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Math Help - Solve this homogeneous type equation

  1. #1
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    Solve this homogeneous type equation

    I can't seem to solve this question, please tell me what I am doing wrong:

    q1.
    2xyy' = y^2 - x^2

    y' = (y^2-x^2)/(2xy)

    divide through by x^2 to get:
    y' = ((y/x)^2-1)/(2(y/x)) = f(y/x)

    let y/x = v, thus y = v + xv'
    1/x dx = 1/(f(v)-v) dv

    f(v)-v= (v^2-1)/2v - v
    = (-2v^2+v^2-1)/2v
    = -(v^2+1)/2v

    integrating:

    int(dx/x) = int(2v/-(v^2+1))
    ln|x| + c = -ln|v^2-1|

    x + c = -(v^2-1)
    x + v^2 -1 = c
    v = +-(c-x+1)
    y = xv
    y = +-x(c-x+1)
    ..

    fairly sure i am correct up until i take the logs out.
    The answer book says the answer is
    x^2 + y^2 = cx

    I am unsure how to get this.
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  2. #2
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    Re: Solve this homogeneous type equation

    Quote Originally Posted by 99.95 View Post
    I can't seem to solve this question, please tell me what I am doing wrong:

    q1.
    2xyy' = y^2 - x^2

    y' = (y^2-x^2)/(2xy)

    divide through by x^2 to get:
    y' = ((y/x)^2-1)/(2(y/x)) = f(y/x)

    let y/x = v, thus y = v + xv'
    You mean y'= v+ xv'

    1/x dx = 1/(f(v)-v) dv
    First, I would have written y'= v+ xv'= (v^2- 1)/(2v)= v/2- 1/(2v) so that xv'= -1/(2v)- v/2.

    f(v)-v= (v^2-1)/2v - v
    = (-2v^2+v^2-1)/2v
    = -(v^2+1)/2v
    Okay, yes, that is what I got.

    integrating:

    int(dx/x) = int(2v/-(v^2+1))
    ln|x| + c = -ln|v^2-1|
    You mean "ln|v^1+ 1|" don't you?

    x + c = -(v^2-1)
    This is wrong or two reasons. First, it should be "-(v^2+ 1)" on the right side.
    More importantly, e^(ln(x)+ C)= e^C e^(ln(x))= cx where c= e^C.
    That is, cx= -(v^2+ 1)

    x + v^2 -1 = c
    v = +-(c-x+1)
    y = xv
    y = +-x(c-x+1)
    ..

    fairly sure i am correct up until i take the logs out.
    The answer book says the answer is
    x^2 + y^2 = cx

    I am unsure how to get this.
    From
    Thanks from 99.95 and topsquark
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  3. #3
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    Re: Solve this homogeneous type equation

    Quote Originally Posted by HallsofIvy View Post
    You mean y'= v+ xv'
    This is wrong or two reasons. First, it should be "-(v^2+ 1)" on the right side.
    More importantly, e^(ln(x)+ C)= e^C e^(ln(x))= cx where c= e^C.
    That is, cx= -(v^2+ 1)
    Ah yes, how could i make such a silly mistake. That definitely makes things more easier.

    Could you explain what to do from there though? I am not 100% sure.

    thanks, you are a gun ivy!
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