215_pdfsam_math 54 differential equation solutions odd

215_pdfsam_math 54 differential equation solutions odd -...

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Exercises 4.6 Substitution into the original equation yields 27( A cos 6 t + B sin 6 t )=27cos6 t A = 1 ,B =0 y p ( t )= cos 6 t. Therefore, a general solution has the form y ( t )= c 1 cos 3 t + c 2 sin 3 t cos 6 t. In (a)–(c), we have the same boundary condition at t =0 ,thatis , y (0) = 1. This yields 1= y (0) = c 1 1 c 1 =0 . Hence, all the solutions satisfying this condition are given by y ( t )= c 2 sin 3 t cos 6 t. (4.4) (a) The second boundary condition gives 3 = y ( π/ 6) = c 2 +1 c 2 = 2, and the answer is y =2s in3 t cos 6 t . (b) This time we have 5 = y ( π/ 3) = c 2 · 0 1 5= 1, and so there is no solution of the form (4.4) satisfying this second boundary condition. (c) Now we have 1= y ( π/ 3) = c 2 · 0 1 ⇒− 1=
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Unformatted text preview: EXERCISES 4.6: Variation of Parameters, page 197 1. The auxiliary equation in this problem is r 2 + 4 = 0, which has the roots r = ± 2 i . Therefore, y 1 ( t ) = cos 2 t and y 2 ( t ) = sin 2 t are two linearly independent solutions, and a general solution to the corresponding homogeneous equation is given by y h ( t ) = c 1 cos 2 t + c 2 sin 2 t. Using the variation of parameters method, we look for a particular solution to the original nonhomogeneous equation of the form y p ( t ) = v 1 ( t ) y 1 ( t ) + v 2 ( t ) y 2 ( t ) = v 1 ( t ) cos 2 t + v 2 ( t ) sin 2 t. 211...
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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 Berkeley.

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