Question

9) Given dx/dt = ax(x – K) & dy/dt = - Bxy, solve for:

Answer 1

Answer:

Step-by-step explanation:

$x = \dfrac{K}{1-Ce^{akt}}$ and $y =Pe^{\dfrac{-B}{a}(ln|\dfrac{ 1}{ 1-Ce^{-aKt}}|)}$

Step-by-step explanation:

One approach to solve this problem is first to solve  $\dfrac{dx}{dt} = ax(x-K)$ for x in terms of t and then substitute this answere in $\dfrac{dy}{dt}=- Bxy$.

Starting with

 $\dfrac{dx}{dt} =ax(x-K)$, we can use the separation of variables method.

$\dfrac{dx}{ax(x-K)} = dt$, integrating

$\int \dfrac{dx}{ax(x-K)} = \int dt$

$\int \dfrac{dx}{ax(x- K)} = t + c$

For solving the integral $\int \dfrac{dx}{ax(x- K)}$ we use partial fraction decomposition method:

$\int \dfrac{1}{ax(x- K)}dx = \int (\frac{A}{ax} +\frac{B}{x- K} ) dx$

this leads us to the equation:

$1=aBx +Ax -AK$, as it can be seen, we can use this to construc a system of equations

1. $1 =-AK$ and,
2. $0=aBx+AX =aB + A$

from 1. we get $A=\frac{-1}{K}$ and, from 2. we get $B= \dfrac{1}{aK}$.

Now

$\int \dfrac{1}{ax(x- K)}dx =\int (\frac{-1/K}{ax} + \frac{1/aK}{x- K} ) dx$

so, we get:

$\int (\frac{-1/K}{ax} + \frac{1/aK}{x- K}) dx = t+ c$

$\dfrac{-1}{aK}\int \frac{1}{x}dx + \dfrac{1}{aK}\int\frac{1}{x- K} dx = t+ c$

Now we end up with some easily solving integrals

$\dfrac{-1}{aK}ln|x|+ \dfrac{1}{aK}ln|x-K| + m = t+ c$, m is the integration cosntant.

$\dfrac{1}{aK}ln|\dfrac{x-K}{x}| + m = t+ c$

$ln|\dfrac{x-K}{x}| = aKt+ c_{1}$

$|\dfrac{x-K}{x}| = e^{aKt+ c_{1}}$

$|x-K| = |x|e^{aKt+ c_{1}}$, taking in count only positive values for x and K

$x-K = xe^{aKt+ c_{1}}$

then after solving for x we end up with the answer:

$x = \dfrac{K}{1-Ce^{aKt}}$.

Now for $\dfrac{dy}{dt} = - Bxy$, we subtitute our last result and get

$\dfrac{dy}{dt} = - By \dfrac{K}{1-Ce^{aKt}}$, by using the separation of variables method:

$\dfrac{dy}{y} = \dfrac{-BK}{1-Ce^{aKt}}dt$, integrating

$ln|y| +p = \int \dfrac{-BK}{1-Ce^{aKt}}dt$,

using the substitution method and letting

$u = 1-Ce^{aKt}$, our integral takes the form

$ln|y| + p=\dfrac{-B}{a}\int \dfrac{du}{u(u-1)}$, using partial fraction decomposition method we end up with

$ln|y| + p =\dfrac{-B}{a}\int(\frac{G}{u}+\frac{F}{u-1})du$, leading us to

1. $-G=1$
2. $G+ F =0$

From this: $G=-1$ and, $F=1$, thus

$ln|y| + p =\dfrac{-B}{a}\int(\frac{-1}{u}+\frac{1}{u-1})du$ , integrating

$ln|y| + p_{1} =\dfrac{-B}{a}(-ln|u|+ln|u-1|)$

$ln|y| + p_{1} =\dfrac{-B}{a}(ln|\dfrac{u-1}{u}|)$ , remembering

$u=1-Ce^{aKt}$ , we get

$ln|y|+p_{1} =\dfrac{-B}{a}(ln|\dfrac{ -Ce^{aKt}}{ 1-Ce^{aKt}}|)$

$ln|y|=\dfrac{-B}{a}(ln|\dfrac{ 1}{ 1-Ce^{-aKt}}|)-p_{1}$

$ln|y|=\dfrac{-B}{a}(ln|\dfrac{ 1}{ 1-Ce^{-aKt}}|)-p_{1}$

$|y|=e^{\dfrac{-B}{a}(ln|\dfrac{ 1}{ 1-Ce^{-aKt}}| )-p_{1}$

$|y|=Pe^{\dfrac{-B}{a}(ln|\dfrac{ 1}{ 1-Ce^{-aKt}}|)}$, for y positive the answer is:

$y =Pe^{\dfrac{-B}{a}(ln|\dfrac{ 1}{ 1-Ce^{-aKt}}|)}$