I'm having trouble solving a differential equation I found:
$$ u''(x) + x\int_0^xu(t)dt = f(x) $$
where: $ x\in[0,1], \quad u(0) = 1, \quad u(1) = -1 $, and $f(x)$ any given function.
One of my problems is I don't really understand the term of the integral because with that it doesn't look like a linear second order ordinary differential equation or anything I have ever seen before. What I have tried is converting it to a system of differential equations and also tried to solve it as a homogenous equation, but it didn't turned out into anything useful,
If anyone can help me getting started I would be very grateful.
With $U(x)=\int_0^x u(t)\,dt$ you get $U'=u$, $U(0)=0$ and inserted it gives a third order ODE $$ U'''(x)+xU(x)=f(x) $$ with initial conditions $U(0)=0$, $U'(0)=u(0)=1$ and $U'(1)=u(1)=-1$.