Linear differential equation

  • Thread starter TiberiusK
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  • #1
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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)
 

Answers and Replies

  • #2
HallsofIvy
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You have shown that
[tex]u(t+ 2\pi/\sqrt{7})= e^{-\pi/\sqrt{7}}u(t)[/tex]

What do you get if you take t= 0 in that?
 
  • #3
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u(2pi/sqrt[7])=-e^(-pi/sqrt[7])*u(0),where u(0) = cos[ Q]
 
  • #4
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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)"?
 
  • #5
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Does someone have any advice?
 
  • #6
HallsofIvy
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Now that you know [itex]u(2\pi/\sqrt{7})[/itex] do the same: use
[tex]u(t+2\pi/\sqrt[7])=-e^(-\pi/\sqrt[7])*u(t)[/tex]
setting [itex]t= 2\pi/\sqrt{7}[/itex].

Then with [itex]t= 4\pi/\sqrt{7}[/itex], [itex]t= 6\pi/\sqrt{7}[/itex], etc.
 
  • #7
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just substitute t with the above values in this formula u(t+2pi/sqrt[7])=-e^(-pi/sqrt[7])*u(t)...ok...thank you
 

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