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1st order linear equations

  1. Oct 18, 2009 #1
    1. The problem statement, all variables and given/known data

    (2e^y -x) dy/dx = 1

    2. Relevant equations

    dy/dx + P(x)y = Q(x)

    3. The attempt at a solution

    I know how to solve these equations, but I can't get this into the dy/dx + P(x)y = Q(x) form.

    In addition to this problem, is also this: (x + y^2)dy = ydx (can't get into the form). Help would be appreciated.

    Also, for both of these questions, they are asking to consider x as function of x = x(y).
     
  2. jcsd
  3. Oct 18, 2009 #2

    LCKurtz

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    Use the hint, using dy/dx = 1 / (dx/dy) you can rewrite it as:

    dx/dy + x = 2ey

    It looks strange because you are used to thinking of y as a function of x. If the x and y were switched it would look familiar:

    dy/dx + y = 2ex

    You could even switch them and work it on a sheet of scratch paper, then copy it back with the x's and y's switched back if that helps you visualize it.
     
  4. Oct 19, 2009 #3
    So, if I were to solve, would it be like this:

    dx/dy + x = 2e^y

    I(x) = e^integral dy = e^y

    multiply integrating factor on both sides will give: e^y (x) = e^(2y) + C

    => x = e^y + Ce^-y.

    That was the answer I got for x, but it does not match answer in the back.

    help.
     
  5. Oct 19, 2009 #4

    LCKurtz

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    It's hard to get used to switching the independent and dependent variables y and x. Your integrating factor isn't I(x) now, it's I(y). Your solution looks OK other than that notational thing. And, regardless what the solution manual may say, you can check that your version of the solution solves the above equation.
     
  6. Oct 19, 2009 #5
    Hmmm ok, still kind of confused. this is what I did after your correction:

    dx/dy + x = 2e^y

    I(y) = e^integral (2e^y) dy = e^(2e^y)

    multiply integrating factor on both sides will give: e^(2e^y) x = Integral e^(2e^y) (2e^y) dy......i am unsure on how to solve that integral on the right side.
     
  7. Oct 19, 2009 #6

    LCKurtz

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    No no. I was just pointing out that your integrating factor is a function of y, not x. As I said before, the solution you had works. You don't need to change it.
     
  8. Oct 20, 2009 #7

    HallsofIvy

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    What is the answer in the book? Knowing that might help us!
     
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