nagle_differential_equations_ISM_Part29

# nagle_differential_equations_ISM_Part29 - Chapter 4 and a...

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and a general solution is given by y ( t ) = - t cos t 2 + c 1 cos t + c 2 sin t. The ﬁrst boundary condition, y (0) = 0, yields c 1 = 0. But this implies that y ( π ) = - t cos t 2 + c 2 sin t ± ± ± ± t = π = π 2 6 = 1 , for any constant c 2 . 48. (a) Using the superposition principle (Theorem 3), we conclude that the functions y 1 ( t ) = ( t 2 + 1 + e t cos t + e t sin t ) - ( t 2 + 1 + e t cos t ) = e t sin t, y 2 ( t ) = ( t 2 + 1 + e t cos t + e t sin t ) - ( t 2 + 1 + e t sin t ) = e t cos t are solutions to the corresponding homogeneous equation. These two functions are linearly independent on ( -∞ , ) since neither one is a constant multiple of the other. (b) Substituting, say, y 1 ( t ) into the corresponding homogeneous equation yields ( e t sin t ) 00 + p ( e t sin t ) 0 + q ( e t sin t ) = (2 + p ) e t cos t + ( p + q ) e t sin t = 0 . Therefore, p = - 2, q = - p = 2, and so the equation becomes y 00 - 2 y 0 + 2 y = g ( t ) . (4.3) Another way to recover p and q is to use the results of Section 4.3. The functions y 1 ( t ) and y 2 ( t ) ﬁt the form of two linearly independent solutions in the case when the auxiliary equation has complex roots α ± βi . Here, α = β = 1. Thus, the auxiliary equation must be [ r - (1 + i )] · [ r - (1 - i )] = ( r - 1) 2 + 1 = r 2 - 2 r + 2 , leading to the same conclusion about p and q . To ﬁnd g ( t ), one can just substitute either of three given functions into (4.3). But we can simplify computations noting that, say, y = t 2 + 1 + e t cos t - y 2 ( t ) = t 2 + 1 is a solution to the given equation (by the superposition principle). Thus, we have

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## This note was uploaded on 02/12/2011 for the course MA 221 taught by Professor Mazmani during the Spring '08 term at Stevens.

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nagle_differential_equations_ISM_Part29 - Chapter 4 and a...

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