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Non Separative Diff-EQ problem

  1. Jun 12, 2013 #1
    The problem I'm given is - I think this is a non separable equation
    dy/dx=x-y ...... u=x-y
    I tried Substituting, where
    (remember above statement u=x-y, and dy/dx=x-y)
    therefore, I got 1-u=du/dx and then I solved it from there by integrating since I think once it's in this form it becomes separable. But I get the answer of

    1-(e^x)+x+c=y which when I checked with wolfram differential eq calculator is a little off
    the differential calculator gives me this answer c1(e^-x)+x-1

    I have solved this several times now, I'm so close I just don't know where I'm making the mistake. Any help will be appreciated.
  2. jcsd
  3. Jun 12, 2013 #2


    Staff: Mentor

    The separable equation is
    du/(u - 1) = -dx
    Integrating both sides gives
    ln|u - 1| = -x + C
    Exponentiating (and this is where you made your mistake, I believe) gives

    |u - 1| = e-x + C = eC * e-x = C1e-x
    BTW, homework-type problems should be posted in the Homework & Coursework section, not in the technical math sections. I am moving this thread now.
  4. Jun 12, 2013 #3


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    If you have had linear equations, just rewrite it as ##y'+y = x## and solve it with an integrating factor.
  5. Jun 12, 2013 #4
    How can you do that?? I don't understand what you are saying.
  6. Jun 12, 2013 #5
    I understood what you did there, but it still gives me y= -c*e^(-x)+x-1, the negative in front of the constant still technically makes the answer wrong. But Thanx for your help. I appreciate it.
  7. Jun 12, 2013 #6


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    Have you studied first order linear equations? If not, then you wouldn't be expected to understand. If you have studied them, perhaps you should review them.
  8. Jun 13, 2013 #7


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    It's hard to believe you are serious. Since C can be any constant, [itex]-c e^{-x}+ x- 1[/itex] is exactly the same solution as [itex]c e^{-x}+ x- 1[/itex]. Just with different "c". If, for example, c= 1 in the first, then c= -1 in the second.
    Last edited by a moderator: Jun 13, 2013
  9. Jun 13, 2013 #8


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    Your equation is y'+ y= x. That is "linear" and so it is simple to find an "integrating factor", a function, u(x), such that mutiplying by it makes the left side an "exact derivative", d(uy)/dx.

    By the product rule, (uy)'= uy'+ u'y. Just multiplying the left side of the equation by u would give uy'+ uy. Comparing those, uy'+ u'y= uy'+ uy, reduces to u'= u which has solution [itex]u= e^x[/itex].

    So multiplying the entire equation by [itex]e^x[/itex] gives [itex]e^xy'+ e^xy= (e^xy)'= xe^x.

    Now just integrate both sides.
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