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Differential Equation (help with the algebra)

  1. Apr 7, 2012 #1
    1. The problem statement, all variables and given/known data[/b]
    I need help knowing how my textbook jumped from one equation to the other in the example given.
    [tex]\frac{y-2}{y+2}= e^{4x+c} [/tex]
    They solved for y and got
    [tex]y = 2\frac{1+ce^4x}{1-ce^4x}[/tex]

    What is the intermediate step? I tried but couldn't isolate y.

    And then they do something again that I don't know how they got:

    "Now if we factor out the right hand sign of the differential equation as [tex]dy/dx = (y-2)(y+2)[/tex]

    I couldn't reach that step either.
     
  2. jcsd
  3. Apr 7, 2012 #2

    Office_Shredder

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    [tex] \frac{y-2}{y+2}=A [/tex]
    becomes
    [tex] y-2 = (y+2)A = Ay+2A[/tex]

    putting all the y's together gives
    [tex] y-Ay=2+2A[/tex]
    factor out a y
    [tex] y(1-A)=2+2A[/tex]
    and solve
    [tex] y=\frac{2+2A}{1-A}[/tex]

    Note that this is NOT what you tex'd up, you have a typo on either your first or second equation

    For the differential equation I think you're going to have to tell us what it looked like before they wrote it like that for us to tell you what they did
     
  4. Apr 7, 2012 #3
    I've checked and rechecked meticulously and I can assure you that there were no mistakes. The only thing that I omitted was the ± sign for e.

    [tex]\frac{y-2}{y+2}= ±e^{4x+c} [/tex]
     
  5. Apr 7, 2012 #4
    Here is the whole problem:

    [tex]\frac{dy}{y^2-4}=dx[/tex]
    Using partial fractions
    [tex]\frac{1/4}{y-2}-\frac{1/4}{y+2}dy=dx[/tex]
    Integrating both sides
    [tex]1/4 (ln(y-2) - ln (y+2)) = x[/tex]
    Logarithmic property gives
    [tex]\frac{1}{4} \frac{ln(y-2)}{y+2} = x + c[/tex]
    Raising everything to e we get
    [tex]\frac{y-2}{y+2}= ±e^{4x+c} [/tex]
    Solving for y
    [tex]y = 2\frac{1+ce^4x}{1-ce^4x}[/tex]
    "Now if we factor out the right hand sign of the differential equation as [tex]dy/dx = (y-2)(y+2)[/tex]"
     
  6. Apr 7, 2012 #5

    SammyS

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    There is a typo in the second equation. It should be:
    [itex]\displaystyle y = 2\frac{1+ce^{4x}}{1-ce^{4x}}[/itex]
    as O.S. showed.
     
  7. Apr 7, 2012 #6

    So much for being meticulous-- its a wonder how much your brain can fool you sometime.
     
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