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122_pdfsam_math 54 differential equation solutions odd

# 122_pdfsam_math 54 differential equation solutions odd -...

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Chapter 3 where M ( t ) represents the surrounding temperature which we’ve assumed to be a constant 16 C. This differential equation is linear and is solved using an integrating factor of e Kt . Rewriting the above equation in standard form, multiplying both sides by e Kt and integrating gives dT dt + KT ( t ) = K (16) e Kt dT dt + e Kt KT ( t ) = 16 Ke Kt e Kt T ( t ) = 16 e Kt + C T ( t ) = 16 + Ce Kt . Let us take t = 0 as the time at which the person died. Then T (0) = 37 C (normal body temperature) and we get 37 = 16 + C C = 21 . Now we know that at sometime, say X hours after death, the body temperature was measured to be 34 . 5 C and that at X + 1 hours after death the body temperature was measured to be 33 . 7 C. Therefore, we have 34 . 5 = 16 + 21 e KX and 33 . 7 = 16 + 21
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Unformatted text preview: . Solving the ±rst equation For KX we arrive at KX = − ln ± 34 . 5 − 16 21 ² = 0 . 12675 . (3.8) Substituting this value into the second equation we, can solve For K as Follows: 33 . 7 = 16 + 21 e − . 12675 − K ⇒ K = − ³ . 12675 + ln ± 33 . 7 − 16 21 ²´ = 0 . 04421 . This results in an equation For the body temperature oF T ( t ) = 16 + 21 e − . 04421 t . ²rom equation (3.8) we now ±nd the number oF hours X beFore 12 Noon when the person died. X = . 12675 K = . 12675 . 04421 ≈ 2 . 867 (hours) . ThereFore, the time oF death is 2 . 867 hours (2 hours and 52 min) beFore Noon or 9 : 08 a.m. 118...
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