Hw05_Sol_Su09

Hw05_Sol_Su09 - ECE220 Homework #5 Solutions Solution to...

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ECE220 Homework #5 Solutions Solution to Problem 7.2 In this problem, we assume that f ( t ) is a continuous function of t for t < 0, we have dy ( t ) dt = 0 integrate both side, we get y ( t ) = C 1 from the initial condition, y (0) = 10, we have C 1 = 10 for 0 t < 1, we have dy ( t ) dt = - 1 integrate both side, we get y ( t ) = - t + C 2 from the initial condition, y (0) = 10, we have C 2 = 10 and we have y (1) = 9 for 1 t < 2, we have dy ( t ) dt = 1 integrate both side, we get y ( t ) = t + C 3 from the initial condition, y (1) = 9, we have C 3 = 8 and we have y (2) = 10 for t 2, we have dy ( t ) dt = 0 integrate both side, we get y ( t ) = C 4 from the initial condition, y (2) = 10, we have C 4 = 10 y ( t ) = 10 , t < 0 - t + 10 , 0 t < 1 t + 8 , 1 t < 2 10 , 2 t 1
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Solution to Problem 7.4 The given equation is, dy ( t ) dx = t 3 - 2 sin(10 t ) Integrate both side, we get y ( t ) = 1 4 t 4 + 1 5 cos(10 t ) + C from the initial condition, y (0) = 100, we have C = 100 - 1 5 So, we get y ( t ) = 1 4 t 4 + 1 5 cos(10 t ) + 99 4 5 Solution to Problem 7.5 Given dv ( t ) dt + 2 . 0 × 10 6 v ( t ) = 10 u ( t ) , v (0) = 0 (1) (a) We obtain the complementary solution v c ( t ) by setting the driving force (10 u ( t )) to zero: dv c ( t ) dt + 2 . 0 × 10 6 v c ( t ) = 0 Put the v c ( t ) terms to the left hand side of the equation and other terms to the right hand side, we have dv c ( t ) v c ( t ) = - 2 × 10 6 dt, (2) integrating to both sides: ± dv c ( t ) v c ( t ) = ± - 2 × 10 6 dt, (3) ln v c ( t ) = - 2 × 10 6 t + C 1 , (4) v c ( t ) = e - 2 × 10 6 t + C 1 u ( t ) = Ce - 2 × 10 6 t u ( t ) (5) (b) Guessing the particular solution v p ( t ) to have the same form as the the driving force, let v p ( t ) = Au ( t ) , and substitute it into the original diFerential equation: 0 + 2 × 10 6 Au ( t ) = 10 u ( t ) (6) A = 10 2 × 10 6 = 0 . 5 × 10 - 5 (7) v p ( t ) = 0 . 5 × 10 - 5 u ( t ) (8) 2
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(c) v
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Hw05_Sol_Su09 - ECE220 Homework #5 Solutions Solution to...

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