Integration: Formulate an Expression for $T$ in terms of $k$ and $T_0$

Question

$T$ and $T_0$ are related through the equation

$$ \frac{dT}{dt}=-k (T-T_{0})$$ where $K>0.$

These are the steps I have worked through so far. Am I correct?

$$ \frac{dT}{dt}=-kT+kT_{0}$$

$$ \frac{dT}{T}=-kdt+kT_{0}$$

$$ \frac{dT}{T}=kT_{0}-kdt$$

$$\int\frac{dT}{T}=\int kT_{0}-kdt$$

$$\log_{e}T+C=kT_{0}-kt$$

Then take Exponentials on each side to get rid of the log.

$$T+C = e^{kT_{0}-kt+C}$$

$$T=Ae^{kT_{0}-kt}$$

Answer 1

You are confusing $\color{red}{T}$ with $\color{blue}{t}$

$$ \frac{{\rm d}\color{red}{T}}{{\rm d}\color{blue}{t}} = -k (\color{red}{T} - T_0) ~~~\Rightarrow~~~ \frac{{\rm d}\color{red}{T}}{ \color{red}{T}- T_0} = -k{\rm d}\color{blue}{t} $$

Integrate on both sides

$$ \ln (\color{red}{T} - T_0) = -k \color{blue}{t} + C $$

Can you take it from here?

Answer 2

You are considering $$\frac{dT}{dt}=-k (T-T_{0})$$ For the time being, make your life simpler defining $Z=T-T_0$ which makes $$\frac{dZ}{dt}=-kZ$$ which looks more familiar.

Solve it for $Z$ and, when done, reuse $T=Z+T_0$ to get the final result.