Taking Laplace on both sides we get

$L(y") + 2L(y') + 5L(y) = Le^{-t}sint$

From the formula

$L(y") = s^2\bar{y} - sy(0) - y'(0) = s^2\bar{y} - 1$

$\therefore L(y') = s\bar{y} - y(0) = s\bar{y}$

$L(y) = \bar{y}$

$L(sint) = \frac{1}{s^2 + 1}$

$L(e^{-t}sint) = \frac{1}{(s + 1)^2 + 1}$

(By first shifting)

$\therefore s^2\bar{y} - 1 + 2s\bar{y} + 5\bar{y} = \frac{1}{(s + 1)^2 + 1}$

$\therefore \bar{y}(s^2 + 2s + 5) = \bigg(\frac{1}{(s + 1)^2 + 1} + 1\bigg)$

$\therefore \bar{y} = \frac{(s + 1)^2 + 2}{[(s + 1)^2 + 1][s^2 + 2s + 5]}$

$\therefore$ Taking Inverse Laplace on both side we get

$L^{-1}\bar{y} = e^{-t}L^{-1}\bigg[\frac{s^2 + 2}{(s^2 + 1)(s^2 + 4)}\bigg]$

Now by partial fraction

$\frac{s^2 + 2}{(s^2 + 1)(s^2 + 4)} = \frac{As + b}{s^2 + 1} + \frac{Cs + D}{s^2 + 4}$

$s^2 + 2 = (As + B)(s^2 + 4) + (Cs + D)(s^2 + 1)$.......(1)

Let $s^2 = -4 \therefore s = 2i$

$-4 + 2 = [C(2i) + D](-3)$

$2Ci + D = \frac{2}{3}$

Equating Real and Imaginary we get

$C = 0$ & $D = \frac{2}{3}$

Now let $s^2 = -1 \therefore s = i$

$\therefore -1 + 2 = (Ai + B)(3)$

$\therefore \frac{-1}{3} = -Ai + B$

Equating Real and Imaginary we get

$A = 0$ & $B = \frac{1}{3}$

$\therefore y = e^{-t}L^{-1}\bigg[\frac{1}{3} * \frac{1}{s^2 + 1} + \frac{2}{3(s^2 + 4)}\bigg]$

$\therefore y = e^{-t}\bigg[\frac{1}{3}L^{-1}\bigg(\frac{1}{s^2 + 1}\bigg) + \frac{1}{3}L^{-1}\bigg(\frac{2}{s^2 + 4}\bigg)\bigg]$

$\therefore y = \frac{e^{-t}}{3}[sint + sin2t]$