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Show the Exact Differential Equation solution is also a solution to another equation

  1. Feb 12, 2012 #1
    1. The problem statement, all variables and given/known data
    Consider the equation (y^2+2xy)dx-x^2dy=0 (a) Show that this equation is not exact. b) Show that multiplying both sides of the equation by y^-2 yields a new equation that is exact. C) use the solution of the resulting exact equation to solve the origional equation. d) Were any solutions lost in the process?

    2. Relevant equations
    How do you show that the solution in part b is a solution to the original?


    3. The attempt at a solution
    answer to part b: (x+x^2y^-1=c)
    I tried (y^2+2xy)dx - x^2dy=(x+x^2y^-1) and integrated both sides to get
    xy^2 + x^2y - x^2y = xy+x^2 ln y

    I tried solving (x+x^2y^-1=c) for x: getting x=-x^2y^-1 +c and plugging in into the original equation : (y^2+2(-x^2y^-1 +c)y)dx - (-x^2y^-1 +c)^2dy=0 and that just got me a mess. I'm not sure what else to try.
    I'm not sure how to do part d, either, but I figured I needed part c first.
     
  2. jcsd
  3. Feb 12, 2012 #2

    tiny-tim

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    welcome to pf!

    hi sunnyceej! welcome to pf! :smile:

    (try using the X2 button just above the Reply box :wink:)
    yes :smile:

    but after that, i don't understand what you're doing :confused:

    just rewrite x + x2/y = C in the form y = … :wink:
     
  4. Feb 12, 2012 #3
    Re: Show the Exact Differential Equation solution is also a solution to another equat

    Thanks! I got y=x^2/(c-x)

    Every time I use the x2 button at the top I get SUP/SUP

    I thought I also had to plug it back into the original to verify, but I found out solving for y is good! :)
     
    Last edited: Feb 12, 2012
  5. Feb 12, 2012 #4

    SammyS

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    Re: Show the Exact Differential Equation solution is also a solution to another equat

    As a result, your x[ SUP]2[ /SUP] looks like x2 as it should.
     
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