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Solve this Differentail equation.HELP!

  1. Oct 18, 2009 #1
    1.Solve this differential equation



    2.(x^2+y^2)y'=xy



    3. Thanks
     
  2. jcsd
  3. Oct 18, 2009 #2

    Mark44

    Staff: Mentor

    What have you tried? The rules in this forum state that you must show some effort before we'll help you out.
     
  4. Oct 18, 2009 #3
    Ok mate I have tried it
    (x^2+y^2)dy/dx=xy

    Divide by [tex]y*(x^2+y^2)[/tex] : 1/y dy/dx= [tex]x/(x^2+y^2)[/tex].

    [tex]\int {1/y}[/tex] dy= 1/2[tex] \int {x/(x^2+y^2)}[/tex]

    My answer is y=A [tex]\sqrt{x^2+y^2} [/tex].
     
    Last edited: Oct 18, 2009
  5. Oct 18, 2009 #4
    I don't see how you get that answer from the equation above it.
    There are various classes of differential equations that can be solved.
    This one is in one of those classes.
    It's probably more instructive to search for which class it is in yourself.
     
  6. Oct 18, 2009 #5
    Can anyone help me plz its very important!
     
  7. Oct 18, 2009 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You don't have a "dx" on the right

    That may seem like a technical point but it would have reminded you that you are integrating on the right with respect to x and y is some (unknown) function of x, not a constant. This is NOT a separable equation- that is you cannot get only x on one side of the equation and only y on the other.

    It is, however, a "homogeneous" first order equation. If you divide both sides of the equation by [itex]x^2[/itex], you get
    [tex](1+ \left(\frac{y}{x}\right)^2)dy= \frac{y}{x}dx[/tex]
    Let u= y/x. Don't for get to replace dy: y= ux so dy= xdu+ udx.

    Why was this posted under "Precalculus"?
     
  8. Oct 18, 2009 #7
    I forgot to write the dx on the right.
    Thanks a lot mate!Really appreciate ur effort!

    Why was this posted under "Precalculus"? Where should I post these kind of questions...
     
  9. Oct 18, 2009 #8

    Mark44

    Staff: Mentor


    In Calculus and Beyond...
     
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