# Math Help - Differential equation / Seperable

1. ## Differential equation / Seperable

Hello!

I have been given an equation as following;

$\frac{dy}{dt}=ay(A-y)$

where
$y>0$
$A$ and $a$ are positive constants

1)
I am then to find the constants $\alpha$ and $\beta$ so the equation can be given as this:

$\frac{1}{y(A-y)}= \frac{\alpha}{y}+ \frac{\beta}{A-y)}$

2)
And at last find a general solution when $y(0)= \frac{A}{2}$

Don't know where to start actually...

Any help is greatly appreciated!

2. This is just the partial fraction decomposition you learned in integral calculus.

For

Multiplty both sides by y(A-y)

Then let y = A to find beta =1/A and y = 0 to find alpha = 1/A
To solve separate the variables and integrate

3. If we separate variables and integrate, we get:

$y=\frac{Ae^{aAt}}{e^{aAt}+AC}$

Using the initial condition, y(0)=A/2, gives C=1/A.

Which results in:

$y=\frac{Ae^{aAt}}{e^{aAt}+1}$

4. A bit heavy for me atm, but I think I see how you two did it. Struggling a bit with DE...

If I could trouble you with an addition to this calculation, as for what happens to $y$ when $t\rightarrow \infty$?

How to explain/show is my concern...

5. Hello,
Originally Posted by jokke22
A bit heavy for me atm, but I think I see how you two did it. Struggling a bit with DE...

If I could trouble you with an addition to this calculation, as for what happens to $y$ when $t\rightarrow \infty$?

How to explain/show is my concern...
$y=\frac{Ae^{aAt}}{e^{aAt}+1}=\frac{A}{1+e^{-aAt}}$, by dividing both numerator and denominator by $e^{aAt}$

And since aA is positive, -aA is negative.

We know that $\lim_{x\to\infty} e^{-x}=0$

Hence the limit is...

6. This is the logistic equation, correct?. Here is a general tutorial to discuss the solutions of the logistic equation.

Let's start from scratch. Okey-doke.

$\frac{dy}{dt}=a(A-y)y$ for y>0. A and a are constants.....[1]

Separate variables:

$\frac{dy}{(A-y)y}=adt$

Integrate:

$\int\frac{1}{(A-y)y}dy=a\int dt+C_{1}$.....[2]

Where $C_{1}$ is a constant of integration. By use of partial

fractions as Calculus26 showed:

$\frac{1}{(A-y)y}=\frac{1}{A}\left(\frac{1}{A-y}+\frac{1}{y}\right)$

Therefore, [2] becomes:

$\frac{1}{A}\left(\int \frac{dy}{A-y}+\int\frac{dy}{y}\right)=a\int dt+C_{1}$

or $\frac{1}{A}\left(-ln|A-y|+ln|y|\right)+at+C_{1}$

when we integrate.

Therefore, $ln\left|\frac{A-y}{y}\right|=-Aat-AC_{1}$.

Exponentiating gives:

$\left|\frac{A-y}{y}\right|=Ce^{-Aat}$

where $C=e^{-AC_{1}}>0$ is now our arbitrary constant.

Since y>0 by our assumption, if we also assume y<A, the absolute

value signs are done with.

$\left|\frac{A-y}{y}\right|=-\left(\frac{A-y}{y}\right)$

We can also consider either case by simply letting the arbitrary cnstant C

be positive or negative.

Therefore, $\frac{A-y}{y}=Ce^{-Aat}$.

Solving for y:

$y(t)=\frac{A}{1+Ce^{-Aat}}$.....[3]

which is the general solution of [1].

To evaluate C we need the initial conditions:

$y(0)=a_{0}=\frac{A}{2}$

Sub into [3]:

$y(0)=\frac{A}{1+Ce^{0}}$

$\frac{A}{2}=\frac{A}{1+C}$

$C=1$

Hence, thus, and therefore,

$y(t)=\frac{A}{1+e^{-Aat}}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now, you asked what happens to y when t-->infinity.

Suppose a>0 and A>0. If y(t) represents population then certain models

say that $a(A-y)=\text{birth rate-death rate}$

If we have a small population(small y), then $a(A-y)>0$

and the birth rate exceeds the death rate. Thus, the population is growing.

We take note that:

$\lim_{t\to \infty}y(t)=\frac{A}{1+(\frac{A}{a_{0}}+1)\display style\lim_{t\to \infty}e^{-Aat}}=A$

Since $\lim_{t\to \infty}e^{-Aat}=0$

This is a curious result since it says the population approaches a limiting

value A independent of the initial population size we call

$a_{0}$

A is called the limiting population that the syatem can support. If

$a_{0}>A$, the population decreases.
\
If $a_{0} the population increases.

We can also take note that when y=A, [1] becomes y'=0 and we are at

equilibrium(birth rate=death rate).

Here is a haphazard 'paint' graph and I hope this little tutorial helps some to understand it.

7. That was some heavy explaining! Appreciate the effort GREATLY and will look it over asap!