We can express the forcing function (the piecewise expression on the right side) in terms of the step function as [tex]e^{-t}(u(t) - u(t-1))[/tex] where
[tex]u(t) = \begin{cases}1&\text{for }t\ge0\\0&\text{otherwise}\end{cases}[/tex]
Let F(s) be the Laplace transform of a function f(t). Now recall the transform pair
[tex]f(t-c) u(t-c) \mapsto e^{-cs} F(s)[/tex]
This means
[tex]e^{-t} u(t) \mapsto \dfrac1{s+1}[/tex]
[tex]e^{-t} u(t-1) = \dfrac1e \times e^{-(t-1)} u(t-1) \mapsto \dfrac{e^{-(s+1)}}{s+1}[/tex]
I assume you're familiar with the transform rule for derivatives of y(t). Now we're ready to take the transform of both sides of the ODE:
[tex]y'' + 3y' + 2y = e^{-t}(u(t) - u(t-1))[/tex]
[tex]\implies \left(s^2 Y(s) - s y(0) - y'(0)\right) + 3 \left(s Y(s) - y(0)\right) + 2 Y(s) = \dfrac{1 - e^{-(s+1)}}{s+1}[/tex]
Plug in the initial values and solve for Y(s) :
[tex]\left(s^2 Y(s) - s + 1\right) + 3 \left(s Y(s) + 1\right) + 2 Y(s) = \dfrac{1 - e^{-(s+1)}}{s+1}[/tex]
[tex](s^2 + 3s + 2) Y(s) - s + 4 = \dfrac{1 - e^{-(s+1)}}{s+1}[/tex]
[tex]Y(s) = \dfrac{1 - e^{-(s+1)} + (s-4)(s+1)}{(s+1)(s^2 + 3s + 2)}[/tex]
[tex]Y(s) = \dfrac{1 - e^{-(s+1)} + (s-4)(s+1)}{(s+1)^2 (s+2)}[/tex]
Consider the partial fraction expansion
[tex]\dfrac1{(s+1)^2(s+2)} = \dfrac a{s+1} + \dfrac b{(s+1)^2} + \dfrac c{s+2}[/tex]
Solve for the coefficients:
[tex]1 = a(s+1)(s+2) + b(s+2) + c(s+1)^2[/tex]
[tex]s = -1 \implies b = 1[/tex]
[tex]s = -2 \implies c = 1[/tex]
[tex]1 = (a+c)s^2 + \cdots \implies a+c = 0 \implies a = -1[/tex]
Hence we can expand Y(s) as
[tex]Y(s) = \dfrac1{(s+1)^2} + \dfrac1{s+2} + \dfrac{e^{-(s+1)}}{s+1} - \dfrac{e^{-(s+1)}}{(s+1)^2} - \dfrac{e \times e^{-(s+2)}}{s+2}[/tex]
The last transform pair we need is
[tex]e^{ct} f(t) \mapsto F(s - c)[/tex]
Now, taking inverse transforms of everything yields
[tex]\dfrac1{(s+1)^2} \mapsto te^{-t}[/tex]
[tex]\dfrac1{s+2} \mapsto e^{-2t}[/tex]
[tex]\dfrac{e^{-(s+1)}}{s+1} \mapsto e^{-t} u(t-1)[/tex]
[tex]\dfrac{e^{-(s+1)}}{(s+1)^2} \mapsto e^{-t} (t-1) u(t-1)[/tex]
[tex]\dfrac{e \times e^{-(s+2)}}{s+2} \mapsto e^{-(2t-1)} u(t-1)[/tex]
and putting everything together gives the same solution as the one provided.
