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Unformatted text preview: Chapter 4 We find c 1 and c 2 by using the initial conditions. We first differentiate y ( t ) to get y ( t ) = 5 c 1 sin 5 t + 5 c 2 cos 5 t. Substituting y ( t ) and y ( t ) into the initial conditions, we obtain the system y (0) = 0 . 3 = c 1 y (0) = . 1 = 5 c 2 . Solving, we find that c 1 = 0 . 3 and c 2 = . 02. Therefore the equation of motion is given by y ( t ) = 0 . 3 cos 5 t . 02 sin 5 t (m) . (b) In part (a) we found that β = 5. Therefore the frequency of oscillation is 5 / (2 π ). 34. For the specified values of the inductance L , resistance R , capacitance C , electromotive force E ( t ), and initial values q and I , the initial value problem (20) becomes 10 dI dt + 20 I + 6260 q = 100; q (0) = 0 , I (0) = 0 . In particular, substituting t = 0, we conclude that 10 I (0) + 20 I (0) + 6260 q (0) = 100 ⇒ I (0) = 10 . Differentiating the above equation, using the relation I = dq/dt , and simplifying, yields an initial value problem I 00 + 2 I + 626 I = 0; I (0) = 0 , I (0) = 10 . The auxiliary equation for this homogeneous second order equation, r 2 + 2 r + 626 = 0, has roots r = 1 ± 25 i . Thus, a general solution has the form I ( t ) = e t ( c 1 cos 25 t + c 2 sin 25 t ) . Since I ( t ) = e t [( c 1 + 25 c 2 ) cos 25 t + ( 25 c 1 c 2 ) sin 25 t ] , for c 1 and c 2 we have a system of equations I (0) = c 1 = 0 116 Exercises 4.3 I (0) = c 1 + 25 c 2 = 10 . Hence, c 1 = 0, c 2 = 0 . 4, and the current at time t is given by I ( t ) = 0 . 4 e t sin 25 t....
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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.
 Spring '08
 MAZMANI
 Equations

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