Quiz tut0702
QUESTION: $$y'+y=te^{-t}+1$$
follow the form $$y'+p(t)y=g(t)$$
$$p(t)=1$$
find the integrating factor: $$u(t)=e^{\int1dt}=e^t$$
mutiply u(t) on both sides $$e^ty'+e^ty=te^te^{-t}+e^t$$
integrating both sides $$\int{(e^ty')}=\int{te^0+e^t dt}$$
$$e^ty=\int{t dt}+\int{e^t dt}$$
$$e^ty=\frac{1}{2}t^2+e^t+c$$
$$y=\frac{1}{2}t^2e^{-t}+1+ce^{-t}$$
So, as $t\rightarrow \infty$, $y \rightarrow 1$, since the term with $e^{-t}$ goes to zero.
