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Solve this diff eq.

  1. Oct 6, 2013 #1
    1. The problem statement, all variables and given/known data
    determine whether the given differential equations are homogenous and, if so, solve them.

    2. Relevant equations

    dy/dx = (( x^2 -2y^2)/xy )

    3. The attempt at a solution

    i assumed it was homoG.

    Then i replaced ( x*v) for each Y
    and made dy/dx --> v+x*(dv/dx)

    i simplify to x*(dv/dx) = (1/v) + v

    And i know here that i made a mistake . because the solution is


    Can someone help me please.
  2. jcsd
  3. Oct 6, 2013 #2


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    Try reading this.
  4. Oct 6, 2013 #3


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    You need to show your work so we can see where you went wrong, if you did. And since you haven't finished solving for v and hence y, how do you know it is wrong in the first place?
  5. Oct 7, 2013 #4
    Y' = (x^2+2y^2)/xy

    make y =xv , y'=v + xv'

    simplify to xv'= (1+v^2)/v
    rearrange to 1/x dx = (v/(1+v^2))dv

    Integrate ---> ln ( 1+v^2) ^ (1/2) = ln (x) + C

    simplify ----> v = ( x^2 + C) ^ (1/2)

    Y = x*( x^2 + C) ^ (1/2)

    Y^2= x^4 + (x^2*C)

    But the solution in the back is y^2 = kx^4 -x^2

    Where did i go wrong?
  6. Oct 7, 2013 #5


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    It's somewhere between your "Integrate ---> ln ( 1+v^2) ^ (1/2) = ln (x) + C" (which looks correct) and the "simplify ----> v = ( x^2 + C) ^ (1/2)" (which looks incorrect). How did you do that? Might also note you've changed your problem statement, which may have caused some confusion.
    Last edited: Oct 7, 2013
  7. Oct 8, 2013 #6


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    Here is your error. Removing the logarithms gives (1+ v^2)^{1/2}= Cx, NOT what you have.
    Squaring, 1+ v^2= 1+ y^2/x^2= Cx^2

  8. Oct 8, 2013 #7
    Ahh, there it is. Thank You !

    I still don't have a 100% handle on why the removal of logarithms creates a multiplication of the inside terms . is there a theorem or rule for this process?
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