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Having an issue differentiating the following equation for purposes of reducing into two 1st order ODE's. Where: $y(1)=0$, $y'(1) = 2/3$ and $h = 0.5$

$$3x^2y'' - 5xy'+5y = 0$$

when making $y'=z$.

Trying to find $dy/dx$ and $dz/dx$, then put it into a 4th order Runge-Kutta where:

$$dy/dx=f(x,y,z)$$ $$dz/dx=g(x,y,z)$$

This is my process (roughly): Help with using the Runge-Kutta 4th order method on a system of 2 first order ODE's.

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  • $\begingroup$ Setting aside your immediate question, this is an example of a Cauchy-Euler equation. As such, it will have solutions of the form $y=x^m$ for appropriate $m$. (The substitution $u=\ln x$ is also effective.) $\endgroup$
    – Semiclassical
    Feb 8, 2021 at 16:31
  • $\begingroup$ @Semiclassical Ah that's super helpful, thanks! $\endgroup$
    – user502961
    Feb 8, 2021 at 16:33
  • $\begingroup$ Glad to help. Of course, if you’re trying to solve numerically then the analytical solution isn’t directly relevant. But it does let you test the numerical agreement afterwards. $\endgroup$
    – Semiclassical
    Feb 8, 2021 at 16:36
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    $\begingroup$ Please do not delete your question immediately after receiving an answer. This is rude to the person who took the time to answer your question, and is rude to potential future readers who might find the Q&A helpful. $\endgroup$
    – Xander Henderson ♦
    Feb 10, 2021 at 23:34

1 Answer 1

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If $z = y'$ then $z' = y''$, and you get \begin{cases} 3x^2z' - 5xz+5y = 0\\ \phantom{3x^2z' - 5xz+5}y' = z. \end{cases}

We can rewrite this by solving for $z'$ and $y'$ as \begin{cases} z' = \frac{5xz - 5y}{3x^{2}}\\ y' = z. \end{cases}

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  • $\begingroup$ Hello, I have made an edit which explains this a bit better. Thanks. $\endgroup$
    – user502961
    Feb 8, 2021 at 16:29
  • $\begingroup$ @BC939422 I've added to my answer. $\endgroup$
    – DMcMor
    Feb 8, 2021 at 16:33
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    $\begingroup$ I see that now thank you! I had something similar but was not sure how right it was. $\endgroup$
    – user502961
    Feb 8, 2021 at 16:34

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