$\dfrac {d^2y}{dt^2} + 2 \dfrac {dy}{dt}+ y = 3te^{-t} \\ \text{Taking Laplace Transform on both sides,}$

$ \\ L[\dfrac {d^2y}{dt^2} + 2 \dfrac {dy}{dt}+ y ]= L[3te^{-t}] \\ s^2 Y(s)-sy(0)-y'(0) + 2(sY(s)-y(0))+Y(s) = 3 \dfrac{1}{(s+1)^2} ... \text{using frequency shift} $

$\\Y(s)[s^2+2s+1] -4s-2-2\times 4 = \dfrac{3}{(s+1)^2}...\text{ Pluggingin the given initial values} \\Y(s)(s+1)^2 = \dfrac{3}{(s+1)^2} +4s+10$

$ \\ Y(s) = \dfrac{3+(4s+10)(s+1)^2}{(s+1)^4} = \dfrac{A}{(s+1)}+\dfrac{B}{(s+1)^2}+\dfrac{C}{(s+1)^3}+\dfrac{D}{(s+1)^4}$

$ \\ Y(s) = \dfrac{3+(4s+10)(s+1)^2}{(s+1)^4} = \dfrac{A(s+1)^3+B(s+1)^2+C(s+1)+D}{(s+1)^4} \\ \text{Comparing the numerators, we get}$

$ \\ 3 +(4s+10)(s+1)^2 = A(s+1)^3+B(s+1)^2+C(s+1)+D $

$ \\ Now, 3+(4s+10)(s^2+2s+1) = A(s^3+3s^2 +3s+1)+B(s^2+2s+1)+C(s+1)+D \\ 3+4s^3+8s^2+4s+10s^2+20s+10 = As^3 +3As^2+3As+A+Bs^2+2Bs+B+Cs+C+D $

$\\ \text{Comparing the co-efficients of } s^3,s^2, s, \text {and constants,} \\ \rightarrow A=4, \\ \rightarrow 8+10 = 3A+B $

$\\ 18 = 12+B \\ B= 6 $

$\\ \rightarrow 4+20= 3A+2B+C \\ 24 = 12+12+C \\ C=0 $

$\\ \rightarrow 3+10 = A+B+C+D \\ 13 = 4+6+0+D \\ D=3 $

$ \\ \therefore, Y(s) = \dfrac{4}{(s+1)}+\dfrac{6}{(s+1)^2}+\dfrac{3}{(s+1)^4}$

$\text{Taking Inverse Laplace Transform,} \\ y = L^{-1}[\dfrac{4}{(s+1)}+\dfrac{6}{(s+1)^2}+\dfrac{3}{(s+1)^4}]$

$ \\ y = e^{-t} L^{-1}[\dfrac{4}{(s)}+\dfrac{6}{(s)^2}+\dfrac{3}{(s)^4}] \\ [From \: Frequency \: Shifting: \: L^{-1}[f(s-a)] = e^{at}L^{-1}[f(s)], here \: a =-1] $

$\\ y = e^{-t}[4 + 6t+3\dfrac{t^3}{3!}] = e^{-t}[4 + 6t+\dfrac{t^3}{2}] \\ \bbox[3pt, yellow] {y =\dfrac{e^{-t}}{2}[t^3+12t+8]}$