If I write it as:

and apply the operator to both sides to annihilate the right side, and work through it, I get:

.

In general if then:

however, the solution above already contains both roots of the square root so that in the solution, , the principal value.

Also, if you're into Mathematica, here's the code to study a particular IVP numerically:

Code:

eqn = Derivative[2][u][t] + (k/m)*u[t] ==
(a/m)*Exp[I*w*t] //. {k -> 2 - I,
m -> 2, w -> 1, a -> -1}
sol = NDSolve[{eqn, u[0] == 1 + I,
Derivative[1][u][0] == 2 - I}, u,
{t, 0, 5}]
p1 = Plot[{Re[Evaluate[u[t] /. sol]],
Im[Evaluate[u[t] /. sol]]},
{t, 0, 5}, PlotStyle -> {Red, Blue}]

And then the code to solve for the two constants and plot the analytic solution:

Code:

clist = First[{c1, c2} /. FullSimplify[
Solve[{c1 + c2 + a/(k - m*w^2) == u,
c1*I*Sqrt[k/m] - c2*I*Sqrt[k/m] +
(I*w*a)/(k - m*w^2) == v},
{c1, c2}] //. {k -> 2 - I, m -> 2,
w -> 1, a -> -1, u -> 1 + I,
v -> 2 - I}]]
solution = k1*Exp[I*Sqrt[k/m]*t] +
k2*Exp[(-I)*Sqrt[k/m]*t] +
(a/(k - m*w^2))*Exp[I*w*t] /.
{k -> 2 - I, m -> 2, w -> 1, a -> -1,
k1 -> clist[[1]], k2 -> clist[[2]]}
p2 = Plot[{Re[solution], Im[solution]},
{t, 0, 5}, PlotRange ->
{{0, 5}, {-3, 3}}, PlotStyle ->
{Red, Blue}]
Show[{p1, p2}]

and then plots of the real component of the solution in red and complex component in blue (pretty neat I think-- I never worked one like this before ):