Differential equation: chain rule

Question

I've the following system: $$ r^2 \psi' = \frac{|L|}{m_p} $$ $$ (r')^2 + r^2(\psi')^2 - 2 \frac{G_m}{r} = 2 \frac{E}{m_p}$$

How can I find a differential equation for $r(\psi)$ using the chain rule $r'=\frac{\partial r}{\partial \psi} \psi'$ ?

Answer 1

$$ r^2 \psi' = \frac{|L|}{m_p} \quad\to\quad \psi' =\frac{d\psi}{dt}=\frac{|L|}{m_pr^2}$$ $$ (r')^2 + r^2(\psi')^2 - 2 \frac{G_m}{r} = 2 \frac{E}{m_p}$$ $$ (r')^2 + \frac{L^2}{m_p^2r^2} - 2 \frac{G_m}{r} = 2 \frac{E}{m_p}$$ $$r'=\frac{dr}{dt}=\pm\sqrt{-\frac{L^2}{m_p^2r^2} + 2 \frac{G_m}{r} + 2 \frac{E}{m_p}}$$ $$\begin{cases} dr=\pm\sqrt{-\frac{L^2}{m_p^2r^2} + 2 \frac{G_m}{r} + 2 \frac{E}{m_p}}\:dt \\ d\psi=\frac{|L|}{m_pr^2}dt \end{cases}$$ A differential equation for $r(ψ)$ is : $$\frac{dr}{d\psi}=\frac{\pm\sqrt{-\frac{L^2}{m_p^2r^2} + 2 \frac{G_m}{r} + 2 \frac{E}{m_p}}}{\frac{|L|}{m_pr^2}}$$ The solution on the form of the inverse function defined by an integral is : $$\psi(r)=\pm\frac{|L|}{m_p}\int \frac{ dr}{r^2\sqrt{-\frac{L^2}{m_p^2r^2} + 2 \frac{G_m}{r} + 2 \frac{E}{m_p}}}+\text{constant}.$$ You can solve it on closed form with change of variable $r=\frac{1}{\xi}$.