Answer
See below
Work Step by Step
We are given that $y''+y'-6y=30u_1(t)e^{-(t-1)}$
and $y(0)=3\\
y'(0)=-4$
We take the Laplace transform of both sides of the differential
equation to obtain:
$[s^2Y(s)-sY(0)-Y'(0)]+[sY(s)-y(0)]-6Y(s)=\frac{30e^{-s}}{s+1}$
Substituting in the given initial values and rearranging terms yields
$Y(s)(s^2+s-6)-3s+1=\frac{30e^{-s}}{s+1}\\
Y(s)(s^2+s-6)=\frac{30e^{-s}}{s+1}+3s-1$
Thus, we have solved for the Laplace transform of $y(t)$. To find y itself, we must take the inverse Laplace transform. We first decompose the right-hand side into partial fractions to obtain:
$Y(s)=\frac{30e^{-s}}{(s+1)(s+3)(s-2)}+\frac{3s}{(s+3)(s-2)}-\frac{1}{(s+3)(s-2)}\\
=e^{-s}(-\frac{5}{s+1}+\frac{3}{s+3}+\frac{2}{s-2})+(\frac{9}{s(s+3)}+\frac{6}{s(s-2)})+\frac{1}{s(s+3)}-\frac{1}{s(s-2)}$
Now,$f(t)= L^{-1}[e^{-s}(-\frac{5}{s+1}+\frac{3}{s+3}+\frac{2}{s-2})+(\frac{9}{s(s+3)}+\frac{6}{s(s-2)})+\frac{1}{s(s+3)}-\frac{1}{s(s-2)}]\\
=e^{-s}(-\frac{5}{s+1}+\frac{3}{s+3}+\frac{2}{s-2})+(\frac{2}{s+3}+\frac{1}{s-2})]\\
=[-se^{-(t-1)}+3e^{-3(t-1)}+2e^{2(t-1)}]u_1(t)+2e^{-3t}+e^{2t}$
