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28_pdfsam_math 54 differential equation solutions odd

# 28_pdfsam_math 54 differential equation solutions odd -...

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Chapter 1 = (1 + h ) 1 /h d dh 1 1 2 h + 1 3 h 2 1 4 h 3 + · · · = (1 + h ) 1 /h 1 2 + 2 3 h 3 4 h 2 + · · · . Hence lim h 0 g ( h ) = lim h 0 (1 + h ) 1 /h 1 2 + 2 3 h 3 4 h 2 + · · · = lim h 0 (1 + h ) 1 /h · lim h 0 1 2 + 2 3 h 3 4 h 2 + · · · . From calculus we know that e = lim h 0 (1 + h ) 1 /h , which gives lim h 0 g ( h ) = e 1 2 = e 2 . So we have lim n →∞ e y n 1 /n = e 2 . 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,
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