Answer:
Step-by-step explanation:
[tex]x = \dfrac{K}{1-Ce^{akt}}[/tex] and [tex]y =Pe^{\dfrac{-B}{a}(ln|\dfrac{ 1}{ 1-Ce^{-aKt}}|)}[/tex]
Step-by-step explanation:
One approach to solve this problem is first to solve [tex]\dfrac{dx}{dt} = ax(x-K)[/tex] for x in terms of t and then substitute this answere in [tex]\dfrac{dy}{dt}=- Bxy[/tex].
Starting with
[tex]\dfrac{dx}{dt} =ax(x-K)[/tex], we can use the separation of variables method.
[tex]\dfrac{dx}{ax(x-K)} = dt[/tex], integrating
[tex]\int \dfrac{dx}{ax(x-K)} = \int dt[/tex]
[tex]\int \dfrac{dx}{ax(x- K)} = t + c[/tex]
For solving the integral [tex]\int \dfrac{dx}{ax(x- K)} [/tex] we use partial fraction decomposition method:
[tex]\int \dfrac{1}{ax(x- K)}dx = \int (\frac{A}{ax} +\frac{B}{x- K} ) dx[/tex]
this leads us to the equation:
[tex]1=aBx +Ax -AK[/tex], as it can be seen, we can use this to construc a system of equations
from 1. we get [tex]A=\frac{-1}{K}[/tex] and, from 2. we get [tex]B= \dfrac{1}{aK}[/tex].
Now
[tex]\int \dfrac{1}{ax(x- K)}dx =\int (\frac{-1/K}{ax} + \frac{1/aK}{x- K} ) dx [/tex]
so, we get:
[tex]\int (\frac{-1/K}{ax} + \frac{1/aK}{x- K}) dx = t+ c[/tex]
[tex]\dfrac{-1}{aK}\int \frac{1}{x}dx + \dfrac{1}{aK}\int\frac{1}{x- K} dx = t+ c[/tex]
Now we end up with some easily solving integrals
[tex]\dfrac{-1}{aK}ln|x|+ \dfrac{1}{aK}ln|x-K| + m = t+ c[/tex], m is the integration cosntant.
[tex]\dfrac{1}{aK}ln|\dfrac{x-K}{x}| + m = t+ c[/tex]
[tex]ln|\dfrac{x-K}{x}| = aKt+ c_{1}[/tex]
[tex]|\dfrac{x-K}{x}| = e^{aKt+ c_{1}}[/tex]
[tex]|x-K| = |x|e^{aKt+ c_{1}}[/tex], taking in count only positive values for x and K
[tex]x-K = xe^{aKt+ c_{1}}[/tex]
then after solving for x we end up with the answer:
[tex]x = \dfrac{K}{1-Ce^{aKt}}[/tex].
Now for [tex]\dfrac{dy}{dt} = - Bxy[/tex], we subtitute our last result and get
[tex]\dfrac{dy}{dt} = - By \dfrac{K}{1-Ce^{aKt}}[/tex], by using the separation of variables method:
[tex]\dfrac{dy}{y} = \dfrac{-BK}{1-Ce^{aKt}}dt[/tex], integrating
[tex]ln|y| +p = \int \dfrac{-BK}{1-Ce^{aKt}}dt[/tex],
using the substitution method and letting
[tex]u = 1-Ce^{aKt}[/tex], our integral takes the form
[tex]ln|y| + p=\dfrac{-B}{a}\int \dfrac{du}{u(u-1)}[/tex], using partial fraction decomposition method we end up with
[tex]ln|y| + p =\dfrac{-B}{a}\int(\frac{G}{u}+\frac{F}{u-1})du[/tex], leading us to
From this: [tex]G=-1[/tex] and, [tex]F=1[/tex], thus
[tex]ln|y| + p =\dfrac{-B}{a}\int(\frac{-1}{u}+\frac{1}{u-1})du[/tex] , integrating
[tex]ln|y| + p_{1} =\dfrac{-B}{a}(-ln|u|+ln|u-1|)[/tex]
[tex]ln|y| + p_{1} =\dfrac{-B}{a}(ln|\dfrac{u-1}{u}|)[/tex] , remembering
[tex]u=1-Ce^{aKt}[/tex] , we get
[tex]ln|y|+p_{1} =\dfrac{-B}{a}(ln|\dfrac{ -Ce^{aKt}}{ 1-Ce^{aKt}}|)[/tex]
[tex]ln|y|=\dfrac{-B}{a}(ln|\dfrac{ 1}{ 1-Ce^{-aKt}}|)-p_{1}[/tex]
[tex]ln|y|=\dfrac{-B}{a}(ln|\dfrac{ 1}{ 1-Ce^{-aKt}}|)-p_{1}[/tex]
[tex]|y|=e^{\dfrac{-B}{a}(ln|\dfrac{ 1}{ 1-Ce^{-aKt}}| )-p_{1}[/tex]
[tex]|y|=Pe^{\dfrac{-B}{a}(ln|\dfrac{ 1}{ 1-Ce^{-aKt}}|)}[/tex], for y positive the answer is:
[tex]y =Pe^{\dfrac{-B}{a}(ln|\dfrac{ 1}{ 1-Ce^{-aKt}}|)}[/tex]
