Chapter 4.6, Problem 25E

In Problems 22 through 25, use variation of parameters to find a general solution to the differential equation given that the functions ${\text{y}}_1$ and ${\text{y}}_2$ are linearly independent solutions to the corresponding homogeneous equation for $t>0$.

$t{y}^{\prime \prime }+\left(5t-1\right)y^{\prime }+5y=t^2{e}^{-5t}$

$y_1=5t-1,y_2=e^{-5t}$