Given, $(D^2 + 3D + 2)y = e^{-2t} sint$ where y(0) = 0, y'(0) = 0

Using Laplace Transform method for "Linear Differentiation Equation" with constant coefficients

Let L[u} = y(s)

Hence applying Laplace transform in the above equation we get

L[y"] + 3L[y'] + 2L[y] = L[$e^{-2t} sint$]

Consider L.H.S.

L[y"] = $s^2 y(s) - sy(0) - y'(0)$

L[y'] = sy(s) - y(0)

Now, consider R.H.S.

L[$e^{-2t} sint$]

We shall first find L[sint] = $\frac{1}{s^2 + 1}$

Then L[$e^{-2t} sint$] = $\frac{1}{(s + 1)^2 + 1}$ [by first shifting property]

Now the equation becomes

$s^2 y(s) - sy(0) - y'(0) + 3[sy(s) - y(0)] + 2y(s) = \frac{1}{(s + 1)^2 + 1}$

Given y(0) = 0 & y'(0) = 0, substituting the given values the above equation becomes

$s^2y(s) + 3sy(s) + 2y(s) = \frac{1}{(s + 1)^2 + 1}$

i.e. ($s^2 + 3s + 2)y(s) = \frac{1}{(s + 1)^2 + 1}$

$\therefore y(s) = \frac{1}{s^2 + 3s + 2} * \frac{1}{(s + 2)^2 + 1}$

$\therefore y(s) = \frac{1}{(s + 2) (s + 1)} * \frac{1}{(s^2 + 4s + 5)}$

$\therefore y(s) = \frac{1}{(s + 2) (s + 1)(s^2 + 4s + 5)}$

By using partial fraction method we get

$ \frac{1}{(s + 2) (s + 1)(s^2 + 4s + 5)} = \frac{A}{s + 2} + \frac{B}{s + 1} + \frac{Cs + D}{s^2 + 4s + 5}$

$\therefore 1 = A(s + 1) (s^2 + 4s + 5) + B(s + 2)(s^2 + 4s + 5) + (Cs + D)((s + 2) (s + 1)$ ------(1)

Now put s = -2

$\therefore 1 = A(-2 + 1)(4 + 4(-2) + 5) = A(-1) (1)$

$\therefore A = -1$

Now put s = -1

$\therefore 1 = B(-1 + 2)(1 + 4(-1) + 5) = B(1) (2)$

$\therefore B = 1/2$

In order to find the values of C & D

From equation (1),

We get A + B + C = 0

i.e. -1 + $\frac{1}{2}$ + C = 0

$\therefore C = \frac{1}{2}$

Now, equating '0' coefficient from equation (1)

$\therefore SA + 10B + 2D = 1$

$\therefore S(-1) + 10 * \frac{1}{2} + 2D = 1$

$\therefore D = \frac{1}{2}$

Hence, A = -1, B = $\frac{1}{2}$, C = $\frac{1}{2}$, D = $\frac{1}{2}$

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

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

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

Taking inverse Laplace transform we get,

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

$\therefore y(t) = -e^{-2t} + \frac{1}{2}e^{-t} + \frac{1}{2}$ {$e^{-2t} sint + e^{-2t} cost$}

$\therefore y(t) = \frac{1}{2} e^{-t} + \frac{1}{2} e^{-2t}$ {$cost - sint - 2$}

Rearranging the common terms,

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