1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Surplus of Y on the wrong side

  1. Jan 25, 2005 #1
    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 :confused:
  2. jcsd
  3. Jan 25, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    Separation of Variables:

    [tex]\frac{dY}{dx} = \left (\frac{a}{b}\right ) \left (Y(1 - Y)\right )[/tex]

    [tex]\frac{dY}{Y(1 - Y)} = \frac{a}{b}dx[/tex]

    [tex]\int \frac{dY}{Y(1 - Y)} = \int \frac{a}{b}dx[/tex]

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

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

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

    [tex]\int \left (\frac{1}{Y} - \frac{1}{Y - 1}\right )dY = \frac{a}{b}x + C[/tex]

    [tex]\ln (Y) - \ln (Y - 1) = \frac{a}{b}x + C[/tex]

    [tex]\ln \right (\frac{Y}{Y - 1}\right ) = \frac{a}{b}x + C[/tex]

    [tex]\frac{Y}{Y - 1} = \exp \left (\frac{a}{b}x + C\right )[/tex]

    [tex]\frac{Y}{Y - 1} = De^{\frac{a}{b}x}[/tex]

    where [itex]D = e^C[/itex]. Some algebra gets you to:

    [tex]Y = \frac{D}{D - \exp (-\frac{ax}{b})}[/tex]

    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).
  4. Jan 26, 2005 #3
    But it is not...
    [tex]\frac{dY}{dx} = \left a \left (Y\frac{(1 - Y)}{b}\right )[/tex]

    It is...
    [tex]\frac{dY}{dx} = \left a \left (Y \left (1 - \frac{Y}{b} \right ) \right )[/tex]

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

    [tex]\int \frac{dY}{Y(1 - Y)} = \int \left (\frac{A}{Y} + \frac{B}{1 - Y}\right )dY [/tex]

    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 ?
  5. Jan 26, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper

    Your equation is
    [tex] \frac{dY}{dx}=aY-\frac{a}{b}Y^{2} [/tex]
    and can be inetgrated by separation of variables
    [tex] \frac{b}{a}\int \frac{dY}{Y(b-Y)} =\int dx [/tex]

    Pay attention with dividing through 0...

  6. Jan 26, 2005 #5


    User Avatar
    Science Advisor
    Homework Helper

    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.
  7. Jan 27, 2005 #6
    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 ?
  8. Jan 27, 2005 #7


    User Avatar
    Science Advisor
    Homework Helper

    MS Word has an equation editor. You might have to download and/or buy it if it doesn't already come with your version of Word.
  9. Jan 28, 2005 #8
    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.

    [tex]\frac{dY}{dx} = \left a \left (Y \left (1 - \frac{Y}{b} \right ) \right )[/tex]

    [tex]\frac{dY}{\left Y (1 - \frac{Y}{b} \right )} \right = adx[/tex]

    [tex]\int \frac{dY}{\left Y (1 - \frac{Y}{b} \right )} \right = a \int dx[/tex]

    [tex]\int \frac{1}{Y} - \frac{1}{Y-b}dY \right = ax + C[/tex]

    [tex]\int \frac{dY}{Y} - \int \frac{dY}{Y-b} \right = ax + C[/tex]

    [tex]\ln Y - \ln (Y-b) \right = ax + C[/tex]

    [tex]\ln \frac{Y}{Y-b} \right = ax + C[/tex]

    If when [tex]X = 0, Y = Y_{0}[/tex] then;

    [tex]C = \ln \frac{Y_{0}}{Y_{0} - b}[/tex]

    [tex]\ln \frac{Y}{Y-b} \right = ax + \ln \frac{Y_{0}}{Y_{0} - b}[/tex]

    [tex]e^{\ln \frac{Y}{Y-b}} \right = e^{ax + \ln \frac{Y_{0}}{Y_{0} - b}}[/tex]

    [tex]\frac{Y}{Y-b} \right = e^{ax}+ \frac{Y_{0}}{Y_{0} - b}[/tex]

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

    [tex]Y(x) = \frac {Y_{0} b e^{ax}}{b + Y_{0} \left (e^{ax} - 1 \right )}[/tex]
  10. Jan 28, 2005 #9


    User Avatar
    Science Advisor
    Homework Helper

    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:

    [tex]e^{ax}\mathbf{+}\frac{Y_0}{Y_0 - b}[/tex]

    rather, it should be:

    [tex]e^{ax}\mathbf{\times }\frac{Y_0}{Y_0 - b}[/tex]


    [tex]Y = \frac{b}{1 - \exp (-ax)(\frac{Y_0 - b}{Y_0})}[/tex]
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Surplus of Y on the wrong side
  1. One sided limits (Replies: 4)

  2. Why is this wrong? (Replies: 24)

  3. Something's wrong (Replies: 4)

  4. Y,y' domain (Replies: 2)