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Dy/dx + dx/dy = 1

  1. Jul 2, 2010 #1
    1. The problem statement, all variables and given/known data
    Solve for either x or y:
    [tex]\frac{dy}{dx} + \frac{dx}{dy} = 1[/tex]

    2. Relevant equations
    I don't know any.

    3. The attempt at a solution

    [tex]y' = \frac{dy}{dx}[/tex]

    so then the problem becomes

    [tex]y' + \frac{1}{y'} = 1[/tex]

    [tex]y'^2 + 1 = y'[/tex]

    [tex]y'^2 - y' + 1 = 0[/tex]

    So then I thought, why not use the quadratic equation? I get

    [tex]y' = \frac{1 \pm \sqrt{-3}}{2}[/tex]

    And indeed, adding this to its reciprocal gives a sum of 1:

    [tex]\frac{1 \pm \sqrt{-3}}{2} + \frac{2}{1 \pm \sqrt{-3}} = 1[/tex]

    What does not sit well with me, though, is that complex numbers are involved. Is that allowed? Furthermore, would y = Ax + C be a solution (where A is the complex number from two lines up)?
    Last edited: Jul 2, 2010
  2. jcsd
  3. Jul 3, 2010 #2
    EDIT: oh wow...I just realized that we're dealing with [tex] (y')^2 [/tex] and not [tex] y'' .[/tex] Please ignore all of the following then haha...I'll leave it up in case you're interested in at all. But it won't help you solve your problem.

    First, we solve the homogeneous case of this second-order differential equation with constant coefficients.

    [tex] y'' - y' = 0 [/tex]

    We easily find two solutions of this homogeneous case:

    [tex] y_1= c [/tex] and [tex] y_2=e^x ,[/tex]

    where c is just a constant. Thus, we now look for a particular solution of the nonhomogeneous case

    [tex] y'' - y' = -1. [/tex]

    Since the right-hand side is a constant, we look for particular solutions of the form

    [tex] y_p = kx, [/tex]

    where k is simply a constant. Differentiating twice and substituting into our equation, we can solve for k. We find

    [tex] k = 1, [/tex]

    meaning that

    [tex] y_p = x.[/tex]

    Thus, the general solution to this second-order differential equation is:

    [tex] y_g = y_c + y_p = c_1 + c_2e^x + x [/tex].

    I hope this helped.
    Last edited: Jul 3, 2010
  4. Jul 3, 2010 #3
    Yes, this is the solution.
  5. Jul 3, 2010 #4


    Staff: Mentor

    Yes, that works.
  6. Jul 3, 2010 #5


    User Avatar
    Science Advisor

    "[tex]y' = \frac{1 \pm \sqrt{-3}}{2}[/tex]"

    is NOT the solution because the problem asked you to solve for y, not y'.

    The solution is
    [tex]y = \frac{1 \pm \sqrt{-3}}{2}x+ C[/tex]
  7. Jul 3, 2010 #6
    Thanks a lot everyone! :D
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