The given differential equation is:

$\frac{dQ}{dt} = 33 - 2Q^{\beta}$

Where $\beta = 2.1$. To find the equilibrium solution $Q_S$, we set $\frac{dQ}{dt} = 0$, as equilibrium occurs when the rate of change of $Q$ is zero.

So, the equation becomes:

$0 = 33 - 2Q_S^{\beta}$

Substitute $\beta = 2.1$:

$0 = 33 - 2Q_S^{2.1}$

Solve for $Q_S$:

$2Q_S^{2.1} = 33$

$Q_S^{2.1} = \frac{33}{2} = 16.5$

Now, to solve for $Q_S$, take the power of $\frac{1}{2.1}$ on both sides:

$Q_S = (16.5)^{\frac{1}{2.1}}$

Let me calculate that for you.The equilibrium solution $Q_S$ is approximately:

$Q_S \approx 3.80$

This is the value of $Q_S$ when $\beta = 2.1$.

Would you like further details on the calculation or have any other questions?

Here are five related questions for further practice:

1. How does changing $\beta$ affect the equilibrium solution $Q_S$?
2. What is the equilibrium solution for $\beta = 2$?
3. How do you solve a similar differential equation if the function is non-linear?
4. Can you analyze the stability of this equilibrium solution?
5. What is the significance of equilibrium solutions in dynamic systems?

Tip: Equilibrium solutions represent steady states where no further change occurs in the system over time.