2302f11-mid2-soln - 1. (40) This question has four parts....

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Unformatted text preview: 1. (40) This question has four parts. a. Solve the differential equation y- 14 y + 53 y = 0. Solution. The characteristic polynomial is r 2- 14 r + 53, which has roots r = 14 ± √ 196- 212 2 = 7 ± 2 i, so the solution is y ( t ) = e 7 t ( C cos 2 t + D sin 2 t ) . b. What form of particular solution y p would you guess in order to solve the differential equation y- 14 y +53 y = e 2 t cos t + t 2 using the method of undetermined coefficients? Do not solve this differential equation! Solution. Our first guess would be y p = Ae 2 t cos t + Be 2 t sin t + Dt 2 + Et + F, because we need to have e 2 t cos t and t 2 , together with all their derivatives. We need to check this against the homogenous solution to see if there is any overlap. From part a above, we see that the homogeneous solution is y h = e 7 t ( C cos 2 t + D sin 2 t ) . Therefore, because there is no overlap, our first guess for y p is correct. c. Solve the differential equation y + 6 y + 9 y = 0. Solution. The characteristic polynomial is r 2 + 6 r + 9 = ( r + 3) 2 , which has a repeated root at r =- 3. Therefore the solution is y ( t ) = Ce- 3 t + Dte- 3 t ....
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This note was uploaded on 12/10/2011 for the course MAP 2302 taught by Professor Tuncer during the Fall '08 term at University of Florida.

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2302f11-mid2-soln - 1. (40) This question has four parts....

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