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I have a second order diff eqn that is in the form of

\begin{equation} y^{\prime\prime} + \Big(\frac{3}{x} - \frac{3x^{-4}}{2[x^{-3} + B]}\Big)y^{\prime} - \frac{3}{2}\frac{1}{x^5[x^{-3} + B]}y = \mathcal{F}\,. \end{equation}

where $B$ and $\mathcal{F}$ are known constant. I have two initial conditions given as:

$$y(x=x_{\rm ini}) = \mathcal{C}x_{\rm ini}, ~~ y'(x_{\rm ini})=\mathcal{C}$$

where $\mathcal{C}$ is a constant number.

Note that this initial condition is for $y(x) \ll 1$ where $B\sim 0$ and $\mathcal{F} = 0$.

Is there a general (known) numeric method to find the $\mathcal{C}$, which satisfies the above equation? I have an algorithm but it's like brute force. It iterates every number and tries to find the root. Note that function is always positive for given $x>0$.

Note that: $0 \leq x \leq 1$ and $y(x) \geq 0$.

seVenVo1d
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  • What is none linear about $y''(x)=0$, why is it complicated and why isn't the solution $y=Ax+b$? – Paul Mar 13 '24 at 14:37
  • @Paul Hey paul, thanks for the comment. I have added the equation. It's about a physical process so you can assume $b=0$. – seVenVo1d Mar 13 '24 at 14:42
  • Are your initial conditions meant to be at $x=0$? – Sal Mar 13 '24 at 14:51
  • Plugging $y=Cx$ and its derivative changes the differential equation into an algebraic equation in $x$. You don't need a differential equation solver, you need one that can solve equations. There are many options there. Newton's Method is often a good approach. – TurlocTheRed Mar 13 '24 at 15:03
  • This is a linear differential equation. Why do you say it's non-linear? – Robert Israel Mar 13 '24 at 17:12
  • If they are initial conditions, they should be at a particular value of $x$, not for all $x$. The function $y(x) = Cx$ does not satisfy the differential equation unless $F = 0$ and $BC = 0$. – Robert Israel Mar 13 '24 at 17:32
  • @TurlocTheRed That is a great idea but these initial conditions only work for $y(x) \ll 1$. In that case does your approach still works? – seVenVo1d Mar 14 '24 at 08:32
  • @RobertIsrael That is true yes. They start from some $x_{\rm ini}$. Yes you are also right that this solution only works when $y(x) \ll 1$, where $B \sim 0$. So thats is why I have to use the brute force approach but maybe theres another way ? – seVenVo1d Mar 14 '24 at 08:35
  • You should label your task more traditionally. You want to implement a single-shooting method to solve a BVP with a regular singular point at $x=0$. What you get in a first step is a function that maps $C$ to $x(1)$. Seen abstractly, this is just another scalar function so that you can apply any scalar root-finding method. As this equation is linear, theoretically one should find the solution in the first secant step, practically a few more steps might be necessary to reach numerical stabilization of the result. – Lutz Lehmann Mar 14 '24 at 08:52

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