Here is a road map for you to follow to the problem's solution.
Newton's law of cooling states that the rate at which a body cools is proportional to the difference between its temperature and the surrounding temperature. Assuming a very large room, we have the following differential equation:
$$\frac{dT}{dt}=k(T-30)$$
Now, you want to separate the variables and integrate. This gives:
$$\frac{dT}{T-30}=kdt\quad\Longrightarrow\quad\ln(T-30)=kt+C\quad\Longrightarrow\quad T(t)=Ae^{kt}+30$$
where $A=e^C$. Now, plug in the two ordered pairs you know, $(10,-10)$ and $(20,-5)$, to obtain a system of two equations in the two variables $A$ and $k$.
$$-10=Ae^{10k}+30\quad\Longrightarrow\quad-40=Ae^{10k}\\-5=Ae^{20k}+30\quad\Longrightarrow\quad-35=Ae^{20k}$$
Solve this system to find the values of your constants. To find $k$, divide the two equations:
$$\frac{35}{40}=e^{10k}\quad\Longrightarrow\quad k=\frac{1}{10}\ln\left(\frac{7}{8}\right)$$
Use this to find $A$. Plug $A$ and $k$ back into $T(t)$. Then just plug in $t=0$ to find the initial temperature.
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 $k\neq 0$...
Best Answer
Here you have two sets of data: T(0)=100 and T(15)=90 while $T_a=25$. So you have two equations:$$100=c*e^{k*0}+25 (1)$$$$90=c*e^{15k}+25 (2)$$ From (1) find c, from (2) find k. Now you have equation in the form $$T=c*e^{k*t}+T_a$$ with c and k known. Here $T_a$ = 100, because "plateau" happens when $T=T_a = 100$. Maybe this problem lacks info about time of heating. If you'll have time of heating, you'll easily solve it.