My teacher gave this question to solve using Laplace Transform
\begin{cases} 2y''+5y'+2y={e^{-2t}}\\ y(0)=1, y'(0)=1 \end{cases}
Taking Laplace transform on both sides we get:
$$L\{2y''(t)+5y'(t)+2y(t)\}=L\{e^{-2t}\}$$
Using laplace transform in ODE
$$L\{y'(t)\}=sL\{y(t)\}-y(0)$$
and
$$L\{y''(t)\}=s^{2}L\{y(t)\}-sy(0)-y'(0)$$
I got
$$L\{y(t)\}=\frac{7}{\left(2s^2+5s+2\right)^{\:}}+\frac{2s}{\left(2s^2+5s+2\right)^{\:}}+\frac{1}{\left(s+2\right)\left(2s^2+5s+2\right)^{\:}}$$
I can easily solve the inverselaplace of the equation and get the solution for y.
But my answer is not matching with the one given by the teacher.
So could anyone tell me had I made any mistake finding the value of L{y}?
The final answer I got is $$ -\frac{e^{-2t}t}{3}+\frac{20}{9}e^{-\frac{t}{2}}-\frac{11}{9}e^{-2t}$$
PLEASE DO NOT UPLOAD THE SOLUTION.
EDIT: I am extremely sorry, the initial values should be $y(0)=1, y'(0)=1$.