Solution:
\( \displaylines{y=\frac{1}{\mu\left(t\right)}\int_{}^{}\mu\left(t\right)g\left(t\right)dt\\ \text{where }\mu=e^{\int_{}^{}p\left(t\right)dt}\text{ in the equation }y^{\prime}+p\left(t\right)y=g\left(t\right)\\ \text{In this case }\mu=e^{\int_{}^{}-4\differentialD t}=e^{-4t}\\ \int_{}^{}\mu g=\int_{}^{}e^{-4t}\left(-5e^{4t}\right)dt=-5t\\ y=-5te^{4t}} \)