# differential eq.

• August 1st 2007, 02:32 PM
viet
differential eq.
Let y be the solution of the initial value problem $y''+2y'+2y=0, y(0)=0, y'(0)=6$

The maximum value of $y$ for $t>0$ is?
• August 1st 2007, 03:26 PM
galactus
Using the characteristic equation we have:

$m^{2}+2m+2=0$

$m_{1}=-1+i, \;\ m_{2}=-1-i$

This gives:

$e^{-x}(C_{1}cos(x)+C_{2}sin(x))$

Now, use your initial conditions and get $C_{1}=0, \;\ C_{2}=6$

$y=6e^{-x}sin(x)$

Now, what's the max value for x>0?. Graph it may help.

Differentiate, set to 0 and solve for x.