1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: 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](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]


    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]

    [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

    User Avatar
    Gold Member

    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


    User Avatar
    Homework Helper

    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|.

  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:


    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


    User Avatar
    Homework Helper

    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.

  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


    User Avatar
    Science Advisor

    [tex]\left|\frac{3+2}{1- 2}\right|[/tex]

  9. Feb 4, 2012 #8


    User Avatar
    Homework Helper

    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.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook