1(c)

For a particular solution, take $y = \mbox{Re} z$ where $$z'' + 4 z = {\rm e}^{2ix}$$. Using Exponential Shift, $$z = u {\rm e}^{2ix}$$ with $((D+2i)^2 + 4)u = (D^2 + 4iD) u = 1$. Taking $v = Du$, $(D+4i) v = 1$, which has a constant solution $v = 1/(4i) = -i/4$. Thus $u = -ix/4$, $z = (-i/4) x {\rm e}^{2ix} = (-i/4) x \cos(2x) +(x/4) \sin(2x)$, and $y = (x/4) \sin(2 x)$. Since $\cos(2x)$ and $\sin(2x)$ are a fundamental set of solutions of the homogeneous equation, we get the general solution

$$y = \left(\frac{x}{4}+ c_1\right) \sin(2 x) + c_2 \cos(2 x)$$