MHB Partial fractions ( part of a logistic equation)

adzix
Messages
1
Reaction score
0
Hi everyone, I am stuck on a problem. I need to give a partial fraction of 1/N(k-N). I have tried every method so far ( plotting roots, systems of equations). I think I found A=1/k but I have no clue how to find B value. I would really appreciate any help as I am a desperate student trying to pass the class. (the initial problem is dN/dt=(r/K)N(K-N), N(0)=No)
 
Physics news on Phys.org
adzix said:
Hi everyone, I am stuck on a problem. I need to give a partial fraction of 1/N(k-N). I have tried every method so far ( plotting roots, systems of equations). I think I found A=1/k but I have no clue how to find B value. I would really appreciate any help as I am a desperate student trying to pass the class. (the initial problem is dN/dt=(r/K)N(K-N), N(0)=No)

$\displaystyle \begin{align*} \frac{A}{N} + \frac{B}{K - N} &\equiv \frac{1}{N \left( K - N \right) } \\ A \left( K - N \right) + B \,N &\equiv 1 \end{align*}$

Let $\displaystyle \begin{align*} N = 0 \end{align*}$ to find $\displaystyle \begin{align*} A\,K = 1 \implies A = \frac{1}{K} \end{align*}$

Let $\displaystyle \begin{align*} N = K \end{align*}$ to find $\displaystyle \begin{align*} B \, K = 1 \implies B = \frac{1}{K} \end{align*}$

so the partial fraction decomposition is $\displaystyle \begin{align*} \frac{1}{K \, N } + \frac{1}{K \left( K - N \right) } \end{align*}$.
 
Although not asked for, it might be insightful for the OP to notice that the logistic equation is in fact a special case of the Bernoulli equation, and can be solved without partial fractions! However it would be advisable to ignore if you (the OP) have not yet covered this :).

The Bernoulli equation is of the form:

$\begin{equation}y' = f(x)y^a - g(x)y \end{equation}$

For convenience, let's jot down the logistic equation also,

$$x' = rx - rx^2$$.

Quite a resemblance, isn't there? Indeed, the Bernoulli equation with $a = 2$ and $f(x) = g(x) = -r$ is the logistic equation, also known as the Riccati equation.

Returning to the logistic equation and dividing through by $x^2$ gives,

$$x' = rx - rx^2 \implies x^{-2} x' = rx^{-1} - r$$.

Now we can make a (perhaps) devious substitution, $u = x^{-1}$.

It follows that, $u' = -x^{-2}x'$.

But we know what $x'$ is, so we can expand this differential equation in $u$,

$u' = -x^{-2}(rx - rx^2) = -r(x^{-1} - 1)$.

Thus, $u' + ru = r ... (1)$

Aha! The differential equation is now in it's linear form. We can use an integrating factor to finish off.

$$I(t) = e^{\int r dt} = e^{rt}.$$

Multiply (1) by $I(t)$,

$e^{rt}u' + e^{rt} ru = e^{rt} r$

We notice the following identity on the left hand side,

$\dfrac{d}{dt}(ue^{rt}) \equiv u'e^{rt} + rue^{rt}$, now we can integrate.$\int \frac{d}{dt}(ue^{rt}) dt = \int re^{rt} dt$

$ue^{rt} = e^{rt} + c$

$u = 1 + ce^{-rt}$

Recall that $u = x^{-1}$,

$x = \dfrac{1}{1 + ce^{-rt}}.$
 
For original Zeta function, ζ(s)=1+1/2^s+1/3^s+1/4^s+... =1+e^(-slog2)+e^(-slog3)+e^(-slog4)+... , Re(s)>1 Riemann extended the Zeta function to the region where s≠1 using analytical extension. New Zeta function is in the form of contour integration, which appears simple but is actually more inconvenient to analyze than the original Zeta function. The original Zeta function already contains all the information about the distribution of prime numbers. So we only handle with original Zeta...

Similar threads

Replies
8
Views
2K
Replies
1
Views
2K
Replies
2
Views
2K
Replies
1
Views
2K
Replies
3
Views
6K
Replies
13
Views
2K
Back
Top