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nagle_differential_equations_ISM_Part26

nagle_differential_equations_ISM_Part26 - Exercises 4.4...

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Exercises 4.4 = ( - A 1 t - A 0 + B 1 ) cos t + ( - B 1 t - B 0 - 2 A 1 ) sin t. Substituting these expressions into the original differential equation, we get θ p - θ p = ( - A 1 t - A 0 + 2 B 1 ) cos t + ( - B 1 t - B 0 - 2 A 1 ) sin t - ( A 1 t + A 0 ) cos t - ( B 1 t + B 0 ) sin t = - 2 A 1 t cos t + ( - 2 A 0 + 2 B 1 ) cos t - 2 B 1 t sin t + ( - 2 A 1 - 2 B 0 ) sin t = t sin t. Equating the coefficients, we see that - 2 A 1 = 0 - 2 A 0 + 2 B 1 = 0 - 2 B 1 = 1 - 2 A 1 - 2 B 0 = 0 A 1 = 0 A 0 = B 1 = - 1 / 2 B 1 = - 1 / 2 B 0 = - A 1 = 0 . Therefore, a particular solution of the nonhomogeneous equation θ - θ = t sin t is given by θ p ( t ) = - ( t sin t + cos t ) / 2 . 18. Solving the auxiliary equation, r 2 +4 = 0, yields r = ± 2 i . Therefore, we seek a particular solution of the form (15) with m = 0, α = 0, β = 2, and take s = 1 since α + = 2 i is a root of the auxiliary equation. Hence, y p = A 0 t cos 2 t + B 0 t sin 2 t , y p = (2 B 0 t + A 0 ) cos 2 t + ( - 2 A 0 t + B 0 ) sin 2 t , y p = ( - 4 A 0 t + 4 B 0 ) cos 2 t + ( - 4 B 0 t - 4 A 0 ) sin 2 t ; y p + 4 y p = 4 B 0 cos 2 t - 4 A 0 sin 2 t = 8 sin 2 t . Equating coefficients yields A 0 = - 2, B 0 = 0. Hence, y p ( t ) = - 2 t cos 2 t . 20. Similarly to Problem 18, we seek a particular solution of the form y p = t ( A 1 t + A 0 ) cos 2 t + t ( B 1 t + B 0 ) sin 2 t = ( A 1 t 2 + A 0 t ) cos 2 t + ( B 1 t 2 + B 0 t ) sin 2 t . Differentiating, we get y p = 2 B 1 t 2 + (2 A 1 + 2 B 0 ) t + A 0 cos 2 t + - 2 A 1 t 2 + ( - 2 A 0 + 2 B 1 ) t + B 0 sin 2 t , y p = - 4 A 1 t 2 + ( - 4 A 0 + 8 B 1 ) t + (2 A 1 + 4 B 0 ) cos 2 t 121
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Chapter 4 + - 4 B 1 t 2 + ( - 8 A 1 - 4 B 0 ) t + ( - 4 A 0 + 2 B 1 ) sin 2 t. We now substitute y p and y p into the given equation and simplify. y p + 4 y p = [8 B 1 t + (2 A 1 + 4 B 0 )] cos 2 t + [ - 8 A 1 t + ( - 4 A 0 + 2 B 1 )] sin 2 t = 16 t sin 2 t.
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