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Seperable Differential Equation Invovling Partial Fractions

  1. Feb 4, 2012 #1
    [itex]y' = (3+y)(1-y)[/itex]
    [itex]y(0) = 2[/itex]

    Attempt to solve:
    [itex]\frac{dy}{dx}=(3+y)(1-y)[/itex]

    [itex](3+y)(1-y)dy = dx[/itex]

    [itex]\int(3+y)(1-y)dy = \int dx[/itex]

    Partial Fractions:
    [itex]\frac{A}{(3+y)} + \frac{B}{(1-y)}[/itex]

    [itex]A(1-y)+B(3+y)[/itex]

    Let y= 1 or -3

    A = 1/4
    B = 1/4

    [itex]\frac{1}{4}\int \frac{1}{(3+y)} dy + \frac{1}{4}\int \frac{1}{(1-y)} dy[/itex]

    [itex]1/4(ln(3+y)+ln(1-y)) + C = x[/itex]

    Re-arranging.
    [itex]ln(3+y)+ln(1-y) = 4x+C[/itex]

    [itex](3+y)+(1-y) = e^{(4x+C)}[/itex]

    Did I do it right so far? How do figure out y if that is the case?
     
  2. jcsd
  3. Feb 4, 2012 #2

    Char. Limit

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    You have an error here. e^(ln(3+y) + ln(1 - y)) is not equal to (3 + y) + (1 - y). Remember your properties of exponentials. From there, if you really want y as an explicit function of x, you'll need to solve the resulting quadratic.
     
  4. Feb 4, 2012 #3

    ehild

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    You have a sign error: the integral of 1/(1-y) is -ln|1-u|
    Moreover: the argument of ln has to be written as |1-y| and |3+y|.

    ehild
     
  5. Feb 4, 2012 #4
    Okay I think I fixed the mistakes mentioned.. I think but I am still unsure how to do solve for y. I tried looking up properties of exponents but not sure if I did it all right.


    My new equation result is:
    [itex]\frac{1}{4}(ln(3+y)-ln(1-y)+c) = x[/itex]

    [itex](ln(3+y)-ln(1-y)) = 4x+c[/itex]

    I know I have to use the exp operator but not sure how you do:

    [itex]e^{ln(3+y)-ln(1-y)}[/itex]

    I know I need to use:
    [itex]\frac{a^{b}}{a^{c}} = a^{b-c}[/itex]

    so is the answer:
    [itex]\frac {e^{ln(3+y)}} {e^{ln(1-y)}} = e^{(4x+c)}[/itex] ??

    [itex]\frac {(3+y)} {(1-y)} = e^{(4x+c)}[/itex]
    How do I get y from this mess?
     
    Last edited: Feb 4, 2012
  6. Feb 4, 2012 #5

    ehild

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    The absolute value signs can not be omitted. The solution of the de is [itex]|\frac{3+y}{1−y}|=e^{4x+c}[/itex]

    Use the initial condition y=2 at t=0 to find c for the case y>1.

    ehild
     
  7. Feb 4, 2012 #6
    So I substitute in y=2 and x=0?

    [itex]|\frac{3+2}{1-3}| = e^{4x+c}[/itex]

    [itex]5 = e^{4x+c}[/itex]

    [itex]ln(5) = 4x+c[/itex]

    since x = 0
    [itex]c = ln(5)[/itex] ?
     
  8. Feb 4, 2012 #7

    HallsofIvy

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    No,
    [tex]\left|\frac{3+2}{1- 2}\right|[/tex]

     
  9. Feb 4, 2012 #8

    ehild

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    Yes, c=ln5. But e4x+ln5=5e4x, so you can write the solution for the given initial condition as [itex]\frac{y+3}{y-1}=5e^{4x}[/itex] if y>1, as |1-y|=y-1.
    If you want an explicit expression for y, then multiply the equation by y-1 and isolate y.

    ehild
     
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