hw4_solution

hw4_solution - a) dc dt =-kc 3 (6) Separate c, t : d c c 3...

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Numerical Analysis in Engineering ME 140A, Fall 2007 HW #4 solutions 1. Problem 1. Consider the frst order linear equation λ dy dt = 1 - y (1) SOLUTION: y 0 + y λ = 1 λ (2) Multiply with an integrating ±actor μ ( t ) : μy 0 + μ y λ = μ λ (3) Add and subtract μ 0 y ( μy 0 + μ 0 y ) + ( μ λ - μ 0 ) y = μ λ (4) Need ( μ λ - μ 0 ) = 0 μ 0 μ = 1 λ ln ( μ ) = t λ + c 1 μ ( t ) = e t λ + c 1 = c 2 e t λ Having ±ound μ ( t ), now solve (4), ( μy 0 + μ 0 y ) + 0 = μ λ ( μy ) 0 = c 2 λ e t/λ μy = c 2 e t/λ + c 3 μy = c 2 e t/λ + c 3 y = c 2 μ e t/λ + c 3 μ y = 1 + c 4 e - t/λ 1
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Initial condition: y (0) = 0 c 4 = - 1 y ( t ) = 1 - e - t λ (5) b) fnd time t * when y ( t ) = 0 . 99 : e - t * λ = 0 . 01 e t * λ = 100 t * λ = ln (100) t * = λln (100) 2. Problem 2.
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Unformatted text preview: a) dc dt =-kc 3 (6) Separate c, t : d c c 3 =-k dt (7) Integrate:-1 2 c-2 =-kt + c 1 c-2 = 2 kt + c 2 c = 1 2 kt + c 2 Initial condition: c (0) = 2 : c 2 = 1 4 c = 1 2 kt + 1 4 b) The equation is nonlinear, homogeneous, constant coeFcient. c) 1 2 t * + 1 4 = 10-3 q 2 t * + 1 4 = 10 3 2 t * + 1 4 = 10 6 t * = 499 , 999 . 875 2...
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hw4_solution - a) dc dt =-kc 3 (6) Separate c, t : d c c 3...

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