Laplace Transform of DE with Discontinuous Forcing Function

[lobsterback]

This is my first time posting on these boards, so forgive me if I am posting incorrectly or in the wrong forum.

I have been able to successfully work through problems with forcing functions of constant value "steps", but am having trouble working with those with linear portions of the discontinuous function.

A problem I am attempting reads:
y'' + 2y' + 5y = G(t); y(0) = -2, y'(0) = 4
and G(t) = (t-2)u₂(t) + (-3t + 18)u₆(t) - (-3t + 18)u₈(t)

I understand that I take the laplace transform of all terms, leaving me with
(s² + 2s + 5) L{y} + 2s = L{(t-2)u₂(t)} + L{(-3t + 18)u₆(t)} - L{(-3t + 18)u₈(t)}

I found L{(t-2)u₂(t)} to be (e^-2s/s²)
and L{(-3t + 18)u₆ to be (-1/3)(e^-6s/s²) by factoring out (-1/3) from the above unit step and translation.

What I cannot figure out is how to take the Laplace Transform of the last term of the forcing function, (-3t + 18)u₈(t), as I cannot conceive a way to transform (-3t + 18) into (t - 8).

Any help is appreciated.

 

[lurflurf]

Don't fool with all that use the convolusion theorem.