Answer:
Step-by-step explanation:
and ![y =Pe^{\dfrac{-B}{a}(ln|\dfrac{ 1}{ 1-Ce^{-aKt}}|)}](https://tex.z-dn.net/?f=y%20%3DPe%5E%7B%5Cdfrac%7B-B%7D%7Ba%7D%28ln%7C%5Cdfrac%7B%201%7D%7B%201-Ce%5E%7B-aKt%7D%7D%7C%29%7D)
Step-by-step explanation:
One approach to solve this problem is first to solve
for x in terms of t and then substitute this answere in
.
Starting with
, we can use the separation of variables method.
, integrating
![\int \dfrac{dx}{ax(x-K)} = \int dt](https://tex.z-dn.net/?f=%5Cint%20%5Cdfrac%7Bdx%7D%7Bax%28x-K%29%7D%20%3D%20%5Cint%20dt)
![\int \dfrac{dx}{ax(x- K)} = t + c](https://tex.z-dn.net/?f=%5Cint%20%5Cdfrac%7Bdx%7D%7Bax%28x-%20K%29%7D%20%3D%20t%20%2B%20c)
For solving the integral
we use partial fraction decomposition method:
![\int \dfrac{1}{ax(x- K)}dx = \int (\frac{A}{ax} +\frac{B}{x- K} ) dx](https://tex.z-dn.net/?f=%5Cint%20%5Cdfrac%7B1%7D%7Bax%28x-%20K%29%7Ddx%20%3D%20%5Cint%20%28%5Cfrac%7BA%7D%7Bax%7D%20%2B%5Cfrac%7BB%7D%7Bx-%20K%7D%20%29%20dx)
this leads us to the equation:
, as it can be seen, we can use this to construc a system of equations
and,![0=aBx+AX =aB + A](https://tex.z-dn.net/?f=0%3DaBx%2BAX%20%3DaB%20%2B%20A)
from 1. we get
and, from 2. we get
.
Now
![\int \dfrac{1}{ax(x- K)}dx =\int (\frac{-1/K}{ax} + \frac{1/aK}{x- K} ) dx](https://tex.z-dn.net/?f=%5Cint%20%5Cdfrac%7B1%7D%7Bax%28x-%20K%29%7Ddx%20%3D%5Cint%20%28%5Cfrac%7B-1%2FK%7D%7Bax%7D%20%2B%20%5Cfrac%7B1%2FaK%7D%7Bx-%20K%7D%20%29%20dx%20)
so, we get:
![\int (\frac{-1/K}{ax} + \frac{1/aK}{x- K}) dx = t+ c](https://tex.z-dn.net/?f=%5Cint%20%28%5Cfrac%7B-1%2FK%7D%7Bax%7D%20%2B%20%5Cfrac%7B1%2FaK%7D%7Bx-%20K%7D%29%20dx%20%3D%20t%2B%20c)
![\dfrac{-1}{aK}\int \frac{1}{x}dx + \dfrac{1}{aK}\int\frac{1}{x- K} dx = t+ c](https://tex.z-dn.net/?f=%5Cdfrac%7B-1%7D%7BaK%7D%5Cint%20%5Cfrac%7B1%7D%7Bx%7Ddx%20%2B%20%5Cdfrac%7B1%7D%7BaK%7D%5Cint%5Cfrac%7B1%7D%7Bx-%20K%7D%20dx%20%3D%20t%2B%20c)
Now we end up with some easily solving integrals
, m is the integration cosntant.
![\dfrac{1}{aK}ln|\dfrac{x-K}{x}| + m = t+ c](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7BaK%7Dln%7C%5Cdfrac%7Bx-K%7D%7Bx%7D%7C%20%2B%20m%20%3D%20t%2B%20c)
![ln|\dfrac{x-K}{x}| = aKt+ c_{1}](https://tex.z-dn.net/?f=ln%7C%5Cdfrac%7Bx-K%7D%7Bx%7D%7C%20%3D%20aKt%2B%20c_%7B1%7D)
![|\dfrac{x-K}{x}| = e^{aKt+ c_{1}}](https://tex.z-dn.net/?f=%7C%5Cdfrac%7Bx-K%7D%7Bx%7D%7C%20%3D%20e%5E%7BaKt%2B%20c_%7B1%7D%7D)
, taking in count only positive values for x and K
![x-K = xe^{aKt+ c_{1}}](https://tex.z-dn.net/?f=x-K%20%3D%20xe%5E%7BaKt%2B%20c_%7B1%7D%7D)
then after solving for x we end up with the answer:
.
Now for
, we subtitute our last result and get
, by using the separation of variables method:
, integrating
,
using the substitution method and letting
, our integral takes the form
, using partial fraction decomposition method we end up with
, leading us to
![-G=1](https://tex.z-dn.net/?f=-G%3D1)
![G+ F =0](https://tex.z-dn.net/?f=G%2B%20F%20%3D0)
From this:
and,
, thus
, integrating
, remembering
, we get
![ln|y|+p_{1} =\dfrac{-B}{a}(ln|\dfrac{ -Ce^{aKt}}{ 1-Ce^{aKt}}|)](https://tex.z-dn.net/?f=ln%7Cy%7C%2Bp_%7B1%7D%20%3D%5Cdfrac%7B-B%7D%7Ba%7D%28ln%7C%5Cdfrac%7B%20-Ce%5E%7BaKt%7D%7D%7B%201-Ce%5E%7BaKt%7D%7D%7C%29)
![ln|y|=\dfrac{-B}{a}(ln|\dfrac{ 1}{ 1-Ce^{-aKt}}|)-p_{1}](https://tex.z-dn.net/?f=ln%7Cy%7C%3D%5Cdfrac%7B-B%7D%7Ba%7D%28ln%7C%5Cdfrac%7B%201%7D%7B%201-Ce%5E%7B-aKt%7D%7D%7C%29-p_%7B1%7D)
![ln|y|=\dfrac{-B}{a}(ln|\dfrac{ 1}{ 1-Ce^{-aKt}}|)-p_{1}](https://tex.z-dn.net/?f=ln%7Cy%7C%3D%5Cdfrac%7B-B%7D%7Ba%7D%28ln%7C%5Cdfrac%7B%201%7D%7B%201-Ce%5E%7B-aKt%7D%7D%7C%29-p_%7B1%7D)
![|y|=e^{\dfrac{-B}{a}(ln|\dfrac{ 1}{ 1-Ce^{-aKt}}| )-p_{1}](https://tex.z-dn.net/?f=%7Cy%7C%3De%5E%7B%5Cdfrac%7B-B%7D%7Ba%7D%28ln%7C%5Cdfrac%7B%201%7D%7B%201-Ce%5E%7B-aKt%7D%7D%7C%20%29-p_%7B1%7D)
, for y positive the answer is:
![y =Pe^{\dfrac{-B}{a}(ln|\dfrac{ 1}{ 1-Ce^{-aKt}}|)}](https://tex.z-dn.net/?f=y%20%3DPe%5E%7B%5Cdfrac%7B-B%7D%7Ba%7D%28ln%7C%5Cdfrac%7B%201%7D%7B%201-Ce%5E%7B-aKt%7D%7D%7C%29%7D)