293_pdfsam_math 54 differential equation solutions odd

293_pdfsam_math 54 differential equation solutions odd -...

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Exercises 5.3 Table 5–B : Approximations of the solutions to Problem 18. t n x 1; n x 1; n x x 2; n x 2; n 2; n 01 0 1 5 0.1 3.124 9.353 0.2 1.381 7.254 0.3 0.707 6.256 0.4 0.389 5.726 0.5 0.223 5.428 21. First, we convert given initial value problem to an initial value problem for a normal system. Let x 1 ( t )= H ( t ), x 2 ( t )= H 0 ( t ). Then H 0 ( t )= x 0 2 ( t ), x 1 (0) = H (0) = 0, x 2 (0) = H 0 (0) = 0, and we get x 0 1 = x 2 , 60 x 1 =(77 . 7) x 0 2 +(19 . 42) x 2 2 , x 1 (0) = x 2 (0) = 0 x 0 1 = x 2 , x 0 2 =[60 x 1 (19 . 42) x 2 2 ] / 77 . 7 , x 1 (0) = x 2 (0) = 0 . Thus f 1 ( t, x 1 ,x 2 )= x 2 , f 2 ( t, x 1 ,x 2 )=[ 6 0 x 1 (19 . 42) x 2 2 ] / 77 . 7, t 0 =0 , x 1;0 =0 ,and x 2;0 =0 . W ith h =0 . 5, we need (5 0) / 0 . 5 = 10 steps to approximate the solution over the
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Unformatted text preview: solution at t = 0 . 5. k 1 , 1 = hx 2;0 = 0 . 5(0) = 0 , k 2 , 1 = h ± 60 − x 1;0 − (19 . 42) x 2 2;0 ² / 77 . 7 = 0 . 5 ± 60 − (0) − (19 . 42)(0) 2 ² / 77 . 7 = 0 . 38610 , k 1 , 2 = h ³ x 2;0 + k 2 , 1 2 ´ = 0 . 5 ³ (0) + . 38610 2 ´ = 0 . 09653 , k 2 , 2 = h 60 − ³ x 1;0 + k 1 , 1 2 ´ − (19 . 42) ³ x 2;0 + k 2 , 1 2 ´ 2 / 77 . 7 = 0 . 38144 , k 1 , 3 = h ³ x 2;0 + k 2 , 2 2 ´ = 0 . 5 ³ (0) + . 38144 2 ´ = 0 . 09536 , 289...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.

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