# Surplus of Y on the wrong side

1. Jan 25, 2005

### Yann

How can the following type of equation revolved ? I don't know how to deal with the surplus of Y on the "wrong" side :grumpy:

dY/dx = aY(1-Y/b)

a and b are constant. And when X = 0, Y = something.

I would really like to know how this kind of equation can be resolved

2. Jan 25, 2005

### AKG

Separation of Variables:

$$\frac{dY}{dx} = \left (\frac{a}{b}\right ) \left (Y(1 - Y)\right )$$

$$\frac{dY}{Y(1 - Y)} = \frac{a}{b}dx$$

$$\int \frac{dY}{Y(1 - Y)} = \int \frac{a}{b}dx$$

$$\int \left (\frac{A}{Y} + \frac{B}{1 - Y}\right )dY = \frac{a}{b}x + C$$

$$\int \left (\frac{A}{Y} + \frac{B}{1 - Y}\right )dY = \frac{a}{b}x + C$$

where A and B are constants that need to be solved for. It's easy to see that A = B = 1, so:

$$\int \left (\frac{1}{Y} - \frac{1}{Y - 1}\right )dY = \frac{a}{b}x + C$$

$$\ln (Y) - \ln (Y - 1) = \frac{a}{b}x + C$$

$$\ln \right (\frac{Y}{Y - 1}\right ) = \frac{a}{b}x + C$$

$$\frac{Y}{Y - 1} = \exp \left (\frac{a}{b}x + C\right )$$

$$\frac{Y}{Y - 1} = De^{\frac{a}{b}x}$$

where $D = e^C$. Some algebra gets you to:

$$Y = \frac{D}{D - \exp (-\frac{ax}{b})}$$

There may be some cases where this solution is not valid, i.e. in the steps above, I may have divided by zero if Y = 1 or Y = 0 in some places, you can check the algebra. In fact, the above solution only holds in the case that Y is neither 0 or 1, since Y = 0 and Y = 1 are solutions on their own. Note:

If Y = 0 or Y = 1, it is a constant, so dY/dx = 0. Also, it is clear that aY(1-Y)/b = 0 for those Y values, as we would expect.

Now, you say that you have, "when X = 0, Y = something." This is an initial value problem. With this, you can solve for D explicitly (in terms of a, b, and other constants).

3. Jan 26, 2005

### Yann

But it is not...
$$\frac{dY}{dx} = \left a \left (Y\frac{(1 - Y)}{b}\right )$$

It is...
$$\frac{dY}{dx} = \left a \left (Y \left (1 - \frac{Y}{b} \right ) \right )$$

...is there a place where i could get explanations on that step;

$$\int \frac{dY}{Y(1 - Y)} = \int \left (\frac{A}{Y} + \frac{B}{1 - Y}\right )dY$$

And thank you for your help, i think theses problems (of mathematical ecology) are a little too advanced for my level but it's very interesting. Is there a good book on calculus (that can explain these kind of problems) you recommend ?

4. Jan 26, 2005

### dextercioby

$$\frac{dY}{dx}=aY-\frac{a}{b}Y^{2}$$
and can be inetgrated by separation of variables
$$\frac{b}{a}\int \frac{dY}{Y(b-Y)} =\int dx$$

Pay attention with dividing through 0...

Daniel.

5. Jan 26, 2005

### AKG

Oops, my mistake. Anyways, if you're familiar with separation of variables, then you should be able to do that.
That uses a technique of integration known as integration by partial fractions. The book I had for class last year was called, I believe, "Calculus" 5th or 6th edition by James Stewart. There are two versions, one of them is multivariable (with a red/purple violin on the front) and a single variable one (with a greenish violin). Get the multivariable one. Chapter 7 contains techniques of integration. That book covers basic introductory calculus, integration, functions of several variables, vector calculus, first and second order homogenous and non-homogeneous linear differential equations, techniques of integration and differentiation, applications, etc. It's a good book.

6. Jan 27, 2005

### Yann

Thank you AKG and dextercioby, i should be ok.

Only now i'm addicted to TeX (which seem to be a good thing), is there a Word-like program to write text along with TeX like on this forum ?

7. Jan 27, 2005

### AKG

8. Jan 28, 2005

### Yann

I'm still stuck... the worst is that i know what the answer is, even if i have no clue how to get to it.

$$\frac{dY}{dx} = \left a \left (Y \left (1 - \frac{Y}{b} \right ) \right )$$

$$\frac{dY}{\left Y (1 - \frac{Y}{b} \right )} \right = adx$$

$$\int \frac{dY}{\left Y (1 - \frac{Y}{b} \right )} \right = a \int dx$$

$$\int \frac{1}{Y} - \frac{1}{Y-b}dY \right = ax + C$$

$$\int \frac{dY}{Y} - \int \frac{dY}{Y-b} \right = ax + C$$

$$\ln Y - \ln (Y-b) \right = ax + C$$

$$\ln \frac{Y}{Y-b} \right = ax + C$$

If when $$X = 0, Y = Y_{0}$$ then;

$$C = \ln \frac{Y_{0}}{Y_{0} - b}$$

$$\ln \frac{Y}{Y-b} \right = ax + \ln \frac{Y_{0}}{Y_{0} - b}$$

$$e^{\ln \frac{Y}{Y-b}} \right = e^{ax + \ln \frac{Y_{0}}{Y_{0} - b}}$$

$$\frac{Y}{Y-b} \right = e^{ax}+ \frac{Y_{0}}{Y_{0} - b}$$

In theory, it's possible to get from there and to go to ;

$$Y(x) = \frac {Y_{0} b e^{ax}}{b + Y_{0} \left (e^{ax} - 1 \right )}$$

9. Jan 28, 2005

### AKG

The first thing to take care of are the "special" solutions, which you can find directly by analyzing the first equation. Note that if Y = b or Y = 0 (as constant functions, so Y(x) = b or 0 for all x), then clearly dY/dx = 0, and also the right side is zero, so these are solutions. Now, consider solutions where Y is never b nor 0.

Now, I'm not sure if you're second last line is a typo or not, but it should not be:

$$e^{ax}\mathbf{+}\frac{Y_0}{Y_0 - b}$$

rather, it should be:

$$e^{ax}\mathbf{\times }\frac{Y_0}{Y_0 - b}$$

So:

$$Y = \frac{b}{1 - \exp (-ax)(\frac{Y_0 - b}{Y_0})}$$