Y" + 6y' + 13y = δ(t-2) + δ(t-5)

solve using Dirac Delta (transforms)

Note: solve without using partial fractions

Applying L to both sides (with y(0) = A and y'(0) = B) yields
(s^2 F(s) - As - B) + 6 (s F(s) - A) + 13 F(s) = e^(-2s) + e^(-5s).

Solve for F(s):
(s^2 + 6s + 13) F(s) - As - B - 6A = e^(-2s) + e^(-5s)
==> F(s) = (As + B)/(s^2 + 6s + 13) + e^(-2s)/(s^2 + 6s + 13) + e^(-5s)/(s^2 + 6s + 13)
==> F(s) = (As + B)/((s+3)^2 + 4) + e^(-2s)/((s+3)^2 + 4) + e^(-5s)/((s+3)^2 + 4)
==> F(s) = A(s+3)/((s+3)^2 + 4) + (B - 3A)/((s+3)^2 + 4)
+ e^6 * e^(-2(s+3))/((s+3)^2 + 4) + e^15 * e^(-5(s+3))/((s+3)^2 + 4)

Invert term by term:
y(t) = Ae^(-3t) cos(2t) + (1/2)(B - 3A) e^(-3t) sin(2t)
+ (1/2) e^6 e^(-3(t - 3)) sin(2(t - 3)) u(t - 3) + (1/2) e^15 e^(-3(t - 5)) sin(2(t - 5)) u(t - 5).

One can relabel A and (1/2)(B - 3A) as C₁ and C₂:
y(t) = C₁ e^(-3t) cos(2t) + C₂ e^(-3t) sin(2t)
+ (1/2) e^6 e^(-3(t - 3)) sin(2(t - 3)) u(t - 3) + (1/2) e^15 e^(-3(t - 5)) sin(2(t - 5)) u(t - 5).
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I hope this helps!

Both of the constants ties to A and B zero out; the rest remains.

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