By using the Laplace transform, show that the solution to

$$y^{\prime \prime}-4 y^{\prime}+3 y=t e^{-3 t},$$

subject to the conditions $y(0)=0$ and $y^{\prime}(0)=1$, is given by

$$y(t)=\frac{37}{72} e^{3 t}-\frac{17}{32} e^{t}+\left(\frac{5}{288}+\frac{1}{24} t\right) e^{-3 t}$$

when $t \geqslant 0$.