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Is this a linear or seperable differential equation?

  1. May 29, 2010 #1
    1. The problem statement, all variables and given/known data

    Solve the initial value problem of the following DE, given that y(0) = 0

    dy/dx = (20-2y)/(5-x)

    2. Relevant equations

    Integrating factor for Linear DE's: e ^ ([tex]\int[/tex]P(x)dx) where Linear DE is of form dy/dx + P(x)y = Q(x).

    3. The attempt at a solution

    I'm having trouble understanding why my technique doesen't work. I suspect this differential equation needs to be treated as a linear one, but if this is the case I have no idea why it must be treated this way? Why does my approach not work?

    Instead of solving for a linear DE, I thought that this equation could be separated:

    1/(20-2y) dy = 1/(5-x) dx

    integrate both sides:
    [tex]\int[/tex]1/(20-2y) dy = [tex]\int[/tex]1/(5-x) dx

    -1/2 ln|10-y| = -ln|5-x| + C

    ln|10-y| = 2ln|5-x| - 2C

    10-y = (5-x)^2 + e^(-2C)

    rewrite e^(-2C) as C...

    10-y = (5-x)^2 + C

    sub in x=0 and y=0

    10 = 25 + C

    C = -15

    10-y = (5-x)^2 - 15

    y = 10x - x^2

    However my book states that the answer is actually y = 4x - (2/5)x^2

    Any help anyone could provide will be very much appreciated. I don't need this done for me, I just want to know why my approach does not work (or what trivial mistake I have made and overlooked), and what alternative approach I should use instead.

    Thanks in advance

    PS, apologies for the strange formatting, the latex integral signs are oversized, I do not know how to write latex...
  2. jcsd
  3. May 29, 2010 #2


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    Science Advisor

    Here's your error. [itex]e^{a+b}e^ae^b[/itex], not [itex]e^a+ e^b[/itex]
    You should have [itex]10- y= (5-x)^2e^{-2C}[/itex] and, writing [itex]e^{-2C}[/itex] as "C" as you do below, [/quote]10- y= C(5- x)^2[/quote]

  4. May 29, 2010 #3
    Thank you! I'm glad it was an algebraic error not some kind of misconception. I need to give constants more thought it seems.
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