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.

  • 1
    $\begingroup$ Let $v(x) = \int_0^x u(t) \, \mathrm dt$. Then $v'(x) = u(x)$, and your equation becomes $$ v'''(x) +xv(x) = f(x). $$ $\endgroup$
    – MSDG
    Commented Nov 23, 2018 at 17:57

1 Answer 1


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 $U'(1)=u(1)=-1$.


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