Solving a Linear Differential Equation with Given Conditions

[TiberiusK]

Homework Statement

Find the general solution of the equation
u'' + u' + 2u = 0 of the form u(t) = C*(e^a*t)* cos[ B*t +Q],
1)Verify that it satisfies u(t+2pi/sqrt[7])=-e^(-pi/sqrt[7])*u(t)
2)Consider a solution satisfying u(0) = 1. Determine u(2pi/sqrt[7])
For what other values of t can you determine u(t) given u(0)?

Homework Equations

The Attempt at a Solution

p^2+p+2=0
delta=-1
p1=-1/2+(i*sqrt[7])/2 and p2=-1/2-(i*sqrt[7])/2
=>C*(e^-1/2*t)* cos[(sqrt[7]/2)* t +Q]=>(e^(-t/2)-(pi/sqrt[7]))*C*cos[(sqrt[7]/2)* t +pi+Q],pi+Q=a constant=>-e^(-pi/sqrt[7])*u(t).I hope this is ok.
I also need help with 2)

 

[HallsofIvy]

You have shown that
$$u(t+ 2\pi/\sqrt{7})= e^{-\pi/\sqrt{7}}u(t)$$

What do you get if you take t= 0 in that?

 

[TiberiusK]

u(2pi/sqrt[7])=-e^(-pi/sqrt[7])*u(0),where u(0) = cos[ Q]

 

[TiberiusK]

For Q=0,or 2pi u(0)=1=>u(2pi/sqrt[7])=-e^(-pi/sqrt[7])...
And for the part "For what other values of t can you determine u(t) given u(0)"?

 

[TiberiusK]

Does someone have any advice?

 

[HallsofIvy]

Now that you know $u(2\pi/\sqrt{7})$ do the same: use
$$u(t+2\pi/\sqrt[7])=-e^(-\pi/\sqrt[7])*u(t)$$
setting $t= 2\pi/\sqrt{7}$.

Then with $t= 4\pi/\sqrt{7}$, $t= 6\pi/\sqrt{7}$, etc.

 

[TiberiusK]

just substitute t with the above values in this formula u(t+2pi/sqrt[7])=-e^(-pi/sqrt[7])*u(t)...ok...thank you