217_pdfsam_math 54 differential equation solutions odd

217_pdfsam_math 54 differential equation solutions odd - c...

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Exercises 4.6 Since y 0 1 ( t )=(cos t ) 0 = sin t, y 0 2 ( t )=(s in t ) 0 =cos t, the system (9) on page 195 becomes v 0 1 ( t )cos t + v 0 2 ( t )sin t =0 , v 0 1 ( t )sin t + v 0 2 ( t )cos t =sec t. (4.6) Multiplying the frst equation by sin t and the second equation by cos t yields v 0 1 ( t )sin t cos t + v 0 2 ( t )sin 2 t =0 , v 0 1 ( t )sin t cos t + v 0 2 ( t )cos 2 t =1 . Adding these equations together, we obtain v 0 2 ( t ) ( cos 2 t +sin 2 t ) =1 or v 0 2 ( t )=1 . From the frst equation in (4.6), we can now fnd v 0 1 ( t ): v 0 1 ( t )= v 0 2 ( t ) sin t cos t = tan t. So, v 0 1 ( t )= tan t, v 0 2 ( t )=1 v 1 ( t )= R tan tdt =ln | cos t | + c 3 , v 2 ( t )= R dt = t + c 4 .
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Unformatted text preview: c 3 = c 4 = 0 and get y p ( t ) = cos t ln | cos t | + t sin t. Thus a general solution to the given equation is y ( t ) = y p ( t ) + y h ( t ) = cos t ln | cos t | + t sin t + c 1 cos t + c 2 sin t. 3. First, we can simpli±y the equation by dividing both sides by 2. This yields x − x − 2 x = e 3 t . 213...
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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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