$y\left(t\right)\text{ = }c_{1}e^{ - t} + \frac{1}{2}e^{1 - t}t^{2}\:\begin{matrix} \text{Solve the linear equation }\frac{dy\left(t\right)}{dt} + y\left(t\right)\text{ =}e^{ - t + 1}t\text{}: \\ \begin{matrix} \begin{matrix} \text{Let }\mu \left(t\right)\text{ =}e^{\int 1dt}\text{ =}e^{t}. \\ \text{Multiply both sides by }\mu \left(t\right)\text{}: \\ \end{matrix} \\ e^{t}\frac{dy\left(t\right)}{dt} + e^{t}y\left(t\right)\text{ }\text{ =}et \\ \end{matrix} \\ \begin{matrix} \text{Substitute }e^{t}\text{ =}\frac{d}{dt}\left(e^{t}\right): \\ e^{t}\frac{dy\left(t\right)}{dt} + \frac{d}{dt}\left(e^{t}\right)y\left(t\right)\text{ = }et \\ \end{matrix} \\ \begin{matrix} \text{Apply the reverse product rule }\frac{dg}{dt}f + \frac{df}{dt}g\text{=}\frac{d}{dt}\left(fg\right)\text{ to the left‐hand side:} \\ \frac{d}{dt}\left(e^{t}y\left(t\right)\right)\text{ = }et \\ \end{matrix} \\ \begin{matrix} \text{Integrate both sides with respect to }t: \\ \int \frac{d}{dt}\left(e^{t}y\left(t\right)\right)dt\text{ }\text{ =}\int etdt \\ \end{matrix} \\ \begin{matrix} \text{Evaluate the integrals:} \\ e^{t}y\left(t\right)\text{ }\text{ =}\frac{et^{2}}{2} + c_{1}\text{, where }c_{1}\text{ is an arbitrary constant.} \\ \end{matrix} \\ \begin{matrix} \text{Divide both sides by }\mu \left(t\right)\text{=}e^{t}\text{}: \\ \begin{matrix} \text{Answer:} & \\ & y\left(t\right)\text{ }\text{ =}e^{ - t}\left(\frac{et^{2}}{2} + c_{1}\right) \\ \end{matrix} \\ \end{matrix} \\ \end{matrix}\:$
4.5