I have a simple chemical reaction $A\leftrightarrow B$ with forward rate $k_1$ and backward rate $k_2$. I can now write the differential equation of this system as following.
$ \frac{dA}{dt} = -k_1A + k_2B, \quad \frac{dB}{dt} = k_1A - k_2B$
Assuming that reactant A initial concentration is $A_0$, I took the Laplace transform:
$ sA(s) - A_0 = -k_1 A(s) +k_2B(s), \quad sB(s) - 0 = k_1A(s) - k_2B(s)$
So far so good. I was hoping to solve for $B(t)$ using any of these two equations, with the initial condition, $A(s) = \frac{A_0}{s}$.
Now, for the first equation,
$$ sA(s) - A_0 = -k_1 A(s) + k_2B(s) \\ A_0 - A_0 = -k_1\frac{A_0}{s} + k_2 B(s) \implies B(s) = \frac{k_1A_0}{k_2s}\\ $$ This is not correct!
If I use the second equation,
$$ B(s) = \frac{k_1A(s)}{s+k_2} = \frac{k_1A_0}{s(s+k_2)}$$ This gives me what I was expecting.
Certainly, I missed something somewhere but I can't figure out what!
You may simplify the derivation considerably by noticing that
$$A(t) + B(t) = A_0 \implies \hat{A}(s) + \hat{B}(s) = \frac{A_0}{s} $$
Then
$$s \hat{B}(s) = k_1 \hat{A}(s) - k_2 \hat{B}(s) = \frac{k_1 A_0}{s} - (k_1+k_2) \hat{B}(s)$$
Therefore
$$\hat{B}(s) = \frac{k_1 A_0}{s (s+k_1+k_2)} $$