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HW6soln

HW6soln - Homework 6 solutions Section 4.6 page 197-8...

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Homework 6 solutions Section 4.6 page 197-8 problem 4: In this case y 1 = e - t , and y 2 = e t . W = y 1 y 2 - y 1 y 2 = e - t e t + e - t e t = 2 and v 1 = - (2 t + 4) e t 2 dt = - ( t + 1) e t + C 1 v 2 = (2 t + 4) e - t 2 dt = - ( t + 3) e - t + C 2 so y = v 1 y 1 + v 2 y 2 = - ( t + 1) - ( t + 3) + C 1 e - t + C 2 e t = - 2 t - 4 + C 1 e - t + C 2 e t . Of course this could also have been computed (more easily) using undetermined coefficients. Problem 6: In this case y 1 = e - t and y 2 = te - t . W = y 1 y 2 - y 1 y 2 = e - t ((1 - t ) e - t ) + e - t te - t = e - 2 t and v 1 = - e - t te - t e - 2 t = - t 2 2 + C 1 v 2 = e - t e - t e - 2 t = t + C 2 so y = v 1 y 1 + v 2 y 2 = - t 2 2 e - t + t 2 e - t + C 1 e - t + C 2 te - t = t 2 2 + C 1 + C 2 t e - t This problem could also be done with undetermined coefficients. Problem 8: In this case y 1 = cos 3 t and y 2 = sin 3 t W = y 1 y 2 - y 1 y 2 = cos 3 t 3 cos 3 t + 3 sin 3 t sin 3 t = 3 and v 1 = - sec 2 3 t sin 3 t 3 dt = - sin 3 t 3 cos 2 3 t dt = - 1 9 sec 3 t + C 1 , v 2 = sec 2 3 t cos 3 t 3 dt = sec 3 t 3 dt = 1 9 ln | sec 3 t + tan 3 t | + C 2 so y = v 1 y 1 + v 2 y 2 = - 1 9 sec 3 t cos 3 t + 1 9 (ln | sec 3 t + tan 3 t | ) sin 3 t + C 1 cos 3 t + C 2 sin 3 t = 1 9 ( - 1 + sin 3 t ln | sec 3 t + tan 3 t | ) + C 1 cos 3 t + C 2 sin 3 t This problem could not have been done using undetermined coefficients. Problem 22: In this case y 1 = t 2 and y 2 = t 3 W = y 1 y 2 - y 1 y 2 = t 2 (3 t 2 ) - 2 t ( t 3 ) = t 4

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and recalling that the denominator in the formula is aW , where in this case a = t 2 we have v 1 = - ( t 3 + 1) t 3 t 6 dt = - 1 + t - 3 dt = - 1 + 1 2 t 2 + C 1 = 1 2 t 2 + C 1 , v 2 = ( t 3 + 1) t 2 t 6 dt = t - 1 + t - 4 dt = ln | t | - 1 3 t 3 + C 2 so y = v 1 y 1 + v 2 y 2 = 1 2 + t 3 ln | t | - 1 3 + C 1 t 2 + C 2 t 3 = 1 6 + t 3 ln | t | + C 1 t 2 + C 2 t 3 Section 4.9 Page 227 Problem 1 1 2 3 4 5 Ω 0.2 0.4 0.6 0.8 1.0 M 8 m = 4, k = 1, b = 2 < Section 4.9 Page 227 Problem 2 1 2 3 4 5 Ω 0.1 0.2 0.3 0.4 M 8 m = 4, k = 3, b = 3 < Section 5.6 Page 291 Problem 1: To solve, we can use equation (4) on page 287 with the given values, and then use the same trick used in an earlier homework to get the value of I’(0) using q(0) and equation (3). Alternately, we can solve using equation (3) in the form L d 2 q dt 2 + R dq dt + 1 C q = E ( t ) using the initial conditions for q(0) and q’(0)=I(0). This would give us a formula for q ( t ), which we then differentiate to get I ( t ). In both cases the roots on the left are 25 2 ± 5 21 2 , and on the right 0. We get I = 19 21 e - 5 2 ( 5+ 21 ) t - e - 5 2 ( 5 - 21 ) t
Problem 2: This problem is very similar to problem 1 above, except that we seek q ( t ) not I ( t ). In this case the left roots are

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