solutions4bd

# solutions4bd - Math 33b Quiz 4bd May 3 2007 Name UCLA ID 1...

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Math 33b, Quiz 4bd, May 3, 2007 Name: UCLA ID: 1. Find the general solution to the differential equation y + 2 y + y = sin 3 t. Solution. The general solution y ( t ) is of the form y = y h + y p , where y h is the general solution to the homogeneous equation y + 4 y + 4 y = 0 and y p is one particular solution to the inhomogeneous equation. To solve the homogeneous equation, we write down its characteristic polynomial λ 2 + 2 λ + 1. It has the root λ = - 1 repeated twice, so the general solution to the homogeneous equation is y h = C 1 e - t + C 2 te - t . Given the form of the inhomogeneous term in the differential equation, we try a solution of the form y p = A cos 3 t + B sin 3
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Unformatted text preview: = 3 B cos 3 t-3 A sin 3 t and y 00 p =-9 A cos 3 t-9 B sin 3 t . Adding everything up, we get y 00 p + 2 y p + y p = (-8 A + 6 B ) cos 3 t + (-6 A-8 B ) sin 3 t . This must all equal sin 3 t , so we have (-8 A +6 B ) = 0 and (-6 A-8 B ) = 1. From (-8 A +6 B ) = 0, we get B = 4 3 A . Substituting into the equation (-6 A-8 B ) = 1, we have-6 A-32 3 A = 1, or-50 3 A = 1, so A =-3 50 . Since B = 4 3 A , we have B =-4 50 . . Hence y p =-3 50 cos 3 t-4 50 sin 3 t , and y ( t ) = C 1 e-t + C 2 te-t-3 50 cos 3 t-4 50 sin 3 t. 1...
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• Spring '07
• lee
• Heterogeneity, Homogeneity

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