Help to solve the integro-differential equation $y'(x)=-k\frac{y^2(x)}{x^2}\int_0^xt^2y(t)\,dt$

Question

I have this differential integral equation from a physics problem $y'(x)=-k\frac{y^2(x)}{x^2}\int_0^xt^2y(t)\,dt$ and i dont know how to solve such equation.

Edit 3 (the third time's a charm):

Comment: I was wrong at the time of calculating $g(r)$ so the differential equation changes a lot.

This problem arrives when i want to calculate a radial density of a sphere that compresses by its own gravity

I start with my equation of state where i assume a isotermal process:

$dV = \frac{dV}{dP}dP$ (1)

$\frac{dV}{dP}=-VB$ (2)

B is the reciprocal of bulk modulus, density and volume are inverse proportional then i get from eq (2):

$\frac{d\delta}{dP}=\delta B$ (3)

where $\delta(r)$ is the density, then i know that:

$dP=\delta(r)g(r)dr$ (4)

$g(r)$ the acceleration inside the planet which is:

$g(r)=-G\frac{M(r)}{r^2}$ (5)

To get $M(r)$ i integrate the density multiplied by the volume of a sphere eq (6).

$M(r) =4\pi\int_0^r t^2\delta(t)dt$ (6)

Back to eq (5) using (6) i get:

$g(r)=-\frac{4\pi G}{r^2}\int_0^r t^2\delta(t)dt$ (7)

from eq (4) and (7) i get:

$\frac{dP}{dr}=-\frac{4\pi G}{r^2}\delta(r)\int_0^r t^2\delta(t)dt$ (8)

Since in the eq (3) i've $\delta(P(r))$ i can use the chain rule to get $\frac{d\delta(r)}{dr}$

$\frac{d\delta}{dr}\frac{dr}{dP}=\delta B$ (9)

using a forbidden method i get from eq (9):

$\frac{d\delta}{dr}=\frac{dP}{dr}\delta B$ (10)

Using eq (10) and (8) i finally get:

$\frac{d\delta}{dr}=-\frac{4\pi BG}{r^2}\delta(r)^2\int_0^r t^2\delta(t)dt$ (11)

which is my initial problem if i set $k=\frac{4}{3}\pi BG$, $y=\delta(t)$ and $x=r$.

$y(x)' = -k \frac{y(x)^2}{x^2}\int_0^xt^2y(t)dt$ (12)

Doing what Cameron Williams said in a comment, i can isolate the integral and differentiate in booth sides:

$-\frac{y'x^2}{ky^2}=\int_0^xt^2y(t)dt$

$(-\frac{y'x^2}{ky^2})'=(\int_0^xt^2y(t)dt)'$

$-\frac{(y''x^2+y'2x)y^2-2yy'^2x^2}{ky^4}=x^2y$

$-y''xy+2y'y+2y'^2x=kxy^4$ (13)

Now i get a non linear ordinary differential equation of order 2, and it looks awful so i will use some numeric method to solve it, but i dont lose hope that it might have a weird close form.

Aside from that equation i've another question and its what happen in the case $x$ or $r =0$ because i should get the maximum density, but if i evaluate everything at 0 i get get from eq (13) and (12)

$y'(0)=0$

$y''(0)=-ky^3(0)$

So i dont have a numerical value of the central density for doing a numerical integration, i am missing something?

Answer 1

Correction: $\left( \frac{y' \cdot x^{2}}{-k \cdot y^{2}} \right)^{\prime} = -\frac{x \cdot \left( x \cdot y \cdot y'' - 2 \cdot x \cdot y'^{2} + 2 \cdot y \cdot y' \right)}{k \cdot y^{3}}$, so it's $-x \cdot y \cdot y'' + 2 \cdot x \cdot y'^{2} - 2 \cdot y \cdot y' = x \cdot k \cdot y^{4}$ (equation $13$).

Easy to find closed form solutions

There would be the trivial solution $y\left( x \right) = 0$.

Your ODE only contains monomials, so I would try a solution of the form $y\left( x \right) = a \cdot x^{b} \wedge x \ne 0$: \begin{align*} y'\left( x \right) &= -k \cdot \frac{y^{2}\left( x \right)}{x^{2}} \cdot \int\limits_{0}^{x} t^{2} \cdot y\left( t \right)\, \operatorname{d}t\\ a \cdot b \cdot x^{b - 1} &= -k \cdot \frac{a^{2} \cdot x^{2 \cdot b}}{x^{2}} \cdot \int\limits_{0}^{x} t^{2} \cdot a \cdot t^{b}\, \operatorname{d}t\\ \end{align*}

We know $\frac{a^{2} \cdot x^{2 \cdot b}}{x^{2}} = a^{2} \cdot x^{2 \cdot b - 2}$ and $\int_{0}^{x} t^{2} \cdot a \cdot t^{b}\, \operatorname{d}t = a \cdot \int_{0}^{x} t^{b + 2}\, \operatorname{d}t = \frac{a}{b + 3} \cdot x^{b + 3}$, so $a \cdot b \cdot x^{b - 1} = \frac{a^{3}}{b + 3} \cdot x^{3 \cdot b + 1}$. Comparing the components: $b - 1 = 3 \cdot b + 1 \implies b = -1$ and $a \cdot b = \frac{a^{3} \cdot k}{b + 3} \implies a \in \left\{ 0,\, \pm\sqrt{\frac{2}{k}} \right\}$ (assuming $b = -1$). So there are the monomial solutions $\fbox{$y\left( x \right) = 0$}$, $\fbox{$y\left( x \right) = \sqrt{\tfrac{2}{k}} \cdot x^{-1}$}$ and $\fbox{$y\left( x \right) = -\sqrt{\tfrac{2}{k}} \cdot x^{-1}$}$. You can check that via substituting this into the ODE.

Series solutions

You could also find a series solution. When you add the RHS from equation $\left( 12 \right)$ to $\left( 12 \right)$ and multiply by $x^{2}$, you get (You don't have to multiply by $x^{2}$ but that will make it easier.): \begin{align*} \text{ODE}\left( x \right) := x^{2} \cdot y'\left( x \right) + k \cdot y^{2}\left( x \right) \cdot \int\limits_{0}^{x} t^{2} \cdot y\left( t \right)\, \operatorname{d}t &= 0\\ \end{align*}

Using Taylor series, we get $\text{ODE}\left( x \right) = \sum\limits_{g = 0}^{\infty}\left[ \frac{\text{ODE}^{\left( g \right)}\left( a \right)}{g!} \cdot \left( x - a \right)^{g} \right]$, so $\frac{\text{ODE}^{\left( g \right)}\left( a \right)}{g!} = 0$ would be a solution to the series. Using the general Leibniz rule and Faà di Bruno's formula you would get a closed form for $\text{ODE}^{\left( g \right)}\left( a \right)$, wich you maybe could solve for each $y^{\left( k \right)}\left( a \right)$ (because the term is still equal to zero) wich you could then recombine to get the Taylor series for $y\left( x \right)$ but that would only be a series solution aka no closed form. This would be the first $6$ equations according to Mathematica $y_{1} = y_{3} = y_{5} = 0$, $\frac{k \cdot y_{0}^{3}}{3} + y_{2} = 0$, $\frac{13 \cdot k \cdot y_{0}^{2} \cdot y_{2}}{30} + \frac{y_{4}}{6} = 0$ and $\frac{462 \cdot k \cdot y_{0} \cdot y_{2}^{2} + 85 \cdot k \cdot y_{0}^{2} \cdot y_{4} + 21 \cdot y_{6}}{2520} = 0$ (where $y_{g} = y^{\left( g \right)}\left( 0 \right)$). $y_{0}$ and $y_{2}$ can be seen as the arbitrary constants and all other $y_{g}$ would therefore be able to be found individually.

Mathematica also finds the series $y\left( x \right) = a - \frac{a^{3} kx^{2}}{6}+ \frac{13a^{5} k^{2} x^{4}}{360} - \frac{25a^{7} k^{3} x^{6}}{3024} - \frac{825389 a^{11} k^{5} x^{10}}{1796256000} + \frac{10543a^{9} k^{4} x^{8}}{5443200} + \dots$ where $y\left( 0 \right) = a$.