Solve the initial$-$value problem $y^{\text{'}\text{'}}-2y^{\text{'}}y=t{e}^t,y(0)=1,y^{\text{'}}(0)=3.$

Possible answers.

a$.y(t)=e^{t}2t{e}^t{t}^3{e}^t$

b$.y(t)=e^{t}2t{e}^t-\frac{t^3{e}^t}{6}$

c$.y(t)=e^{t}t{e}^t{t}^3{e}^t$

d$.y(t)=e^{t}2t{e}^t\frac{t^3{e}^t}{6}$

c$.y(t)=e^t-t{e}^t-t^3{e}^t$