Question

Pollution in a Lake. Consider a lake with a stream flowing out of it and a nearby factory dumping pollution into it. Without worrying about the modeling details yet, the mass of pollution in the lake is represented by dP dt = − r/V P + F, where P is the mass of pollution in the lake (kg), r is the flow rate of the stream (m3/sec), V is the volume of the lake (m3) and F is the mass of pollution dumped per time by the factory (kg/sec). Treat r, V and F as constants. Use an integrating factor to solve the ODE in general form. Find the constant of integration if the initial condition is P(0) = A, and write the solution for P(t) in a clean, final form.

Answer 1

Answer:

$P(t) = \displaysyle\frac{VF}{r}\bigg(1 - e^{\frac{-rt}{V}}\bigg) + Ae^{\frac{-rt}{V}}$

Step-by-step explanation:

We are given that pollution in a lake is given by the differential equation:

$\displaystyle\frac{dP(t)}{dt} = -\displaystyle\frac{r}{V}P(t) + F$

where, P(t) is the pollution at time t, r is the flow rate, V is the volume of lake and F is the mass of pollution dumped.

The given differential equation can be written as:

$\displaystyle\frac{dP(t)}{dt} + \displaystyle\frac{r}{V}P(t) = F$

Comparing to linear differential equation:

$\displaystyle\frac{dP(t)}{dt} = a(t)P(t) + b(t)$,

we get,

$a(t) = \displaystyle\frac{r}{V}, b(t) = F$

Integrating factor:

$e^{\int a(t)dt} = e^{\frac{r}{V}dt} = e^{\frac{rt}{V}}$

Solution:

$P(t)\text{Integrating Factor} = \int b(t)\text{Integrating Factor} + C\\\\P(t)e^{\frac{rt}{V}}= \int Fe^{\frac{rt}{V}}dt + C$

$P(t)e^{\frac{rt}{V}}= \displaysyle\frac{VF}{r}e^{\frac{rt}{V}} + C$,

where C is the constant of integration.

Now, we are given that P(0) = A

putting these value in the above equation, we get,

$A = \displaystyle\frac{VF}{r} + C\\\\C = A - \displaystyle\frac{VF}{r}$

Putting this value of C in equation, we get:

$P(t)e^{\frac{rt}{V}} = \displaysyle\frac{VF}{r}e^{\frac{rt}{V}} + A - \displaystyle\frac{VF}{r}$

Dividing the equation by $e^{\frac{rt}{V}}$, we get:

$P(t) = \displaysyle\frac{VF}{r} + \bigg(A - \displaystyle\frac{VF}{r}\bigg)e^{\frac{-rt}{V}}$

$P(t) = \displaysyle\frac{VF}{r}\bigg(1 - e^{\frac{-rt}{V}}\bigg) + Ae^{\frac{-rt}{V}}$