Originally Posted by

**compliant** The problem is as follows:

Solve for f(x), given

$\displaystyle f(x) = x + \int^{x}_{0}{(x-2t)f(t) dt}$

What I did:

1. Differentiate both sides wrt x, and partially differentiate through the integral.

$\displaystyle f'(x) = 1 + \frac{d}{dx} \int^{x}_{0}{(x-2t)f(t) dt}$

$\displaystyle f'(x) = 1 + \int^{x}_{0}\frac{\partial}{\partial x}{(x-2t)f(t) dt}$

$\displaystyle f'(x) = 1 + (x-2x)f(x) + \int^{x}_{0}{f(t) dt}$

$\displaystyle f'(x) = 1 - xf(x) + \int^{x}_{0}{f(t) dt}$

2. Differentiate both sides wrt x again.

$\displaystyle f''(x) = - xf'(x) - f(x) + \frac{d}{dx}\int^{x}_{0}{f(t) dt}$

$\displaystyle f''(x) = - xf'(x) - f(x) + f(x)$

$\displaystyle f''(x) = - xf'(x)$

This is an ODE, yes, but how would I solve for this? (i.e. is there a less messy way to calculate this other than using series solutions, which I hate with a passion?)