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Differential Equ - sub

  1. Sep 14, 2013 #1
    1. The problem statement, all variables and given/known data

    [itex]yy' + x = \sqrt{x^2 + y^2}[/itex]


    2. Relevant equations

    y = vx
    y' = v'x + v

    3. The attempt at a solution

    [itex]y' + \frac{x}{y} = \frac{\sqrt{x^2 + y^2}}{y}[/itex]

    [itex](v'x+v) + v =\frac{ \sqrt{x^2 + (vx)^2}}{vx}[/itex]
     
    Last edited: Sep 14, 2013
  2. jcsd
  3. Sep 14, 2013 #2

    epenguin

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    Stuck? No idea what to do with vx - vx ? Not noticed the whole thing has a factor x?
     
  4. Sep 14, 2013 #3
    Sorry, i typed the problem wrong the first time.
     
  5. Sep 14, 2013 #4

    epenguin

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    Are you still stuck? It seems to me this is fairly easily expressed in terms of y and a variable x/y.
     
  6. Sep 14, 2013 #5
    I am sure it is easy one i figure out what i am trying to accomplish with this substition, I am just not certain where it is i am heading for other than d/dx(y') = and something i can integrate on the right side.
     
  7. Sep 14, 2013 #6


    [itex]v' = \frac{ \sqrt{x^2 + (vx)^2}}{vx^2} - \frac{v}{x}[/itex]
     
  8. Sep 14, 2013 #7

    epenguin

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    The way I look at it if you write it like this

    [itex]y' = -\frac{x}{y} + \frac{\sqrt{x^2 + y^2}}{y}[/itex]

    you see the RHS is just a function of x/y. So let this be one of your variables and keep y as the other.
     
  9. Sep 14, 2013 #8
    Okay, I am a bit dense. I hopefully walking through this will get me going.

    v = y/x, x/y = 1/v

    [itex]v'x + v = -\frac{1}{v} + \frac{\sqrt{x^2 + (vx)^2}}{vx}[/itex]
     
  10. Sep 14, 2013 #9
    After talking with another student, they found out this uses v = x^2 + y^2

    [itex] v' = 2x + 2yy' = 2(x+yy')[/itex]

    [itex]\frac{1}{2} \int \frac{1}{\sqrt{v}} = \int dx[/itex]

    [itex]\sqrt{v} = x + c[/itex]

    [itex] v = (x+c)^2[/itex]

    [itex] x^2 + y^2 = (x+c)^2[/itex]

    [itex] y^2 = -x^2 + x^2 + 2xc + C[/itex]

    [itex] y = +/- \sqrt{2xc + C} [/itex]
     
    Last edited: Sep 14, 2013
  11. Sep 14, 2013 #10
    The answer in the book is:

    [itex] x - \sqrt{x^2 + y^2} = C [/itex]

    And that is not what I got :-(
     
  12. Sep 14, 2013 #11

    epenguin

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    Yes (9) was well spotted, when you see yy' be looking out for (y2)' .

    My thinking was that your d.e. can be written

    [tex] y' = - \frac{x}{y} +\sqrt{\left(\frac{x}{y}\right)^2 + 1} [/tex]

    and the RHS is purely a function of (x/y) which we could call u, which looked simple. But that is overlooking that then you have to change y' into dy/du - in fact the LHS becomes

    Edit something complicated and I think there is no or less than no advantage in this approach which seemed initially attractive after all.
     
    Last edited: Sep 15, 2013
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