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A detective finds a murder victim at 9 am. The temperature of the body is measured at 90.3F90.3^{\circ} \mathrm{F}. One hour later, the temperature of the body is 89.0F89.0^{\circ} \mathrm{F}. The temperature of the room has been maintained at a constant 68F68^{\circ} \mathrm{F}. Solve the differential equation to estimate the time the murder occurred.

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Answered 2 years ago
Answered 2 years ago
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Lets solve the equation

dHdt=(H68)kdH(H68)=kdtln(H68)=kt+CH(t)=68+Cekt\begin{align*} \dfrac{dH}{dt}&=-(H-68)k\\ \\ \dfrac{dH}{(H-68)}&=-kdt\tag{integrate both sides}\\ \ln (H-68)&=-kt +C\\ H(t)&=68+Ce^{-kt} \end{align*}

At the time of murder body temperature would be 98.6°98.6\text{\textdegree}F, therefore

98.6=68+CC=30.6\begin{align*} 98.6&=68+C\\ C&=30.6 \end{align*}

If the murder was xx hours before 99am

90.3=68+(30.6)ekx22.3=30.6ekx\begin{align*} 90.3&=68+(30.6)e^{-kx}\\ 22.3&=30.6e^{-kx}\\ \end{align*}

and after 1 hour.

89=68+(30.6)ek(x+1)21=30.6ekxek1.0626=ekk=0.06\begin{align*} 89&=68+(30.6)e^{-k(x+1)}\\ 21&=30.6e^{-kx}e^{-k}\\ 1.0626&=e^{k}\\ k&=0.06 \end{align*}

Now lets find xx

90.3=68+(30.6)e0.06xe0.06x=1.372x=5.27\begin{align*} 90.3&=68+(30.6)e^{-0.06x}\\ e^{-0.06x}&=1.372\\ x&=5.27 \end{align*}

The murder happened at 3:443:44am

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