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I have the ODE

$\left(\frac{y'}{x^2 \rho(x)} \right)' + \frac{4 \pi G \rho(x)}{x^2 P(x)}\, y - \left( \frac{F(x)}{x^2 \rho(x)} \right)' = 0$

with the boundary conditions $y(x=0)=0, y(x=1)=0$. The independent variable is $x$, the variable I would like to solve for is $y(x)$, the functions $\rho$, $P$, and $F$ are all known as well as the constants $G$ and $4\pi$ (duh).

I would like to solve this equation numerically using scipy's sp.integrate.bvp_solve. For this I need to turn it into a first order system.

This is my attempt:

\begin{align} y_1' &= y_2 \\ y_2' &= \left(\frac{F}{x^2 \rho} \right)' - \frac{4 \pi G \rho}{x^2 P}\, y \end{align}

but I think that it is not correct because the first term in the original equation does not seem properly represented. Can someone please help?

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  • $\begingroup$ The LHside of your second ODE should be $\left( \frac{y_2}{x^2 \rho(x)}\right)^\prime$ instead of $y_2^\prime$; and of course you should have $y_1$ instead of $y$ in the RHside. $\endgroup$
    – Pacciu
    Aug 7 '20 at 18:11
  • $\begingroup$ @Pacciu thanks - that makes a lot more sense - however I am still stuck as at least the examples in scipy all have "pure" variables on the LHS. Can I maybe integrate the second ODE to get something "pure"? Or do you have advice on how to proceed from there? $\endgroup$ Aug 7 '20 at 18:17
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The original ODE can be written also as

$\left(\frac{y'-F(x)}{x^2 \rho(x)} \right)' + \frac{4 \pi G \rho(x)}{x^2 P(x)}\, y = 0$

and making

$$ \cases{y_1 = y\\ y_2=\frac{y_1'-F(x)}{x^2 \rho(x)} } $$

can be represented as

$$ \cases{ y'_1 = x^2\rho(x) y_2+F(x)\\ y'_2 = -\frac{4\pi G \rho(x)}{x^2 P(x)}y_1 }. $$

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  • $\begingroup$ It works, it works, oh my gosh it works! You have made me a very happy person today. Thank you! $\endgroup$ Aug 8 '20 at 10:04

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