 dt
dQ
​
 =33−2Q 
β
 Now take 
β
=
2.1
β=2.1 and determine the new equilibrium solution 
Q
S
Q 
S
​
  of Equation 
(
2
)
(2).

Q
S
=
Q 
S
​
 =

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.

The query submitted by the user: dt/dQ = 33−2Q^β. Now take β = 2.1 and determine the new equilibrium solution Q_S of Equation (2).

Learn how to solve the differential equation dQ/dt = 33 − 2Q^β with β = 2.1 by finding the equilibrium solution Q_S. This example covers key concepts like differential equations and equilibrium conditions, suitable for undergraduate students studying calculus and dynamic systems.

Math Problem Statement

dt dQ =33−2Q β Now take β

Q S

Q S

Solution

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