economics using differential equations

Question

I need help with this calculus problem

The producer of a certain commodity determines that to protect profits, the price p should decrease at a rate equal to half the inventory surplus $S−D$, where $S$ and $D$ are respectively the supply and demand for the commodity. Suppose the supply and demand vary with price in such a way that $S(p)=80+3p$ and $D(p)=140−2p$ and that the price is $3$ dollars per unit when $t=0$. Determine $p(t)$.

Can anyone show me how they solve ?

Answer 1

Let's phrase the problem mathematically. You are being asked to solve::

$\frac{dp}{dt} = -\frac{1}{2} (S(p)-D(t))$

So substituting in the given supply and demand equations:

$\frac{dp}{dt} = -\frac{1}{2} (5p-60)$

Separating the variables:

$\frac{dp}{p-12} = -\frac{5}{2} dt$

Integrating both sides, we get the general result:

$\ln|p-12| = -\frac{5}{2}t+C$

...where C is a constant.

Exponentiate get both sides to tidy things up:

$|p-12|=Ae^{-\frac{5}{2}t}$

For some constant $A$. But you were given that $p$ was $3$ when $t$ was $0$.

So $A=-9$ and our final answer:

$p(t)=-9e^{-\frac{5}{2}t}+12$

Answer 2

First step is to solve for the rate of $p$, which is $p'$. $$S-D=80+3p-(140-2p)=5p-60$$$$\frac{S-D}{2}=\frac{5p-60}{2}$$ Since $p$ is decreasing at the rate, $p'$ needs to be negative.So$$p'=\frac{60-5p}{2}$$ Now it's easy to get $$2p'+5p=60$$It's a non-homogeneous first-order differential equation. $$2D+5=0,D=-\frac52$$Therefore you get the general solution$$p_c(t)=c_1e^{-\frac52t}$$ Now let's do the particular solution: $p_o(t)=A$ where A is a constant,$$A'=0,2(0)+5A=60,A=12$$ So $$p_p(t)=12$$Therefore,$$p(t)=p_c(t)+p_p(t)=c_1e^{-\frac52t}+12$$ Plug in the initial value we get $$p(0)=c_1+12=3$$$$c_1=-9$$ So $$p(t)=-9e^{-\frac52t}+12$$

Answer 3

First we translate:

$$\frac{dP}{dt}=-\frac{1}{2}(S(P)-D(p))$$ Then we substitute: $$\frac{dP}{dt}=-\frac{1}{2}(5P-60) $$

But now you have a first order differential equation.

hint: You can seperate them. you're going to have something involving $e$, and something constant, involving your initial condition.

Here are some notes