IE17

Solve: $u(x) = \int_0^x e^{x-y} u(y) dy, u(0)=0\,$

$u(x) = e^x \int_0^x e^{-y} u(y) dy\,$

$u'(x) = \frac{d}{dx} \left[ e^x \int_0^x e^{-y} u(y) dy \right]\,$

$= e^x \int_0^x e^{-y} u(y) dy + e^x \left[ \frac{d}{dx} \int_0^x e^{-y} u(y) dy \right]\,$

$u'(x) = u(x) + e^x e^{-x} u(x)\,$

$u'(x) = 2u(x)\,$

$u(x) = c_1 e^{2x}\,$

$u(0) = 0 \implies c_1 = 0\,$

Therefore

$u(x) = 0\,$

Note: Infact any homogenous Volterra type equation gives a trivial solution.