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Differential equations

  1. Oct 25, 2009 #1
    find y if

    dy/dx=[tex]\frac{x-y+2}{x-y+3}[/tex]

    what i tried to do was
    u=[tex]\frac{x-y+2}{x-y+3}[/tex]


    ux-uy+3u=x-y+2

    y(1-u)=x+2-u(3+x)

    y=[tex]\frac{x+2-u(3+x)}{(1-u)}[/tex]


    y'=[tex]\frac{(1-u'(3+x)-u)(1-u)+u'(x+2-u(3+x))}{1-2u+u^2}[/tex]

    [tex]\frac{(1-u'(3+x)-u)(1-u)+u'(x+2-u(3+x))}{1-2u+u^2}[/tex]=u

    (1-u'(3+x)-u)(1-u)+u'(x+2-u(3+x))=u-2u^2+u^3

    from here try get u's one side anx x's the other
    surely this isnt the way to do this ??
     
  2. jcsd
  3. Oct 25, 2009 #2
    Try [tex]u=x-y[/tex].
     
  4. Oct 25, 2009 #3
    perfect thanks
     
  5. Oct 25, 2009 #4
    what would i do in a problem where the numerator and denominator have different x and y's


    for example

    y'=(3x-y-9)/(x+y+1)
     
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