Newton's Law of Cooling- Differential Equations

I am working on a differential equations problem that involves Newton's Law of Cooling. It is:

Joe and Dave sit for coffee. Joe adds the cream right away and Dave waits 5 mins before adding the cream. Who has the hotter coffee? (Assume that the cream was kept at constant temperature , which is less than room temperature. You may ignore any chemical reactions, and treat the thermal properties of the cream as being the same as coffee)

Any help is great, thanks :)

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Joe does. Perhaps you should post some equations and some discussion of what you've done. –  rogerl May 26 '13 at 0:54
@rogerl: good point. I will undelete my answer then. –  robjohn May 26 '13 at 1:06

Equation of the temperature of the coffee $$\frac{\mathrm{d}}{\mathrm{d}t}T_c=-k(T_c-T_r)\implies T_c=T_r+(T_0-T_r)e^{-kt}$$ Temperature of the coffee after adding cream ($m_c=$ specific heat $\times$ mass of coffee, $m_d=$ specific heat $\times$ mass of dairy) $$\frac{m_cT_c+m_dT_d}{m_c+m_d}$$ Temperature of coffee when cream is added before $$T_r+\left(\frac{m_cT_0+m_dT_d}{m_c+m_d}-T_r\right)e^{-kt}$$ Temperature of coffee when cream is added after $$\frac{m_c\left(T_r+(T_0-T_r)e^{-kt}\right)+m_dT_d}{m_c+m_d}$$ The difference of the temperature of the coffee with the cream added before minus that of the coffee with the cream added after is $$(T_r-T_d)(1-e^{-kt})\frac{m_d}{m_c+m_d}\gt0$$ This is exactly the temperature removed by keeping the cream cool while the coffee was cooling.