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Solution of linear differential equation

  1. Sep 28, 2008 #1
    1. The problem statement, all variables and given/known data

    solve differential equation

    xdy + (xy+y-1)dx=0


    2. Relevant equations

    dy/dx + P(x)y = Q(x)

    u = exp[int(P(x)] - integrating factor

    exp[int(P(x))]*[dy/dx+P(x)y = Q(x)]

    => d/dx[P(x)]y = exp[int(P(x))]Q(x)

    int[d/dx[P(x)]y = exp[int(P(x))]Q(x)]

    solution is with respect to y.

    3. The attempt at a solution

    xdy = -(xy+y-1)dx

    dy/dx = (1-y+xy)/x

    dx/dy = x/(1-y+xy)

    dx/dy = x-x/y+1/y

    dx/dy - x + x/y = 1/y

    I'm stuck here, I'm just not sure who to get the equation into a linear form where I could take the integrating factor and solve for x in the case since I switched the dependent variables to see if that wouldn't make it more complicated, any help from that point on would be greatly appreciated.
     
  2. jcsd
  3. Sep 28, 2008 #2
    divide the whole equation by dx, then u get x(dy/dx)+xy+y-1=0

    move the 1 to other side, and factor a y out , and u get

    x(dy/dx)+y(x+1)=1

    divide by x

    (dy/dx)+[(x+1)/x]y=1/2

    now u in form ,and u have P(x)Y :)

    the rest i think you know how to do it
     
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