solutions2 - 3.6 4. W (x, xex ) = 5. W (et sin t, et cos t)...

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4. W ( x,xe x ) = ± ± ± ± x xe x 1 e x + xe x ± ± ± ± = xe x + x 2 e x - xe x = x 2 e x 5. W ( e t sin t,e t cos t ) = ± ± ± ± e t sin t e t cos t e t sin t + e t cos t e t cos t - e t sin t ± ± ± ± = - e 2 t 7. Rewrite the equation in standard form. y 00 + 3 t y 0 = 1 The coefficient p ( t ) = 3 /t is continuous for all t > 0. Since t 0 > 0, the solution to the IVP has a unique solution for all t > 0. 8. Rewrite the equation in standard form. y 00 - 3 t t - 1 y 0 + 4 t - 1 = sin t t - 1 The coefficients p ( t ) = 3 t/ ( t - 1) and q ( t ) = 4 / ( t - 1) as well as the right hand side g ( t ) = sin t/ ( t - 1) are continuous for t < - 1. Since t 0 < - 1, the IVP has a unique solution for all t < - 1. 14. Direct substitution shows that y 1 ( t ) = 1 and y 2 ( t ) = t - 1 are solutions to the equation and that y = c 1 y 1 ( t ) + c 2 y 2 ( t ) is not a solution. This result does not verify Theorem 3.2.2 because the equation in the problem is nonlinear. 16. Substitute y = sin t 2 into the differential equation. - 4 t 2 sin t 2 + 2 cos t 2 + 2 t cos t 2 p ( t ) + sin t 2 q ( t ) = 0 At t = 0 the equation becomes 2 = 0 (if we suppose that p and q are continuous), which is a contradiction. the function y = sin t 2 cannot be a solution under the given assumptions. 25. Direct substitution shows y 1 = e t and y 2 = te t to be solutions. W ( e t ,te t ) = ± ± ± ± e t te t e t ( t + 1) e t ± ± ± ± = e 2 t Since W ( y 1 ,y 2 ) 6 = 0, these solutions form a fundamental set of solutions. 31. Write the equation in standard form to identify
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This note was uploaded on 12/29/2010 for the course MATH 265 taught by Professor Leahkeshet during the Winter '10 term at The University of British Columbia.

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solutions2 - 3.6 4. W (x, xex ) = 5. W (et sin t, et cos t)...

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