To find the equilibrium price, we need to set the demand function equal to the supply function. The demand function is given by:

$q_D = 12 - p - 3 \frac{dp}{dt} + \frac{d^2p}{dt^2}$

The supply function is:

$q_S(p) = -8 + 3p$

Condition for Equilibrium

At equilibrium, demand equals supply:

$12 - p - 3 \frac{dp}{dt} + \frac{d^2p}{dt^2} = -8 + 3p$

Rearranging this equation gives us:

$\frac{d^2p}{dt^2} - 3 \frac{dp}{dt} + 4p = 20$

This is a second-order linear differential equation.

Finding the Equilibrium Price $p(t)$

We need to solve the differential equation with the initial conditions $p(0) = 7$ and $\frac{dp}{dt}(0) = 9$.

1. Homogeneous Equation: The homogeneous part is:

   $\frac{d^2p}{dt^2} - 3 \frac{dp}{dt} + 4p = 0$

   The characteristic equation is:

   $r^2 - 3r + 4 = 0$

   Solving this using the quadratic formula:

   $r = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 - 16}}{2} = \frac{3 \pm \sqrt{-7}}{2}$

   This gives:

   $r = \frac{3}{2} \pm \frac{\sqrt{7}}{2} i$

   Thus, the general solution to the homogeneous equation is:

   $p_h(t) = e^{\frac{3}{2} t} \left( C_1 \cos\left(\frac{\sqrt{7}}{2} t\right) + C_2 \sin\left(\frac{\sqrt{7}}{2} t\right) \right)$

2. Particular Solution: For the non-homogeneous equation $\frac{d^2p}{dt^2} - 3 \frac{dp}{dt} + 4p = 20$, we can guess a constant particular solution $p_p = k$. Plugging this into the equation gives:

   $0 - 0 + 4k = 20 \implies k = 5$

   Thus, the particular solution is $p_p = 5$.

3. General Solution: Combining the homogeneous and particular solutions:

   $p(t) = e^{\frac{3}{2} t} \left( C_1 \cos\left(\frac{\sqrt{7}}{2} t\right) + C_2 \sin\left(\frac{\sqrt{7}}{2} t\right) \right) + 5$

Applying Initial Conditions

1. At $t=0$: $p(0) = C_1 + 5 = 7 \implies C_1 = 2$

2. Finding $C_2$: The derivative $\frac{dp}{dt}$ at $t=0$:

   $\frac{dp}{dt} = \frac{3}{2} e^{\frac{3}{2} t} (C_1 \cos(\frac{\sqrt{7}}{2} t) + C_2 \sin(\frac{\sqrt{7}}{2} t)) + e^{\frac{3}{2} t} \left( -\frac{\sqrt{7}}{2} C_1 \sin\left(\frac{\sqrt{7}}{2} t\right) + \frac{\sqrt{7}}{2} C_2 \cos\left(\frac{\sqrt{7}}{2} t\right) \right)$

   Evaluating this at $t=0$:

   $\frac{dp}{dt}(0) = \frac{3}{2} (2) + 0 + 0 = 3 + \frac{\sqrt{7}}{2} C_2 = 9$

   $\frac{\sqrt{7}}{2} C_2 = 6 \implies C_2 = \frac{12}{\sqrt{7}} = \frac{12\sqrt{7}}{7}$

Final Solution

Thus, the equilibrium price function is:

$p(t) = e^{\frac{3}{2} t} \left( 2 \cos\left(\frac{\sqrt{7}}{2} t\right) + \frac{12\sqrt{7}}{7} \sin\left(\frac{\sqrt{7}}{2} t\right) \right) + 5$

Summary

The equilibrium price $p(t)$ is determined as above. If you have any questions or need further details, feel free to ask!

Related Questions

1. What is the significance of the parameters in the demand function?
2. How does the supply function influence equilibrium?
3. What happens to equilibrium if the supply function changes?
4. Can you explain the implications of price trends on consumer behavior?
5. What methods can be used to analyze stability around the equilibrium price?

Tip

To analyze the stability of the equilibrium, consider examining the second derivative of the demand and supply functions to assess whether the equilibrium is stable or unstable.