## Related questions with answers

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

Solution

VerifiedLets solve the equation

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

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

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

If the murder was $x$ hours before $9$am

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

and after 1 hour.

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

Now lets find $x$

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

The murder happened at $3:44$am

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