Unformatted text preview: The next step is to redo the above work with K = 0 . 4. That is, we substitute K = 0 . 4 and h = 2 / 3 ≈ . 6667 into equations (3.17) and (3.18) above. This yields F = 135 . 1 − 8 cos ± πt 12 ² − 1 . 9 T, G = 135 . 1 − 8 cos ³ π ( t + 0 . 6667) 12 ´ − 1 . 9( T + 0 . 6667 F ) , and T = T + (0 . 3333)( F + G ) . Then, using these equations, we go through the process of Frst Fnding F , then using this result to Fnd G , and Fnally using both results to Fnd T . (This process must be done for 152...
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Tn, 12hour clock, a.m., 24 hr, 3g, 0.6667 hours

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