Use the Laplace transform to solve the initial$-$value problem $y^{\text{'}\text{'}}{y}^{\text{'}}-2y=4,y(0)=1,y^{\text{'}}(0)=4.$

Possible answers.

a$.y(t)=-2\frac{10}{3}{e}^t-\frac{1}{3}{e}^{-2t}$

b$.y(t)=-2\frac{10}{3}{e}^t\frac{1}{3}{e}^{2t}$

c$.y(t)=-2\frac{10}{3}{e}^{-t}-\frac{1}{3}{e}^{-2t}$

d$.y(t)=-2\frac{10}{3}{e}^{-t}-\frac{1}{3}{e}^{2t}$

e$.y(t)=2\frac{10}{3}{e}^t\frac{1}{3}{e}^{-2t}$