## Elementary Differential Equations and Boundary Value Problems 9th Edition

(a) $u = T+(u_{0}-T)e^{-kt}$. (b) $k\tau = \ln2$.
1.2.15 Newton's Law of Cooling: $du/dt = -k(u-T)$ Initial Condition: $u(0) = u_{0}$. (a) Find the temperature, $u$, of the object at any time, $t$, Where $T$ is the ambient temperature. $\textit{Solution}$(a): We must find the solution to the differential equation, (1) $du/dt = -k(u-T)$, having initial condition, (2) $u(0) = u_{0}$. Let us transform (1): $\frac{du/dt}{u-T} = -k$. Integrating, $\ln(u-T) = -kt +C$ $u-T = e^{-kt+C}$ $= e^{C}e^{-kt}$ $= ce^{-kt}$. (3) At initial condition, $t =0$, $u-T = c$. So. (4) $u-T = (u_{0}-T)e^{-kt}$, or (5) $u = T+(u_{0}-T)e^{-kt}$. (b) Let $\tau$ be the time at which the initial temperature has been reduced by one half. Find the relation between $k$ and $\tau$. $\textit{Solution}$(b): From (4), $\frac{1}{2}(u_{0}-T) = (u_{0}-T)e^{-k\tau}$. $\frac{1}{2} = e^{-k\tau}$., $2 = e^{k\tau}$, $k\tau = \ln2$.