**Unformatted text preview: **Chapter 5
Thus, I (t) satisﬁes I (0) = 0, I (0) = −95. Next, we compute √ √ c1 (−25 − 5 21) (−25−5√21)t/2 c2 (−25 + 5 21) (−25+5√21)t/2 I (t) = e e + , 2 2 substitute t = 0 into formulas for I (t) and I (t), and obtain the system 0 = I (0) = c1 + c2 , √ √ −95 = I (0) = c1 (−25 − 5 21)/2 + c2 (−25 + 5 21)/2 ⇒ √ c1 = 19/ 21 , √ c2 = −19/ 21. So, the solution is
√ √ 19 e(−25−5 21)t/2 − e(−25+5 21)t/2 . I (t) = √ 21 3. In this problem L = 4, R = 120, C = (2200)−1 , and E (t) = 10 cos 20t. Therefore, we see that 1/C = 2200 and E (t) = −200 sin 20t. By substituting these values into equation (4) on page 287 of the text, we obtain the equation 4 By simplifying, we have d2 I dI + 2200I = −200 sin 20t. + 120 dt2 dt dI d2 I + 550I = −50 sin 20t. + 30 2 dt dt (5.36) The auxiliary equation associated with the homogeneous equation corresponding to (5.36) √ above is r 2 + 30r + 550 = 0. This equation has roots r = −15 ± 5 13i. Therefore, the transient current, that is Ih (t), is given by √ √ Ih (t) = e−15t C1 cos 5 13t + C2 sin 5 13t . By the method of undetermined coeﬃcients, a particular solution, Ip (t), of equation (5.36) will be of the form Ip (t) = ts [A cos 20t + B sin 20t]. Since neither y (t) = cos 20t nor y (t) = sin 20t is a solution to the homogeneous equation (that is the system is not at resonance), we can let s = 0 in Ip (t). Thus, we see that Ip (t), the steady-state current, has the form Ip (t) = A cos 20t + B sin 20t. 318 ...

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