328_pdfsam_math 54 differential equation solutions odd

# 328_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Chapter 5 Multiplying the ﬁrst equation by D , the second equation – by 2, and subtracting the results, we eliminate q3 : D 2 + 2D + 4 [I1 ] = −6 sin 3t. √ The roots of the characteristic equation, r 2 + 2r + 4 = 0, are r = −1 ± 3i, and so a general ⇒ solution to the corresponding homogeneous equation is √ √ I1h = C1 e−t cos 3t + C2 e−t sin 3t. A particular solution has the form I1p = A cos 3t + B sin 3t. Substitution into the equation yields (−5A + 6B ) cos 3t + (−6A − 5B ) sin 3t = −6 sin 3t ⇒ Therefore, √ √ 30 36 cos 3t + sin 3t. = C1 e−t cos 3t + C2 e−t sin 3t + 61 61 Substituting this solution into (5.37)we ﬁnd that 1 dI1 I2 = − 2 dt √ √ √ √ 54 C1 3 + C2 −t 45 C1 − C2 3 −t e cos 3t + e sin 3t − cos 3t + sin 3t. = 2 2 61 61 The initial condition, I1 (0) = I2 (0) = 0 yields C1 + 36/61 = 0, √ (C1 − C2 3)/2 − 45/61 = 0 Thus ⇒ C1 = −36/61, √ C2 = −42 3/61. I1 = I1h + I1p −5A + 6B = 0, −6A − 5B = −6 ⇒ A = 36/61, B = 30/61. D 2 + 2(D + 2) [I1 ] = −6 sin 3t √ √ √ 30 36 −t 42 3 −t 36 e sin 3t + cos 3t + sin 3t, I1 = − e cos 3t − 61 61 61 61 √ √ √ 54 45 −t 39 3 −t 45 I2 = e cos 3t − e sin 3t − cos 3t + sin 3t, 61 61 61 61 √ √ √ 24 81 3 3 −t 81 I3 = I1 − I2 = − e−t cos 3t − e sin 3t + cos 3t − sin 3t. 61 61 61 61 324 ...
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