> [!Example] > Suppose that a body is found in a room which is kept at a constant temperature of $24^{\circ}C.$ At 8 a.m. its temperature is $28 ^{\circ}C$; an hour later it is $26 ^{\circ}C.$ Use [[Newton's law of cooling]], to estimate the time of death. > **Solution**: With time $t$ measured in hours, solve $\frac{dT}{dt}+kT=kA.$By [[Implicit Solution to First Order Linear Ordinary Differential Equation Initial Value Problem]], $T(t_{2})e^{kt_{2}} - T(t_{1})e^{kt_{1}} = A(e^{kt_{2}}-ek^{t_{1}})$rearranging this gives the temperature at time $t_{2}$ in terms of the temperature at time $t_{1},$ $T(t_{2})= A+[T(t_{1})-A]e^{-k(t_{2}-t_{1})}$This gives $T(9)=A+[T(8)-A]e^{-k}$Substituting given values gives $e^{-k}=0.5 \implies k=\ln2$If the time of death was $t_{0}$ then $T(8)=A+[T(t_{0})-A]e^{-k(8-t_{0})}$Since $T(t_{0})=37,T(8)=28$ and $A=24,$ $28=24+[37-24]e^{-k(8-t_{0})}$which gives $t_{0} \approx 1.7.$