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Math Help - [SOLVED] homogeneous DEQ

  1. #1
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    [SOLVED] homogeneous DEQ

    obtain the family of solutions
    x^2y' = 4x^2 + 7xy + 2y^2

    let
    y = vx

    dy = vdx + xdv

    and after that i substitute and got a result of

    xdv + vdx = (4 + 7v + v^2)dx

    \frac{dv}{4 + 6v + v^2}= \frac{dx}{x}
    i dont know if what i did is right .... im stuck with integrating the left part of this equation
    thank you

    edit
    xdv + vdx = (4 + 7v + 2v^2)dx
    i think i already know what's wrong with my soln

    how do i delete this thread? thanks
    Last edited by ^_^Engineer_Adam^_^; November 28th 2007 at 07:13 AM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    obtain the family of solutions
    x^2y' = 4x^2 + 7xy + 2y^2

    let
    y = vx

    dy = vdx + xdv

    and after that i substitute and got a result of

    xdv + vdx = (4 + 7v + v^2)dx

    \frac{dv}{4 + 6v + v^2}= \frac{dx}{x}
    i dont know if what i did is right .... im stuck with integrating the left part of this equation
    thank you

    edit
    xdv + vdx = (4 + 7v + 2v^2)dx
    i think i already know what's wrong with my soln

    how do i delete this thread? thanks
    Rather than delete it, why don't you tell us how you solved it? Your answer might be helpful to others.

    -Dan
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  3. #3
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    x^2y' = (4x^2 + 7xy + 2y^2)

    let  y = vx
    dy = vdx + xdv

    x^2dy = (4x^2 + 7xy + 2y^2)dx

    x^2(xdv + vdx) = (4x^2 + 7x^2v + 2x^2v^2)dx

    (xdv + vdx) = (4 + 7v + 2v^2)dx

    \frac{dv}{4 + 6v + 2v^2} = \frac{dx}{x}

    integrate the left side by partial fraction

    \frac{1}{2}(-\ln|v+2| + \ln|v+1|) = \ln{x} + \ln{c}
    -ln|v+2| + \ln|v+1| = 2lnx + c

    \frac{v+1}{v+2} = x^2c

    substitute y = vx

    \frac{y}{x} + 1 = cx^2(\frac{y}{x} + 2)

    y + x = cx^2(y+2x)
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    \frac{1}{2}(-\ln|v+2| + \ln|v+1|) = \ln{x} + \ln{c}
    -ln|v+2| + \ln|v+1| = 2lnx + c

    \frac{v+1}{v+2} = x^2c
    The only thing I have to say about this, and I would imagine it's likely due to a typo, is your treatment of the arbitrary constant in these three lines. I would write something like
    \frac{1}{2}(-\ln|v+2| + \ln|v+1|) = \ln{x} + \ln{c}

    -ln|v+2| + \ln|v+1| = 2lnx + ln(c^{\prime}) <-- Where c^{\prime} is a new constant.

    \frac{v+1}{v+2} = x^2c^{\prime}

    or just keep the original constant and wind up with
    \frac{v+1}{v+2} = x^2c^2
    (This last form might even be seen to be the best because it illustrates that the constant c^{\prime} is required to be positive.)

    -Dan
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