The reaction scheme is:
$$\ce{S + E ->[k_1] C1} \tag{1} $$ $$\ce{C1 ->[k_2] E + P} \tag{2}$$ $$\ce{S + C1 <=>[k_3][k_4]C2} \tag{3} $$
Let's define a bit of nomenclature.
- S = compound S in chemical reaction equation
- s = $[S]_x$, the concentration of S at t=x.
- $S^*$ = ds/dt
The whole set of differential equations has 5 equations with 5 unknowns.
\begin{align} \\
S^* = \frac{\mathrm{d}s}{\mathrm{d}t} &= -k_1se-k_3sc_1+k_4c_2 \tag{4} \\ E^* = \frac{\mathrm{d}e}{\mathrm{d}t} &= -k_1se+k_2c_1\tag{5} \\
C^*_1 = \frac{\mathrm{d}c_1}{\mathrm{d}t} &= k_1se-k_2c_1-k_3sc_1+k_4c_2 \tag{6}
\\ C^*_2 = \frac{\mathrm{d}c_2}{\mathrm{d}t} &= k_3sc_1-k_4c_2 \tag{7}\\ P^* = \frac{\mathrm{d}p}{\mathrm{d}t} &= k_2c_1 \tag{8}\end{align}
What is a conservation equation
In mathematical terms a conservation equation is some linear combination of the time derivatives which add up to zero on the right hand side.
The gist is that the various conservation equations allow some simplification in the various interactions in the system.
Rather than just search blindly, there are a couple of relationships that can be exploited to find time derivatives which add up to zero on the right hand side. First is a sort of mass balance. The initial concentration of the reactants has to be spread as the various intermediate species and the final product. A second line of attack is to exploit and equilibrium conditions such as equation (3).
First conservation equation from "mass balance"
The first conservation equation is found noting that at t=0 then all is [S], a constant. Noting that it takes 2 $\ce{S}$ to make one $\ce{C_2}$, at some later time t=x, thus:
$$[\ce{S}]_{t=0} = [\ce{S}]_{t=x} + [\ce{C_1}]_{t=x} + 2[\ce{C_2}]_{t=x} + [\ce{P}]_{t=x}\tag{9} $$.
Now for some intermediate time we can take derivative of the mass balance equation with respect to time. All the dt's get cumbersome so lets use just $S^*$ from now on to mean ds/dt. From the mass equation then since the initial concentration is a constant we get
$0 = S^* + C^*_1 + 2C^*_2 + P^*\tag{10} $
Most likely solving for either $S^*$, or $P^*$ would give an interesting result.
A second conservation equation from mass balance
The second conservation equation comes from noting the enzyme participation in the reaction.
$$[\ce{E}]_{t=0} = [\ce{E}]_{t=x} + [\ce{C_1}]_{t=x} + [\ce{C_2}]_{t=x}\tag{11}$$.
Taking the derivative we get:
$ 0 = E^* + C^*_1 + C^*_2 \tag{12}$
which can be rearranged to
$E^* = - C^*_1 - C^*_2\tag{13}$
Substituting equation 6 for $C^*_1$ and equation 7 for $C^*_2$ this gives us equation 5 again which isn't helpful.
A third conservation equation from the equilibrium condition
From the third reaction we can derive the equation.
$ \frac{k_3}{k_4}= \frac{[S]_x[C_1]_x}{[C_4]_x} = \frac{sc_1}{c_4}\tag{14}$
Taking the derivative:
$ 0 = \frac{c_1}{c_4}S^* + \frac{s}{c_4}C^*_1 - \frac{sc_1}{c_4^2}C^*_4\tag{15}$
rearranging
$ S^* = -\frac{s}{c_1}C^*_1 + \frac{s}{c_4}C^*_4\tag{16}$
Infinite time as a boundary condition
Obviously at $t=\infty$, all the S has been converted to all P which is a constant. This factoid isn't very useful though in predicting how long that an "effective infinite time would be. Rather it is just a boundary condition.
$$[\ce{P}]_{t=\infty} = [\ce{S}]_{t=0}\tag{17}$$
