## Elementary Differential Equations and Boundary Value Problems 9th Edition

$y=\frac{t^2-1}{2e^{2t}}$
Since in form of $\frac{dy}{dt}+ay=g(t)$ $\mu(t)=e^{2t}$ $e^{2t}y'+e^{2t}2y=e^{2t}e^{-2t}t$ $\frac{d}{dt}(e^{2t}y)=e^{-2t}e^{2t}t$ $\int{\frac{d}{dt}(e^{2t}y)}=\int{e^{0}t}$ $e^{2t}y=\frac{t^2}{2}+\frac{2c}{2}$ Constant 2c/2 can be simplified to c, since it can be any constant. $y=\frac{t^2+c}{2e^{2t}}$ Plug in $y=0, t=1$ $0=\frac{1+c}{2e^2}$ $c=-1$