# Solving second order ordinary differential equation with variable constants

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.

• Let $v(x) = \int_0^x u(t) \, \mathrm dt$. Then $v'(x) = u(x)$, and your equation becomes $$v'''(x) +xv(x) = f(x).$$ – MSobak Nov 23 '18 at 17:57

## 1 Answer

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