac{dy}{dx}$

$$y\frac{dy}{dx}=−\frac{1}{a x + y}$$

Consider the inverse function $x(y)$ $$\frac{dx}{dy}=−(a x + y)y$$ This is a first order linear ODE with respect to the function $x(y)$. The solution is : $$x(y)=-\frac{y}{a}+\sqrt{\frac{\pi}{2a^3}}e^{-ay^2/2}\text{erfi} \left(\sqrt{\frac{a}{2}}\:y\right)+c_1 e^{-ay^2/2}$$

Unfortunately there is no closed form for the inverse function $y(x)$ in general (for arbitrary values of $a$ and $c_1$).

A fortiori no closed form for $t=\int\frac{dx}{y(x)}$.

Of course, the solutions of equation $(1)$ exist but in general they cannot be expressed with a finite number of standard functions.

As a consequence, in practice solve the ODE $(1)$ with numerical calculus. It should be a waste of time to try to solve it analytically in general.

Nevertheless, it might be possible to solve the problem analytically is some particular cases, depending on the initial conditions of the ODE and/or with particular values of the parameter $a$.

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    $\begingroup$ should be $y^{-1}dx/dy=-(ax+y)$ $\endgroup$ – user254433 Jan 6 '19 at 20:22
  • $\begingroup$ You are right. Thank you for your remark. I corrected my calculus. Finally the conclusion remains the same. $\endgroup$ – JJacquelin Jan 6 '19 at 21:41
  • $\begingroup$ @JJacquelin I am much obliged to you for your help. $\endgroup$ – SOQEH Jan 7 '19 at 0:17
  • $\begingroup$ I don't think $x=-y/a$ is a solution $\endgroup$ – Dylan Jan 7 '19 at 9:16
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    $\begingroup$ Hi Jean ! You probably noticed that, assuming $a>0$ and small values of $y$, we have $x(y)=-\frac{y^3}{3}+\frac{a y^5}{15}+O\left(y^7\right)$ and then we could (may be !) get ... something. Cheers. $\endgroup$ – Claude Leibovici Jan 7 '19 at 11:15

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