Another Modeling Growth and Decay Problem

[Burjam]

Homework Statement

Consider an item that is initially sold at a market price of $10 per unit. Over time, market forces push the price toward the equilibrium price, $20, at which supply balances demand. After 6 months, the price has increased to $15. the Evans Price Adjustment model says that the rate at which the market price changes is proportional to the difference between the market price and the equilibrium price.

Homework Equations

N/A

The Attempt at a Solution

dp/dt = 20-p, where p=market price in dollars and t=time in months
∫dp/20-p = ∫1dt
ln (20-p) = t+c
20-p = e^(t+c)
20-p = e^t*e^c
20-p = Ae^t, where A=e^c
p = 20-Ae^t
10 = 20-Ae^0
10 = 20-A
A=10

Now that I've solved for A with the initial condition, I have hit a roadblock. I'm pretty sure my initial setup of this problem is off. When I substitute 15 in for p, I get a negative t, which is impossible. Please help.

 

[LCKurtz]

Homework Statement

Consider an item that is initially sold at a market price of $10 per unit. Over time, market forces push the price toward the equilibrium price, $20, at which supply balances demand. After 6 months, the price has increased to $15. the Evans Price Adjustment model says that the rate at which the market price changes is proportional to the difference between the market price and the equilibrium price.

Homework Equations

N/A

The Attempt at a Solution

dp/dt = 20-p, where p=market price in dollars and t=time in months
∫dp/20-p = ∫1dt
ln (20-p) = t+c
20-p = e^(t+c)
.

$$\int \frac{dp}{20-p}\ne \ln(20-p)$$

 

[Burjam]

Whoops that's a negative. Now this problem makes sense.

 

[Burjam]

Sorry for the double post, but actually it looks like the sign doesn't even matter. Can someone please give me a little more insight? I'm pretty sure at this point it has to do with my initial setup.

 

[Burjam]

Ok so I've become aware that I need to solve for the constant of proportionality k.

dp/dt = k(20-p)

The price rises $5 in 6 months, so dp/dt=5/6 when p=10

5/6=k(20-10)
k=1/12
dp/dt=(20-p)/12

From here, I get what appears to be a faulty equation:

p = 20 - 10e^t/12

The reason I say this is because when I plug in 6 for t I don't get 15.

 

[pasmith]

Ok so I've become aware that I need to solve for the constant of proportionality k.

dp/dt = k(20-p)

The price rises $5 in 6 months, so dp/dt=5/6 when p=10

In general $p'(6) \neq \dfrac{p(6) - p(0)}{6}$.

You need to solve
$$\frac{dp}{dt} = k(20 - p)$$subject to $p(0) = 10$, and then find $k$ from the condition that $p(6) = 15$.