ordinary differential equations
Another problem I'm trying to solve that I have no idea what to do with. Here it goes:
In 1798, Rev. Robert Malthus proposed that the rate of change of a population is proportional to the actual population at any given time.
If the world population was 3.712 billion in 1970 and 4.453 billion in 1980, then what does Malthus's law predict the world population to be in 2008?
Your doubt is lecit: let us check what does it happen if we consider the case $dT=kdt+kT-k\tilde{T}$. A Physicist would probably start with the difference equation (let us put $k=1$ for simplicity)
$$\Delta T(t)=\Delta t+T(t)-\tilde{T},~~(*)$$
for all $t\geq 0$ and check if its solutions are somehow compatible with the physical reality, where $\Delta T(t)=T(t)-T(t_1)$ and $\Delta t=t-t_1$. Choosing for example $t_1=0$, i.e. the starting moment of our analysis of the physical system, we obtain
$$T(t)-T(0)=t+T(t)-\tilde{T},$$
which leads to the "nonsense" $t=T(0)-\tilde{T}$. We can then deduce that the difference equation $(*)$ leads to no sound theory of the evolution of $T(t)$.
We are left to the interpretation $dT=kdt(T-\tilde{T})$ of the text in the OP. Even in this case we have to compare the analytical results with the physical reality: after all, we did not specify the sign of the constant $k\neq 0$...
I shall use $u$ for the temperature as indicated in the problem. Formulating the statement, we have $$\frac{{du}}{{dt}} = - h(u - {u_{{\rm{external}}}})$$ which could easily be solved (first order ODE) to yield $$u(t) = {u_{{\rm{external}}}} + A{e^{ - ht}}$$ where $A$ is a constant. To obtain this constant ($A$) and also the conduction constant ($h$), note that (we have chosen the unit of time to be hour) $${u_{{\rm{external}}}} = 10,u(0) = 70,u(2) = 40$$ so that $$\begin{array}{l}70 = 10 + A\\40 = 10 + A{e^{ - 2h}}\end{array}$$ which would yield $$A = 60,h = \frac{{\ln 2}}{2}$$ Summarizing, the answer could be stated as $$u(t) = 10 + 60{e^{ - \frac{{\ln 2}}{2}t}}$$ Below I have plotted the variation of the inside temperature with time from the moment of heating failure to 20 hours later 
Best Answer
General edit.
If $p(t)$ is the population as a function of the time $t$ and $k$ is the constant of proportionality, the differential equation that models this law is
$$\dfrac{dp}{dt}=kp.\tag{1}$$
To find the solution $p(t)$ integrate $(1)$ and use the given values
$$p(t_0)=p(1970)=3.712\, \mathrm{billion},\tag{2a}$$
$$p(1980)=4.453\, \mathrm{billion},\tag{2b}$$
where $t_0=1970$ is the initial instant. The equation $(1)$ can be integrated by the method of separation of variables, rewriting it as $$ \begin{equation*} \frac{dp}{p}=k\, dt\tag{3} \end{equation*} $$
and integrating both sides $$ \begin{eqnarray*} \int \frac{dp}{p} &=&\int k\, dt \\ \ln p &=&kt+C,\tag{4} \end{eqnarray*} $$ where $C$ is the constant of integration. The equation $(4)$ can be rewritten in different forms, such as
$$ \begin{eqnarray*} p(t)&=&e^{kt+C}=Be^{kt}\tag{5a} \\ p(t) &=&Ae^{k(t-t_0)}=Ae^{k(t-1970)},\tag{5b} \end{eqnarray*} $$
where $A$ and $B$ are constants related to $C$. To proceed we could use either $(5\mathrm{a})$ or $(5\mathrm{b})$. Choosing equation $(5\mathrm{b})$, we find $A$ using the initial population $p(t_0)$ $(2\mathrm{a})$:
$$p(1970)=Ae^{k(1970-1970)}=Ae^0=A= 3.712\text{ billion}.\tag{5b'}$$
Now we can find the constant $k$ using the $1980$ population $(2\mathrm{b})$. We get the equation $$4.453=3.712e^{k(1980-1970)},\tag{5b''}$$
whose solution, by applying logarithms, is
$$k=\frac{1}{10}\ln\frac{4.453}{3.712}\approx 1.8201\times 10^{-2}\tag{5b'''}.$$
We thus have the following solution of $(1)$ that satisfies the initial conditions $(2\mathrm{a,b})$:
$$p(t)=3.712e^{1.8201\times 10^{-2}(t-1970)},\tag{6}$$
from which we obtain the Malthus's law prediction for the world population in $2008$:
$$p(2008)=3.712e^{1.8201\times 10^{-2}(2008-1970)}\approx 7.413\, \text{billion}.\tag{7}$$
Since $k>0$ the function $(6)$ expresses an exponential growth.