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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.

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    $\begingroup$ Hint: $$y'(y''+a/y^2-b)=(\tfrac12(y')^2-a/y-by)'$$ $\endgroup$
    – Did
    Commented Sep 21, 2015 at 12:44

4 Answers 4

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$\textbf{hint}$

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

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 $$ \dfrac{dz}{dy} =b-\frac{a}{y^2} $$

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  • $\begingroup$ 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
    Commented Sep 21, 2015 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
    Commented Sep 21, 2015 at 15:54
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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.

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

$$\int\pm\sqrt{\frac{y}{y^2+Cy+\frac{a}{b}}}dy=\sqrt{2b}t+D.$$

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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You can write $y''+\frac{a}{y^2}=b$ as $y''+\frac{a}{y^2}-b=0$, and you can multiply by $2y'$, hence $2y'y''+\frac{2ay'}{y^2}-2by'=0$. Notice that $2y'y''=(y'^2)'$ and $\frac{y'}{y^2}=(-\frac{1}{y})'$. Hence $(y'^2-\frac{2a}{y}-2by)'=0$. Thanks to the fundamental theorem of calculus, we have $y'(t)^2-\frac{2a}{y(t)}-2by(t)=y'(0)^2-\frac{2a}{y(0)}-2by(0)$, or simply, let $c=y'(0)^2-\frac{2a}{y(0)}-2by(0)$, so $y'^2=\frac{2a}{y}+2by+c$ or $y'=\pm\sqrt{\frac{2a}{y}+2by+c}=\pm\sqrt{\frac{2by^2+cy+2a}{y}}$. So $$\int_{y(0)}^{y(t)}\sqrt\frac{y}{2by^2+cy+a}\,\mathrm{d}y=\pm{t}.$$ The difficult part comes with expressing the integral in terms of some well-known function. This would be much easier if $b=0$, but in the general case, it is more difficult. Anywho, suppose that $$F(s,t,u;x)=\int_0^x\sqrt{\frac{v}{sv^2+tv+u}}\,\mathrm{d}v.$$ Then $$F(2b,c,a;y(t))=F(2b,c,a;y(0))\pm{t}.$$ This is the implicit solution, but there is nothing you can do to go further.

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