Differential equation with twosided decay exponential

[center o bass]

Homework Statement

Solve
$$y''(t) - k^2 y(t) = e^{-a|t|}$$ where a and k are both positive and real.

Homework Equations

The solution was obtained trough a Fourier transform.

The Attempt at a Solution

I got the solution

$$y(t) = \frac{ke^{-at} - ae^{-kt}}{k(a^2 - k^2)}$$

but when i plug it back into the differential equation i just get
$$e^{-at}$$
how could I get the absolute value back in there?

Might there be anything wrong with my solution procedure?

 

[Gib Z]

Start your solution again but branch it into two cases straight away, one for $t\geq 0$ and another for $t < 0$. Solve them separately, then only at the end try to unify them into 1 solution.

 

[center o bass]

Hmm.. Where might this branching take place? I considered to do a branch when i took the Fourier transform of $$e^{-a|t|}$$, but here i found that

$$\int_{-\infty}^{\infty} e^{-a |t|}e^{-i \omega t} dt = \int_{-\infty}^{\infty} e^{-a |t|} ( \cos(\omega t) + i \sin (\omega t)) dt$$

so the sine term drops out and since the rest of the integrand is even we're left with

$$2 \int_{0}^{\infty} e^{-a t } \cos(\omega t) dt$$

which I evaluated, isolated for $$Y(\omega)$$ and took the inverse transform.

 

[Gib Z]

I can't see anything wrong with that so frustratingly I can't see a mathematical reason as to why the solution only ending up solving correctly for half the domain. In any case though, to fix you solution you can just replace all t's with |t|'s .

 

[center o bass]

That's very weird indeed. I know that I got the right transform.. and I don't see how that absolute value can appear in the inverse transform which is just
$$y(t) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} Y(\omega) e^{i \omega} d\omega$$
Well thanks for helping me out anyway! You've been really helpfull :)