Using Laplace Transform , solve the following differential equation {y}text{}-{4}{y}={e}^{{-{3}{t}}},{y}{left({0}right)}={0},{y}'{left({0}right)}={2}

Step 1
As per guidelines, only typed answers are acceptable. so we will give only typed solution.
Given initial value problem is
$y-4y=e^{-3t},y\left(0\right)=0,y^{\prime }\left(0\right)=2$
Step 2
Applying Laplace transform on both sides
$L\{y-4y\}=L\left\{e^{-3t}\right\}$
$\Rightarrow L\left\{y\right\}-4L\left\{y\right\}=\frac{1}{s+3}$
$\Rightarrow s^{2}L\left\{y\right\}-sy\left(0\right)-y^{\prime }\left(0\right)-4L\left\{y\right\}=\frac{1}{s+3}$
$\Rightarrow (s^2-4)L\left\{y\right\}-2=\frac{1}{s+3}$
$\Rightarrow L\left\{y\right\}=\frac{1}{(s+3)(s^2-4)}+\frac{2}{s^2-4}$
$=(\frac{1}{5(s+3)}+\frac{1}{20(s-2)}-\frac{1}{4(s+2)})+(-\frac{1}{2(s+2)}+\frac{1}{2(s-2)})$
$=\frac{1}{5(s+3)}+\frac{11}{20(s-2)}-\frac{3}{4(s+2)}$
Taking inverse Laplace transform on both sides, we get
$y=L^{-1}\{\frac{1}{5(s+3)}+\frac{11}{20(s-2)}-\frac{3}{4(s+2)}\}$
$=L^{-1}\left\{\frac{1}{5(s+3)}\right\}+L^{-1}\left\{\frac{11}{20(s-2)}\right\}-L^{-1}\left\{\frac{3}{4(s+2)}\right\}$
$=\frac{1}{5}{e}^{-3t}+\frac{11}{20}{e}^{2t}-\frac{3}{4}{e}^{-2t}$
Hence, solution is
$y\left(t\right)=\frac{11}{20}{e}^{2t}-\frac{3}{4}{e}^{-2t}+\frac{1}{5}{e}^{-3t}\text{ or }y\left(t\right)=\frac{11}{20}{e}^{2t}-\frac{15}{20}{e}^{-2t}+\frac{4}{20}{e}^{-3t}$
Comparing the answer with options, none of the option is correct.
The option given are all wrong.