Using the transforms listed in your attachment and taking the Laplace transform of both sides of the ODE gives
[tex]L\{y''+y'-2y\} = L\{e^t \sin(t)\}[/tex]
[tex](s^2 Y(s) - s y(0) - y'(0)) + (s Y(s) - y(0)) - 2 Y(s) = L\{e^t \sin(t)\}[/tex]
[tex](s^2 + s - 2) Y(s) = L\{e^t \sin(t)\}[/tex]
where Y(s) denote the Laplace transform of y(t).
For the remaining transform, we use the frequency shifting property,
[tex]L\left\{e^{at} f(t)\right\} = F(s - a)[/tex]
Then
[tex]L\{\sin(t)\} = F(s) = \dfrac1{s^2+1} \\ \implies L\left\{e^t\sin(t)\right\} = F(s - a) = \dfrac1{(s - 1)^2+1} = \dfrac1{s^2-2s+2}[/tex]
Solving for Y(s) yields
[tex]Y(s) = \dfrac1{(s^2-2s+2)(s^2+s-2)} = \dfrac1{(s - 1)(s + 2) (s^2 - 2s + 2)}[/tex]
Break up the right side into partial fractions.
[tex]\dfrac1{(s - 1)(s + 2) (s^2 - 2s + 2)} = \dfrac a{s-1} + \dfrac b{s+2} + \dfrac{cs + d}{s^2 - 2s + 2}[/tex]
[tex]\dfrac1{(s - 1)(s + 2) (s^2 - 2s + 2)} =\\ \dfrac{a(s+2)(s^2-2s+2) + b(s-1)(s^2-2s+2)+(cs+d)(s-1)(s+2))}{(s - 1)(s + 2) (s^2 - 2s + 2)}[/tex]
[tex]1 = a(s+2)(s^2-2s+2) + b(s-1)(s^2-2s+2)+(cs+d)(s-1)(s+2)[/tex]
[tex]1 = 4 a - 2 b + 2 d + (-2 a + 4 b + 2 c - 2 d) s + (-3 b - 2 c + d) s^2 + (a + b + c) s^3[/tex]
Solve for the unknown coefficients:
[tex]\begin{cases}4a-2b+2d=1 \\ -2a+4b+2c-2d = 0 \\ -3b-2c+d = 0 \\ a+b+c=0\end{cases} \implies a=\dfrac13,b=-\dfrac1{30},c=-\dfrac3{10},d=\dfrac15[/tex]
So we have
[tex]Y(s) = \dfrac13\times\dfrac1{s-1} -\dfrac1{30}\times \dfrac1{s+2} - \dfrac3{10} \times \dfrac{s}{s^2-2s+2} + \dfrac15 \times\dfrac1{s^2-2s+2}[/tex]
which we can rewrite as
[tex]Y(s) = \dfrac13\times\dfrac1{s-1} -\dfrac1{30}\times \dfrac1{s+2} - \dfrac3{10} \times \dfrac{s - 1}{(s-1)^2+1} - \dfrac1{10} \times\dfrac1{(s-1)^2+1}[/tex]
Now use the frequency-shifting property to compute the inverse transforms.
[tex]L^{-1}\left\{\dfrac1{s-1}\right\} = e^t L^{-1}\left\{\dfrac1s\right\} = e^t[/tex]
[tex]L^{-1}\left\{\dfrac1{s+2}\right\} = e^{-2t} L^{-1}\left\{\dfrac1s\right\} = e^{-2t}[/tex]
[tex]L^{-1}\left\{\dfrac{s-1}{(s-1)^2+1}\right\} = e^t L^{-1}\left\{\dfrac s{s^2+1}\right\} = e^t\cos(t)[/tex]
[tex]L^{-1}\left\{\dfrac1{(s-1)^2+1}\right\} = e^t L^{-1}\left\{\dfrac1{s^2+1}\right\} = e^t\sin(t)[/tex]
Putting everything together, we end up with
[tex]y(t) = \dfrac{e^t}3 - \dfrac{e^{-2t}}{30} - \dfrac{3e^t\cos(t)}{10} - \dfrac{e^t\sin(t)}{10}[/tex]
