To solve for the new equilibrium quantity $Q_S$ in the given differential equation, let's proceed step by step.

The equation provided is:

$\frac{dQ}{dt} = 35 - 4 Q^\beta$

where $\beta = 1.7$.

The equilibrium solution occurs when the system reaches a steady state, meaning the change in $Q$ with respect to time becomes zero:

$\frac{dQ}{dt} = 0$

So, we set the equation to 0 and solve for $Q_S$:

$0 = 35 - 4 Q_S^{1.7}$

Rearranging the equation:

$4 Q_S^{1.7} = 35$

Now, solve for $Q_S$:

$Q_S^{1.7} = \frac{35}{4} = 8.75$

Take the power of $\frac{1}{1.7}$ on both sides to isolate $Q_S$:

$Q_S = \left( 8.75 \right)^{\frac{1}{1.7}}$

Now, let's compute the numerical value for $Q_S$.The equilibrium solution $Q_S$ is approximately 3.58.

Would you like more details or have any questions?

Here are five related questions to expand on this topic:

1. How does changing the value of $\beta$ affect the equilibrium solution?
2. What happens to the system's dynamics if the constant 35 is altered?
3. How can we interpret this equilibrium in a real-world context (e.g., in economics or biology)?
4. Can the system have multiple equilibrium points under different conditions?
5. What if we introduce a time-dependent factor to the equation?

Tip: In nonlinear equations like this one, equilibrium points give insight into the long-term behavior of the system. Always check their stability to fully understand the system's dynamics.