Answer
See below
Work Step by Step
Given: $\frac{dT}{dt}=-k(T-5\cos 2t)$
and $T(0)=0\\\frac{dT}{dt}(0)=5$
a) Find constant $-k(0-5.1)=5\\ \rightarrow k=1$
b) Substitute: $\frac{dT}{dt}=-(T-5\cos 2t)\\
\rightarrow \frac{dT}{dt}+T=5\cos 2t$
Integrating factor could be: $I(x)=e^{\int dt}=e^{t}$
The equation becomes: $ \frac{d}{dt}(Te^{t})=5\cos2te^t$
Integrating both sides:
$\rightarrow Te^{t}=\int 5\cos2te^t$
After using an integration by part, we have:
$Te^t=e^t(2\sin 2t+\cos 2t)+c \\
\rightarrow T(t)=2\sin 2t+\cos 2t+ce^{-t}$
Find value: $c+0+1=0 \rightarrow c=-1$
Hence, the solution is $T(t)=2\sin 2t+\cos 2t-e^{-t}$
c) For large values of $t$, since $e^{-t} \rightarrow 0, t\rightarrow \infty$, we have: $T(t)=2\sin 2t+\cos 2t$
