- #1
jonnejon
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Homework Statement
Let y be the solution of the initial value problem:
y'' + 2y' + 2y = 0 , y(0) = 0 , y'(0) = 5
The maximum value of y over 0 less than or equal to x less than infinity is ??.
The Attempt at a Solution
r = -2 +/- i
I solved it:
y = c1 e^-2x cos(x) + c2 e^-2x sin(x)
y' = -2c1 e^-2x cos(x) - c1 e^-2x sin(x) - 2 c2 e^-2x sin(x) + c2 e^-2x cos(x)
y = c1 = 0
y' = -2c1 + c2 = 5 => c2 = 5
y = 5 e^-2t sin(x)
But I am terrible at trig. What is the maximum value of y? I got (x = -3pi/2 so therefore max y is 5 e^(6pi/2) but that is incorrect. I have one more attempt and don't want to lose out on this question.
Thanks.
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