246_pdfsam_math 54 differential equation solutions odd

246_pdfsam_math 54 differential equation solutions odd - 5....

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Chapter 4 whe rewetake s = 1 because r =3 i is a simple root of the auxiliary equation. Computing the derivatives y 0 ( t )= A cos 3 t + B sin 3 t + t ( 3 A sin 3 t +3 B cos 3 t ) , y 0 ( t )=6 B cos 3 t 6 A sin 3 t + t ( 9 A cos 3 t 9 B sin 3 t ) , and substituting y ( t )and y 0 ( t ) into the original equation, we get 6 B cos 3 t 6 A sin 3 t + t ( 9 A cos 3 t 9 B sin 3 t )+9 t ( A cos 3 t + B sin 3 t )=2cos3 t 6 B cos 3 t 6 A sin 3 t =2cos3 t A =0 , B =1 / 3 . So, y p ( t )=(1 / 3) t sin 3 t ,and y ( t )= C 1 cos 3 t + C 2 sin 3 t +(1 / 3) t sin 3 t is a general solution. To satisfy the initial conditions, we solve y (0) = C 1 =1 , y 0 (0) = 3 C 2 =0 C 1 =1 , C 2 =0 . So, the solution to the given initial value problem is y ( t )=cos3 t + 1 3 t sin 3 t. The graph of y ( t
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Unformatted text preview: 5. (a) The corresponding homogeneous equation, my + ky = 0, is the equation of a simple harmonic motion, and so its general solution is given by y h ( t ) = C 1 cos t + C 2 sin t, = p k/m . Since 6 = , we look for a particular solution of the form y p ( t ) = A cos t + B sin t y p ( t ) = A sin t + B cos t y p ( t ) = A 2 cos t B 2 sin t. 242...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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