155_pdfsam_math 54 differential equation solutions odd

155_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Exercises 3.6 problem 75 − 20 cos πt 12 − T (t) + 0.1 + 1.5[70 − T (t)], T (0) = 65, with K = 0.2 . Since h = 2/3, it will take 36 steps to go from t = 0 to t = 24. By simplifying the above expression, we obtain dT = (75K + 105.1) − 20K cos dt πt 12 − (K + 1.5)T (t), T (0) = 65. (Note that here t takes the place of x and T takes the place of y .) Therefore, with K = 0.2 the inputs to the subroutine are t0 = 0, T0 = 65, c = 24, and N = 36. For Step 3 of the subroutine we have F = f (t, T ) = (75K + 105.1) − 20K cos G = f (t + h, T + hF ) = (75K + 105.1) − 20K cos For Step 4 in the subroutine we have t = t + h, h T = T + (F + G ). 2 Now, starting with t = t0 = 0 and T = T0 = 65, and h = 2/3 (as specified) we have Step 3 of the subroutine to be F = [75(0.2) + 105.1] − 20(0.2) cos 0 − [(0.2) + 1.5](65) = 5.6 , π (0.6667) G = [75(0.2) + 105.1] − 20(0.2) cos − [(0.2) + 1.5][65 + (0.6667)(5.6)] ≈ −0.6862 . 12 Hence in Step 4 we compute t = 0 + 0.6667 = 0.6667 T = 65 + 0.3333(5.6 − 0.6862) ≈ 66.638 . 151 π (t + h) 12 − (K + 1.5){T + hF }. (3.18) πt 12 − (K + 1.5)T, (3.17) ...
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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 University of California, Berkeley.

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