28_pdfsam_math 54 differential equation solutions odd

28_pdfsam_math 54 - Chapter 1 = −(1 h)1/h = −(1 h)1/h Hence 12 3 lim g(h = lim −(1 h)1/h − h − h2 h→0 h→0 23 4 3 12 = − lim(1

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Unformatted text preview: Chapter 1 = −(1 + h)1/h = −(1 + h)1/h Hence 12 3 lim g (h) = lim −(1 + h)1/h − + h − h2 + · · · h→0 h→0 23 4 3 12 = − lim (1 + h)1/h · lim − + h − h2 + · · · h→0 h→0 23 4 From calculus we know that e = lim (1 + h)1/h , which gives h→0 1 1 d 1 1 − h + h2 − h3 + · · · dh 2 3 4 12 3 − + h − h2 + · · · . 23 4 . h→0 lim g (h) = −e − 1 2 = e . 2 So we have e e − yn =. n→∞ 1/n 2 lim 15. The independent variable in this problem is the time t and the dependent variable is the temperature T (t) of a body. Thus, we will use the recursive formulas (2) and (3) on page 25 with x replaced by t and y replaced by T . In the differential equation describing the Newton’s Law of Cooling, f (t, T ) = K (M (t) − T ). With the suggested values of K = 1 (min)−1 , M (t) ≡ 70◦ , h = 0.1, and the initial condition T (0) = 100◦ , the initial value problem becomes dT = 70 − T, dt and so the recursive formulas are tn+1 = tn + 0.1 , Tn+1 = Tn + 0.1(70 − Tn ). For n = 0, t1 = t0 + 0.1 = 0.1 , 24 T1 = T0 + 0.1(70 − T0 ) = 100 + 0.1(70 − 100) = 97 ; T (0) = 100, ...
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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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