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Homework Help: Solve Homogeneous D.E. need help integrating

  1. Mar 5, 2014 #1
    1. The problem statement, all variables and given/known data

    Dy/Dx = (Y-x)/(Y+x)

    2. Relevant equations


    3. The attempt at a solution

    Dy/Dx = (Y-x)/(Y+x)

    Plug in my substitutions
    udx+xdu(1/dx)=(ux/ux+x) - X/(ux+x)

    u+x(du/dx)=(ux)/x(u+1) - (x)/((x)(u+1))

    u+x(du/dx)=u/(u+1) -(1)/(u+1))


    This is where I think i begin to mess up

    u+du=(u-1)/(u+1) dx/x

    substract (u-1)/(u+1) to the other side

    u-(u-1)/(u+1) du=dx/x

    I know the right side integrates to Lnx +c

    but on the left side if i do

    I split it up into

    the integral (u^2)/(u+1) minus integral of 1/(u+1)
    (u^2)/(u+1)<-use long division

    I get u+(1/u+1) minus the integral of 1/(u+1)
    i am left with just the integral
    of u

    u^2/2= lnx+c

    plug u back in.

    ((y/x)^2)/2 =lnx +c

    is this sufficient of an answer?

    according to the answer key im gonna end up with the arctan somewhere in my answer. so i may have already messed up :(
    Last edited: Mar 6, 2014
  2. jcsd
  3. Mar 6, 2014 #2


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    Homework Helper

    Why are you dividing by [itex]y + x[/itex] here? Is your equation actually
    \frac{dy}{dx} = \frac{y - x}{y + x}
    and not
    \frac{dy}{dx} = (y - x)(y + x)
    as you have written?

    This should be "u dx/x + du" on the left hand side.

    You can't; it's multiplied by dx/x.

    What you have after replacing [itex]y[/itex] is
    x\frac{du}{dx} + u = \frac{u - 1}{u + 1}
    Subtracting [itex]u[/itex] from both sides and then dividing by [itex]x[/itex] puts this in the separable form
    \frac{du}{dx} = \frac1x \left(\frac{u-1}{u+1} - u\right)
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