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Differential eq

  1. Jan 19, 2013 #1
    1. The problem statement, all variables and given/known data

    Given: (x2-y2)dx + (x2-xy)dy=0,

    Verify that the following function is a solution for the given differential equation:

    c1(x+y)2=xey/x

    2. The attempt at a solution

    I've gotten this far:

    1st - I solved for [itex]\frac{dy}{dx}[/itex] in the given equation.

    [itex]\frac{dy}{dx}[/itex]=[itex]\frac{-x^2-y^2}{x^2-xy}[/itex]

    2nd - I used implicit differentiation on the function and got:

    2c1(x+y)(1+y)=xey/x([itex]\frac{xy'-y}{x}[/itex]) + ey/x

    Now....I believe that I can solve for y' in: 2c1(x+y)(1+y)=xey/x([itex]\frac{xy'-y}{x}[/itex]) + ey/x?
     
  2. jcsd
  3. Jan 19, 2013 #2

    ehild

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    Gold Member

    Your method is correct, but there is a sign error in the equation in red, it should be

    [itex]\frac{dy}{dx}[/itex]=[itex]\frac{-x^2+y^2}{x^2-xy}[/itex]

    and there are mistakes also in the blue equation:

    2c1(x+y)(1+y')=xey/x([itex]\frac{xy'-y}{x^2}[/itex]) + ey/x

    ehild
     
  4. Jan 19, 2013 #3
    My mistake. In the original equation it was supposed to be (x2 + y2)dx. So my original [itex]\frac{dy}{dx}[/itex]=[itex]\frac{-x^2-y^2}{x^2-xy}[/itex] equation was okay. And you are correct about the blue equation. I completely forgot the y'.

    Thanks for catching that.
     
    Last edited: Jan 19, 2013
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