42_pdfsam_math 54 differential equation solutions odd

42_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Chapter 2 (a) We are given that M = 100. To find C we must solve the equation T (0) = 40 = C + 100. This gives C = −60. Thus the equation becomes T (t) = −60e−0.02180t + 100. We want to solve for t when T (t) = 90. This gives us 90 = −60e−0.02180t + 100 ⇒ −0.0218t = ln 1 6 ⇒ ⇒ 1 = e−0.02180t 6 0.0218t = ln 6 . Therefore t = ln 6/0.0218 ≈ 82.2 min. (b) Here M = 140, so we solve T (0) = 40 = C + 140 As above, solving for t in the equation T (t) = −100e−0.02180t + 140 = 90 (c) With M = 80, we solve 40 = C + 80, yielding C = −40. Setting T (t) = −40e−0.02180t + 80 = 90 ⇒ 1 − = e−0.02180t . 4 ⇒ t ≈ 31.8 . ⇒ C = −100. This last equation is impossible because an exponential function is never negative. Hence it never attains desired temperature. The physical nature of this problem would lead us to expect this result. A further discussion of Newton’s law of cooling is given in Section 3.3. 37. The differential equation dP r = P dt 100 38 ...
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