How to solve an ODE containing the integral of the unknown Is there a way to get an analytical solution for $\psi(y)$ for this second order ODE that contains the integral of $\psi$ over the domain:
$$\frac{d^2\psi}{dy^2}=a+b\int^{L}_{0}{\psi dy}$$
where $a$ and $b$ are constants, and with boundary conditions:
$\psi(0)=\psi_1$
$\dfrac{d\psi}{dy}(L)=0$
If not, how would you do it numerically? Do Matlab or Maple handle this?
 A: Well, I think that
$$
\frac{d^3 \psi}{dy^3}=0
$$
since it is a definite integral.
Now you only have to write $\psi(y)$ like a degree 2 polynomial and compute coefficients.
Edit:
Elaborating a bit more, $\psi(y)=Ay^2+By+C$. But $C=\psi_1$, because of the initial condition. And
$$
\frac{d\psi}{dy}(L)=2AL+B=0,
$$
and then $B=-2AL$.
Therefore $\psi(y)=Ay^2-2ALy+\psi_1$.
Substituting in the original equation,
$$
\frac{d^2\psi}{dy^2}=2A=a+b\int_0^L Ay^2-2ALy+\psi_1 dy
$$
so
$$
2A=a+b(AL^3/3-2AL^3/2+\psi_1 L)
$$
and you solve for $A$.
A: Let us solve $$\frac{d^2y}{dx}=a+b\int_{0}^{L} y(x) dx, dy/dx, y(0)=y_0, \frac{dy}{dx}(x=L)=0.....(1)$$
Let $$\int_{0}^{L} y dx=c,......(2)$$ then
$$\frac{d^2y}{dx^2}=a+bc \implies \frac{dy}{dx}=(a+bc)x+u \implies y(x)=(a+bc)x^2/2+ux+v......(3)$$
We get $v=y_0$ and $(a+bc)L=-u \implies c=\frac{\frac{-u}{L}-a}{b}.$
Next from (2) and (3),
$$c=\int_{0}^{L} [(a+bc)x^2/2+ux+v] dx,$$
you solve for $c$ and get it, then get $u$ and $v$ is already $y_0$.. Here, $y_0,a,b, L$ are fixed,
