I want to find a solution to a problem I have in physics, the problem is as following:

$$ ma = u\beta - \frac{GMm}{r²} $$

Where $u$, $\beta$, $G$, $M$ and $m$ are all constants. I want to find a function of either $a(t)$, $v(t)$ or $r(t)$.

If you rewrite $a$ to $\frac{\partial^2r}{\partial t²}$, divide by $m$ and move $-\frac{GMm}{r²}$ to the left side, you get this equation:

$$ \frac{\partial^2r}{\partial t²} + \frac{GM}{r²} = \frac{u\beta}{m} $$

which is a second-order nonlinear ordinary differential equation, in the form of:

$$ y'' + \frac{a}{y²} = b $$

Is there a solution to this differential equation?

Edit: I'm not required understand a method for this [it's probably beyond my level anyway], I just want to know if there exists a general solution and if so; what it is.

  • 1
    $\begingroup$ Hint: $$y'(y''+a/y^2-b)=(\tfrac12(y')^2-a/y-by)'$$ $\endgroup$ – Did Sep 21 '15 at 12:44


$y'' = \dfrac{d}{dr}y' = \dfrac{dy}{dr}\dfrac{d}{dy}y' = \frac{1}{2}\frac{d}{dy}y'^2$

thus you equation becomes $$ \frac{1}{2}\frac{d}{dy}y'^2 =b-\frac{a}{y^2} $$ setting $\frac{y'^2}{2} = z$ we get $$ \dfrac{dz}{dy} =b-\frac{a}{y^2} $$

  • $\begingroup$ Isn't this just another second-degree non-linear ODE? I probably should've clarified that this is way beyond my level as well, I'm just looking for a solution out of curiosity. $\endgroup$ – user272755 Sep 21 '15 at 14:03
  • $\begingroup$ I applaud your desire to go off piste with this topic. As for your question, the original ode is a second order nonlinear ode, however this is an example of reduction of order .. So we go from a nonlinea second order to a first order nonhomogenous ode since we have a function $$z' = f(y)$$ where the prime denotes the derivative of the independent variable. $\endgroup$ – Chinny84 Sep 21 '15 at 15:54

You basically have the inhomogeneous differential equation of a radial Kepler orbit. You can just add the particular solution, which satisfies the RHS, to any solution of the homogeneous differential equation.

This particular solution can be found easily by setting $y''=0$, such that

$$ y_p(t) = \sqrt{\frac ab}. $$

However the there is no explicit solution to the homogeneous differential equation, only an implicit out for the time as a function of position.


We have $$y'dy'=(b-ay^{-2})dy,$$ $$y'^2=2(by+\frac{a}{y})+C_1,$$ $$\pm\sqrt{\frac{y}{y^2+Cy+\frac{a}{b}}}dy=\sqrt{2b}dt,\,\,\,C=\frac{C_1}{2b},$$


Integration of LHS can be performed in terms of elliptical functions, but we need to fix the signs of $a$ and $b$, and also give initial conditions $y(0)$ and $y'(0)$ in order to choose the sign of the radical on LHS and calculate $C$ and $D$.


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