nagle_differential_equations_ISM_Part28

nagle_differential_equations_ISM_Part28 - Exercises 4.5 and...

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Exercises 4.5 and a general solution is y ( t ) = y p ( t ) + y h ( t ) = t 2 - 4 t + 7 - e t 4 + c 1 e - t + c 2 te - t . Next, we satisfy the initial conditions. 0 = y (0) = 7 - 1 / 4 + c 1 2 = y (0) = - 4 - 1 / 4 - c 1 + c 2 c 1 = - 27 / 4 c 2 = 25 / 4 + c 1 = - 1 / 2 . Therefore, the solution to the given initial value problem is y ( t ) = t 2 - 4 t + 7 - e t 4 - 27 e - t 4 - te - t 2 . 32. For the nonhomogeneous term, e 2 t + te 2 t + t 2 e 2 t = ( 1 + t + t 2 ) e 2 t , a particular solution has the form y p ( t ) = t s ( A 0 + A 1 t + A 2 t 2 ) e 2 t . Since r = 2 is not a root of the auxiliary equation, r 2 - 1 = 0, we choose s = 0. 34. Neither r = i nor r = 2 i is a root of the auxiliary equation, which is r 2 + 5 r + 6 = 0. Thus, by the superposition principle, y p ( t ) = A cos t + B sin t + C cos 2 t + D sin 2 t . 36. Since the auxiliary equation, r 2 - 4 r + 4 = ( r - 2) 2 = 0 has a double root r = 2 and the nonhomogeneous term can be written as t 2 e 2 t - e 2 t = ( t 2 - 1 ) e 2 t , a particular solution to the given equation has the form y p ( t ) = t 2 ( A 2 t 2 + A 1 t + A 0 ) e 2 t . 38. Since, by inspection, r = i is not a root of the auxiliary equation, which is r 4 - 5 r 2 +4 = 0, we look for a particular solution of the form y p ( t ) = A cos t + B sin t. 131
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Chapter 4 Differentiating y p ( t ) four times, we get y p ( t ) = - A sin t + B cos t , y p ( t ) = - A cos t - B sin t , y p ( t ) = A sin t - B cos t , y (4) p ( t ) = A cos t + B sin t . Therefore, y (4) p - 5 y p + 4 y p = 10 A cos t + 10 B sin t = 10 cos t - 20 sin t .
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