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Quiz-4-Math-20D-W09solutions

# Quiz-4-Math-20D-W09solutions - initial point y 00 y-2 y = 0...

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Quiz #4 Math 20D February 20 SOLUTIONS Two problems ( § 3.2, #21 and § 3.6, #4) taken from homework problems from sections 3.2, 3.4, 3.5. 1. ( § 3.6, #4.) Find the general solution to the following differential equation: y 00 + 2 y 0 = 3 + 4 sin 2 t. Hint. You may use without proof: the general solution to the homogeneous solution y 00 + 2 y 0 = 0 is c 1 + c 2 e - 2 t . Solution : Method of undetermined coefficients: Let y p = At + B cos 2 t + C sin 2 t (subscript p stands for ‘particular’ solution). Then y 0 p = A - 2 B sin 2 t + 2 C cos 2 t, y 00 p = - 4 B cos 2 t - 4 C sin 2 t. For y p to be a solution is equivalent to: 3 + 4 sin 2 t = - 4 B cos 2 t - 4 C sin 2 t + 2 ( A - 2 B sin 2 t + 2 C cos 2 t ) = 2 A + ( - 4 B + 4 C ) cos 2 t + ( - 4 C - 4 B ) sin 2 t. Therefore 3 = 2 A, 0 = - 4 B + 4 C, 4 = - 4 C - 4 B. We conclude A = 3 2 , B = C = - 1 2 . So a particular solution is: y p = 3 2 t - 1 2 cos 2 t - 1 2 sin 2 t. The general solution is: y = c 1 + c 2 e - 2 t + 3 2 t - 1 2 cos 2 t - 1 2 sin 2 t. 2. ( § 3.2, #21.) Find the fundamental set of solutions y

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Unformatted text preview: initial point: y 00 + y-2 y = 0 , t = 0 . The solution y 1 satisﬁes the initial conditions: y 1 (0) = 1 , y 1 (0) = 0 and the solution y 2 satisﬁes the initial conditions: y 2 (0) = 0 , y 2 (0) = 1 . Solution : Characteristic equation is: 0 = r 2 + r-2 = ( r + 2)( r-1) . Roots are r = 1 and r =-2 . General solution is: y = c 1 e t + c 2 e-2 t . Then y = c 1 e t-2 c 2 e-2 t . For y 1 1 = y 1 (0) = c 1 + c 2 , 0 = y 1 (0) = c 1-2 c 2 . So c 1 = 2 3 and c 2 = 1 3 , which imply y 1 = 2 3 e t + 1 3 e-2 t . For y 2 0 = y 1 (0) = c 1 + c 2 , 1 = y 1 (0) = c 1-2 c 2 . So c 1 = 1 3 and c 2 =-1 3 , which imply y 1 = 1 3 e t-1 3 e-2 t ....
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