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?

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- Aug 1st 2007, 02:32 PMvietdifferential eq.
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? - Aug 1st 2007, 03:26 PMgalactus
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.