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221_pdfsam_math 54 differential equation solutions odd

221_pdfsam_math 54 differential equation solutions odd -...

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Exercises 4.6 and, in the variation of parameters method, a particular solution has the form y p ( t )= v 1 ( t )cos2 t + v 2 ( t )sin2 t, where v 0 1 ( t ), v 0 2 ( t ) satisfy the system v 0 1 ( t )cos2 t + v 0 2 ( t )sin2 t =0 , 2 v 0 1 ( t )sin2 t +2 v 0 2 ( t )cos2 t =csc 2 2 t. Multiplying the Frst equation by sin 2 t and the second equation by (1 / 2) cos 2 t , and adding the resulting equations, we get v 0 2 ( t )= 1 2 csc 2 2 t cos 2 t v 2 = 1 2 Z csc 2 2 t cos 2 tdt = 1 4 csc 2 t + c 3 . ±rom the Frst equation in the system above we also Fnd v 0 1 ( t )= v 0 2 ( t )tan2 t = 1 2 csc 2 2 t cos 2 t tan 2 t = 1 2 csc 2 t v 1 ( t )= 1 2 Z csc 2 tdt = 1 4 ln | csc 2 t +cot2 t | + c 4 . With c 3 = c 4 =0, y p ( t )= 1 4 cos 2 t ln | csc 2 t +cot2 t |− 1 4 csc 2 t sin 2 t = 1 4 (cos 2 t ln | csc 2 t +cot2 t |− 1) y ( t )= c 1 cos 2 t + c 2 sin 2 t + 1 4 (cos 2 t ln | csc 2 t +cot2 t |− 1) . 11. This equation is similar to that in Example 1 on page 196 in the text. Only the nonhomoge-
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Unformatted text preview: neous term is di²erent. Thus we will follow steps in Example 1. Two independent solutions to the corresponding homogeneous equation, y + y = 0, are y 1 ( t ) = cos t and y 2 ( t ) = sin t . A particular solution to the original equation is of the form y p ( t ) = v 1 ( t ) cos t + v 2 ( t ) sin t, where v 1 ( t ) and v 2 ( t ) satisfy v 1 ( t ) cos t + v 2 ( t ) sin t = 0 , − v 1 ( t ) sin t + v 2 ( t ) cos t = tan 2 t. 217...
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