I have a differential equation which looks like:
$$ \dfrac{dT}{dt} = \dfrac{P}{\rho Ac_ph} + \dfrac{q (T_{in} - T)}{Ah} - \dfrac{U\pi D(T - T_{a})}{\rho Ac_p} $$
where $P$, $h$, $q$, $ T_{in}$ are functions and everything else is constants; or rather these functions are just defined as having some values at $t=0$ and then at for example $t=600$, and $h$ in this case is constant.
Where would I start if I need to find the Laplace transform for $T$?
Edit: So I realised that $P$, $q$ and $ T_{in}$ are unit step functions, and I know how to find Laplace transforms for them. My problem now is the following term: $ \dfrac{q (T_{in} - T)}{Ah} $ since I don't know if it is possible to find the Laplace transform of a product of two functions ( $ q T_{in} $ and $ qT $).
The Laplace Transform of a product is usually well-defined. There are 2 basic ways of finding it; either through the definition of the Laplace transform, or through the convolution theorem.
For the term $q \ T_{in}$ you have said that both of these are unit step functions, and finding the transform of a product of these is easy from the definition. You can verify that if $q = U(t-t_1)$ and $T_{in} = U(t-t_2)$ then $$ \mathcal{L}(q \ T_{in}) = \int_0^{\infty} U(t-t_1) \ U(t-t_2) \ e^{- s t} dt = \frac{1}{s}e^{-s \ \text{Max}(t_1,t_2)}; \ Re(s)>0, \ t_1>0, \ t_2>0$$
For the term $q \ T$ we don't know the explicit form of $T$ so using the definition will probably not be helpful. However, we can use the convolution theorem instead, one form of which says that, $$\mathcal{L} (f_1(t) f_2(t)) = \frac{1}{2 \pi i} \int_{c-i \infty}^{c+i \infty} F_1(s^{\prime}) F_2(s-s^{\prime}) \ ds^{\prime}$$ where $F(s) = \mathcal{L}f(t)$. In your case let's choose $q = f_2$ and $T = f_1$ to find that $$\mathcal{L}( q \ T) = \frac{1}{2 \pi i}\int_{c-i \infty}^{c+i \infty} \hat{T}(s^{\prime}) \frac{e^{-t_1(s - s^{\prime})}}{s - s^{\prime}} \ ds^{\prime}$$
I'm using $\hat{T}$ to denote $\mathcal{L}T$. We need to choose $c$ so that the Laplace transform is in its domain of analyticity (basically so that the original Laplace transform actually exists). For this case, $c<s$ should suffice.