3(b)

A solution is $y = \mbox{Re} z$ where $$z'' - 2 z' + 2 z = {\rm e}^{(-1+i) x}$$. Since $(-1+i)^2 - 2 (-1+i) + 2 = 1 - 2 i - 1 + 2 - 2 i + 2 = 4 - 4 i$, this has a solution $$z = (1/(4-4i)) {\rm e}^{(-1+i)x} = ((1+i)/8) {\rm e}^{(-1+i) x}$$. The real part is

$$y = {\rm e}^{-x} \frac{\cos x - \sin x}{8}$$