Let y be the solution of the initial value problem $\displaystyle y''+2y'+2y=0, y(0)=0, y'(0)=6 $
The maximum value of $\displaystyle y $ for $\displaystyle t>0 $ is?
Using the characteristic equation we have:
$\displaystyle m^{2}+2m+2=0$
$\displaystyle m_{1}=-1+i, \;\ m_{2}=-1-i$
This gives:
$\displaystyle e^{-x}(C_{1}cos(x)+C_{2}sin(x))$
Now, use your initial conditions and get $\displaystyle C_{1}=0, \;\ C_{2}=6$
$\displaystyle y=6e^{-x}sin(x)$
Now, what's the max value for x>0?. Graph it may help.
Differentiate, set to 0 and solve for x.