Partial fractions ( part of a logistic equation)

[adzix]

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)

 

[Prove It]

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

 

[Joppy]

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,

\(\displaystyle x' = rx - rx^2\).

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,

\(\displaystyle x' = rx - rx^2 \implies x^{-2} x' = rx^{-1} - r\).

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.

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

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 $u = x^{-1}$,

$x = \dfrac{1}{1 + ce^{-rt}}.$