solutions4bd - Math 33b Quiz 4bd February 7 2008 Name UCLA...

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Math 33b, Quiz 4bd, February 7, 2008 Name: UCLA ID: 1. Find the general solution to the differential equation y 00 - 2 y 0 + y = sin 4 t. Solution. The solution has the form y = y h + y p , where y h is the solution to the homogeneous equation y 00 - 2 y 0 + y = 0 and y p is one particular solution to our inhomogeneous equation. To find the homogeneous solution, we write down the characteristic polynomial λ 2 - 2 λ +1 and find its roots. It factors as ( λ - 1) 2 , so it has a double root at λ = 1. The corresponding form of the homogeneous solution is y h = C 1 e t + C 2 te t . The right-hand side of the differential equation suggests trying a particular solution y p of the form y p = A cos 4 t + B sin 4 t . We want to find A, B such that y 00 p - 2 y 0 p + y p = sin 4 t . We compute y 0 p = 4 B cos 4
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Unformatted text preview: t and y 00 p =-16 A cos 4 t-16 B sin 4 t . Crunching everything through, we get (-16 A cos 4 t-16 B sin 4 t ) + (-8 B cos 4 t + 8 A sin 4 t ) + ( A cos 4 t + B sin 4 t ) = sin 4 t (-15 A-8 B ) cos 4 t + (-15 B + 8 A ) sin 4 t = sin 4 t Hence we must solve the simultaneous equations-15 A-8 B = 0 and-15 B + 8 A = 1. From the first equation, we get B =-15 8 A . Plugging this into the second, we have 225 8 A + 8 A = 1, or 289 8 A = 1. Hence A = 8 289 . Since B =-15 8 A , we have B =-15 289 . Putting everything together, we get y = C 1 e t + C 2 te t + 8 289 cos 4 t-15 289 sin 4 t. s 1...
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  • Spring '07
  • staff
  • Math, Sin

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