Respuesta :
Answer:
[tex]P(t) = \displaysyle\frac{VF}{r}\bigg(1 - e^{\frac{-rt}{V}}\bigg) + Ae^{\frac{-rt}{V}}[/tex]
Step-by-step explanation:
We are given that pollution in a lake is given by the differential equation:
[tex]\displaystyle\frac{dP(t)}{dt} = -\displaystyle\frac{r}{V}P(t) + F[/tex]
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:
[tex]\displaystyle\frac{dP(t)}{dt} + \displaystyle\frac{r}{V}P(t) = F[/tex]
Comparing to linear differential equation:
[tex]\displaystyle\frac{dP(t)}{dt} = a(t)P(t) + b(t)[/tex],
we get,
[tex]a(t) = \displaystyle\frac{r}{V}, b(t) = F[/tex]
Integrating factor:
[tex]e^{\int a(t)dt} = e^{\frac{r}{V}dt} = e^{\frac{rt}{V}}[/tex]
Solution:
[tex]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[/tex]
[tex]P(t)e^{\frac{rt}{V}}= \displaysyle\frac{VF}{r}e^{\frac{rt}{V}} + C[/tex],
where C is the constant of integration.
Now, we are given that P(0) = A
putting these value in the above equation, we get,
[tex]A = \displaystyle\frac{VF}{r} + C\\\\C = A - \displaystyle\frac{VF}{r}[/tex]
Putting this value of C in equation, we get:
[tex]P(t)e^{\frac{rt}{V}} = \displaysyle\frac{VF}{r}e^{\frac{rt}{V}} + A - \displaystyle\frac{VF}{r}[/tex]
Dividing the equation by [tex]e^{\frac{rt}{V}}[/tex], we get:
[tex]P(t) = \displaysyle\frac{VF}{r} + \bigg(A - \displaystyle\frac{VF}{r}\bigg)e^{\frac{-rt}{V}}[/tex]
[tex]P(t) = \displaysyle\frac{VF}{r}\bigg(1 - e^{\frac{-rt}{V}}\bigg) + Ae^{\frac{-rt}{V}}[/tex]
