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Quiz-3-Math-20D-W09solutions

# Quiz-3-Math-20D-W09solutions - 00-12 y 4 y = 0 y(0 = 2 y(0...

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Quiz #3 Math 20D February 13 SOLUTIONS Two problems ( § 3.4, #21 and § 3.5, #11) taken from homework problems from sections 3.1, 3.4, 3.5. 1. ( § 3.4, #21. 10 points) Find the solution of the given initial value problem (1 . 25 = 5 4 ): y 00 + y 0 + 1 . 25 y = 0 , y (0) = 3 , y 0 (0) = 1 . Solution : Characteristic equation: r 2 + r + 5 4 = 0 . Roots: r = - 1 ± q 1 1 - 4 · 1 · 5 4 2 = - 1 2 ± i. General solution is: y = c 1 e - t 2 cos t + c 2 e - t 2 sin t. Then y 0 = - c 1 e - t 2 sin t + c 2 e - t 2 cos t - 1 2 c 1 e - t 2 cos t - 1 2 c 2 e - t 2 sin t = c 2 - 1 2 c 1 e - t 2 cos t + - c 1 - 1 2 c 2 e - t 2 sin t. From the initial conditions we have 3 = y (0) = c 1 , 1 = y 0 (0) = c 2 - 1 2 c 1 . That is, c 1 = 3 , c 2 = 5 2 . So the solution is: y = 3 e - t 2 cos t + 5 2 e - t 2 sin t. 2. ( § 3.5, #11. 10 points) Solve the given initial value problem:

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Unformatted text preview: 00-12 y + 4 y = 0 , y (0) = 2 , y (0) =-1 . Solution : Characteristic equation: 0 = 9 r 2-12 r + 4 = (3 r-2) 2 . Repeated root: r = 2 3 . General solution is: y = c 1 e 2 t 3 + c 2 te 2 t 3 . Then y = 2 3 c 1 e 2 t 3 + 2 3 c 2 te 2 t 3 + c 2 e 2 t 3 = 2 3 c 1 + c 2 e 2 t 3 + 2 3 c 2 te 2 t 3 . Initial conditions imply 2 = y (0) = c 1 ,-1 = y (0) = 2 3 c 1 + c 2 . That is, c 1 = 2 , c 2 =-7 3 . So the solution is: y = 2 e 2 t 3-7 3 te 2 t 3 ....
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