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Math Help - Method of characteristics and general solution for partial differential equations

  1. #1
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    Method of characteristics and general solution for partial differential equations

    Find the general solution of the linear equation , x2ux + y2uy = (x+y)u .

    I found out the first constant C1 = x-1 - y-1 ,

    using dx/x2 = dy/y2 = du/(x+y)u. (1)

    According to my book, the second constant C2 = (x -y)/u , using (dx -dy)/ (x2-y2) = du/(x+y)u.

    I don't understand how did they derive (dx -dy)/(x2 - y2) from equation (1)?

    Can someone please help me out with this ?
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  2. #2
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    Re: Method of characteristics and general solution for partial differential equations

    (x2 - y2) = (x-y)* (x+y) and then (dx -dy)/(x-y)(x+y)=du/(x+y)u.

    simplify (x+y) in the left side with (x+y) in the right side

    Then (dx -dy)/(x-y) = du/u.

    => d(x-y)/(x-y) = du/u.

    Integrating => ln(x-y) = ln U + ln C
    => ln[(x-y)/u] = ln C

    =>(x-y)/u = c
    Last edited by julihasanaj; September 14th 2014 at 04:46 AM.
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