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Seperable Differential Eq

  1. Oct 30, 2008 #1
    1. The problem statement, all variables and given/known data
    x' - x^2 = 1, x(0) = 1
    find x(t)


    2. Relevant equations



    3. The attempt at a solution

    This was for an assignment due today, I kept trying but had to give up after 2 hours. I don't get the idea of seperable equations when there is just freaking x involved, what is it that I seperate? And all the examples in my calculus made sure they did have a y and x, geez must be hard to seperate those..

    I just took a wild shot and arrived at: arctan x + 1
     
  2. jcsd
  3. Oct 30, 2008 #2

    Dick

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    You separate x and t. x' must mean d(x(t))/dt. There certainly is an arctan in the problem. Take the examples in your calculus book and replace y and x by x and t.
     
  4. Oct 30, 2008 #3
    ok so I end up with

    dx/dt = 1+ x^2

    and then I invert it and get dt/dx = 1/(1+x^2)
    and then I get dt = 1/(1+x^2) dx
    and take the integral and get x(t) = arctanx + C
    and then 1 = arctan 0 + C

    So I end up with x(t) = arctanx + 1 ?

    I tried drawing it in python using eulers, and apparently what I have arrived at here is wrong
     
  5. Oct 30, 2008 #4

    Dick

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    If you integrate dt=1/(1+x^2) dx you get t+C=arctan(x). You don't get an x on both sides. Integral dt=t+C. Not x+C.
     
  6. Oct 30, 2008 #5
    Meaning that t+ C = arctan(x)

    tan(t + C) = tan(arctan(x))
    x(t) = tan(t + C) ?

    So the solution is x(t) = tan(t + 0.25pi) ?

    edit: It would have to be tan(t - 0.25pi) right?
     
    Last edited: Oct 30, 2008
  7. Oct 30, 2008 #6

    Dick

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    Yes, x(t)=tan(t+pi/4). Just because the calculus book always uses y as the dependent variable and x as the independent variable doesn't mean they aren't allowed to throw a problem at you where t is the independent variable and x is the dependent. Just switch the letters around.
     
  8. Oct 30, 2008 #7

    Dick

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    No, tan(0-pi/4)=(-1). Your initial condition said x(0)=+1, not -1.
     
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