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Differential equation first, first degree help

  1. Jul 11, 2011 #1
    Differential equation first, first degree help!!

    1. The problem statement, all variables and given/known data

    solve: x.dy + y.dx + (x.dy - y.dx)/(x^2 + y^2) = 0

    2. Relevant equations

    if M.dx + N.dy = 0 has to be exact then

    equation 1: partial derivative of M w.r.t y (keeping x constant) = partial derivative of N w.r.t x (keeping y constant)

    3. The attempt at a solution

    the idea is to find if this equation is exact because once you do that the integration is easy..

    but

    now i first i simplify the equation by multiplying throughout by (x^2 + y^2)

    simplified form: (x^3 + x.y^2 - y).dx + (y^3 + y.x^2 + x).dy = 0
    try to check if it satisfies equation 1: partial derivative of M w.r.t y (keeping x constant) = partial derivative of N w.r.t x (keeping y constant)

    here M = (x^3 + x.y^2 - y) & N = (y^3 + y.x^2 + x)

    we get 2.x.y - 1 is not equal to 2.x.y + 1

    now i try another method i.e. group the terms without multiplying throughout by (x^2 + y^2)

    simplified equation: [x - (y/(x^2 + y^2))].dx + [y + (x/(x^2 + y^2))].dy = 0
    now when we apply equation 1 criteria we get:

    partial derivative of M w.r.t y (keeping x constant) = (y^2 - x^2)/(x^2 + y^2)^2 = partial derivative of N w.r.t x (keeping y constant) = (y^2 - x^2)/(x^2 + y^2)^2

    my question is - why is it that i am getting the two methods to be different???
     
  2. jcsd
  3. Jul 12, 2011 #2
    Re: Differential equation first, first degree help!!

    eh... anyone???? wold really help.. this has been bugging me...
     
  4. Jul 12, 2011 #3
    Re: Differential equation first, first degree help!!

    Have you seen integrating factors yet? That whole method is based on the idea of multiplying the equation by a function to put it in a 'nice' form. When you have an equation M(x,y)dx + N(x,y)dy=0, multiplying by anything is going to change the derivatives. It's very likely that it will change the partial derivatives of M and N in different ways, which may or may not help you make the equation into an exact equation. For example, if I multiply the equation by x, that's going to change the x partial derivative of N, but it's not going to change the y partial of M.
     
  5. Jul 12, 2011 #4

    SammyS

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    Re: Differential equation first, first degree help!!

    Notice that, d(xy) = x.dy + y.dx .

    Also, d(y/x) = (xdy - ydx)/x2

    So, you can rewrite your equation as: (x2 + y2)d(xy) + x2(d(y/x)) = 0

    This suggests that you let u = xy and v = x/y.

    Can you take it from here?
     
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