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321_pdfsam_math 54 differential equation solutions odd

321_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Exercises 5.6 This equation has the auxiliary equation mr 2 + with roots ±i and mg +k l 2 − k 2 = mr 2 + mg l mr 2 + mg + 2k = 0 l g /l g /l and ±i (g/l) + (2k/m). As discussed on page 211 of the text (g/l) + (2k/m) are the normal angular frequencies. To ﬁnd the normal frequencies we divide each one by 2π and obtain 1 2π EXERCISES 5.6: g l and 1 2π g 2k + . l m Electrical Circuits, page 291 1. In this problem, R = 100 Ω, L = 4 H, C = 0.01 F, and E (t) = 20 V. Therefore, the equation (4) on page 287 of the text becomes 4 d(20) dI d2 I + 100I = =0 + 100 2 dt dt dt ⇒ d2 I dI + 25I = 0. + 25 2 dt dt The roots of the characteristic equation, r 2 + 25r + 25 = 0, are r= and so a general solution is I (t) = c1 e(−25−5 √ 21)t/2 −25 ± √ (25)2 − 4(25)(1) −25 ± 5 21 = , 2 2 + c2 e(−25+5 √ 21)t/2 . To determine constants c1 and c2 , ﬁrst we ﬁnd the initial value I (0) using given I (0) = 0 and q (0) = 4. Substituting t = 0 into equation (3) on page 287 of the text (with dq/dt replaced by I (t)), we obtain L d[I (t)] 1 + RI (t) + q (t) = E (t) dt C 1 (4) = 20 ⇒ 4I (0) + 100(0) + 0.01 ⇒ I (0) = −95. 317 ...
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