Y" + 6y' + 13y = δ(t-2) + δ(t-5) with y(0)=y'(0)=0

solve using Dirac Delta (transforms)

Note: we cannot solve using partial fractions!!

What do you mean by "we cannot solve using partial fractions"? You don't need partial fraction decomposition.

Take the Laplace transform of both sides using the initial conditions. Letting Y(s) denote the transform of y(t)

(s² + 6s + 13)Y(s) = e^(-2s) + e^(-5s) ==>

Y(s) = e^(-2s)/(s² + 6s + 13) + e^(-5s)/(s² + 6s + 13)

The denominator is s² + 6s + 13 = (s + 3)² + 4.

To take the inverse transform, use the shifting theorem:

If L{f(t)} = F(s) and a ≥ 0, then L{f(t - a)U(t - a)} = e^(-as) F(s),

where U(t - a) is the Heaviside step function centered at a.

y(t) = ½ e^(-3(t - 2)) sin(2(t - 2)) U(t - 2) + ½ e^(-3(t - 5)) sin(2(t - 5)) U(t - 5).