Question

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

A) x in terms of t. Ans. B) y in terms of t. Ans.

Answer 1

Explanation:

`x = (K)/(1-Ce^(akt))` and
`y =Pe^)`

Explanation:

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

Starting with

`(dx)/(dt) =ax(x-K)`, we can use the separation of variables method.

`(dx)/(ax(x-K)) = dt`, integrating

`\int (dx)/(ax(x-K)) = \int dt`

`\int (dx)/(ax(x- K)) = t + c`

For solving the integral
`\int (dx)/(ax(x- K))` we use partial fraction decomposition method:

`\int (1)/(ax(x- K))dx = \int ((A)/(ax) +(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=(-1)/(K)` and, from 2. we get
`B= (1)/(aK)`.

Now

`\int (1)/(ax(x- K))dx =\int ((-1/K)/(ax) + (1/aK)/(x- K) ) dx`

so, we get:

`\int ((-1/K)/(ax) + (1/aK)/(x- K)) dx = t+ c`

`(-1)/(aK)\int (1)/(x)dx + (1)/(aK)\int(1)/(x- K) dx = t+ c`

Now we end up with some easily solving integrals

`(-1)/(aK)ln|x|+ (1)/(aK)ln|x-K| + m = t+ c`, m is the integration cosntant.

`(1)/(aK)ln|(x-K)/(x)| + m = t+ c`

`ln|(x-K)/(x)| = aKt+ c_(1)`

`|(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 = (K)/(1-Ce^(aKt))`.

Now for
`(dy)/(dt) = - Bxy`, we subtitute our last result and get

`(dy)/(dt) = - By (K)/(1-Ce^(aKt))`, by using the separation of variables method:

`(dy)/(y) = (-BK)/(1-Ce^(aKt))dt`, integrating

`ln|y| +p  = \int (-BK)/(1-Ce^(aKt))dt`,

using the substitution method and letting

`u = 1-Ce^(aKt)`, our integral takes the form

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

`ln|y| + p =(-B)/(a)\int((G)/(u)+(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 =(-B)/(a)\int((-1)/(u)+(1)/(u-1))du` , integrating

`ln|y| + p_(1) =(-B)/(a)(-ln|u|+ln|u-1|)`

`ln|y| + p_(1) =(-B)/(a)(ln|(u-1)/(u)|)` , remembering

`u=1-Ce^(aKt)` , we get

`ln|y|+p_(1) =(-B)/(a)(ln|( -Ce^(aKt))/( 1-Ce^(aKt))|)`

`ln|y|=(-B)/(a)(ln|( 1)/( 1-Ce^(-aKt))|)-p_(1)`

`ln|y|=(-B)/(a)(ln|( 1)/( 1-Ce^(-aKt))|)-p_(1)`

`|y|=e^{(-B)/(a)(ln|( 1)/( 1-Ce^(-aKt))| )-p_(1)`

`|y|=Pe^(-B)/(a)(ln`, for y positive the answer is:

`y =Pe^( 1)/( 1-Ce^(-aKt))`