Answer
$t=1.364,\ y=0.820$
Work Step by Step
Finding the integrating factor:
$\mu(t)=\exp(\int 1/2\ dt)$
$\mu(t)=e^{t/2}$
Finding the y-function:
$y=\frac{1}{\mu(t)}[\int \mu(s)(2cos(s))ds+c]$
$y=e^{-t/2}[4/5\ e^{t/2}(2\sin t+\cos t)+c]$
$y=4/5(2\sin t+\cos t)+ce^{-t/2}$
$y(0)=4/5+c=-1$
$c=-9/5$
Local maximum,
$y'=0\rightarrow 4/5(2\cos t-\sin t)+9/10\ e^{-t/2}=0$
The first solutions is $t=1.364,\ y=0.820$
