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Homework Help: Homogenous Diff. Eqn

  1. Apr 15, 2008 #1
    1. The problem statement, all variables and given/known data

    dy/dx = (x + 3y)/(x - y)

    A) Solve the Differential Eqn
    B) Draw a Direction Field and some integral curves. Are they symmetric w/ respect to the origin?

    2. Relevant equations

    I believe i solved the equation correctly, but i dont know how to draw the direction fields and integral curves. I tried plotting y v. y' in order to create the resulting direction field and integral curves, but i dont know what it looks like.

    *Also, how do I integrate the left side? I just used an online calculator to get the answer, but i would like to know how to solve this

    3. The attempt at a solution

    A) After dividing by x and substituting v = y/x:

    (1 + 3v)/(1 - v) = xv' + v

    v' = (1/x)( (1 + 3v)/(1 - v) - v )

    * (1-3v)/(1+3v) - 1/v dv = dx/x

    After integrating and solving for c:

    C = (2/3)ln(3y/x + 1) - y/x -ln(y/x) - ln(x)
    Also, y = -x
     
  2. jcsd
  3. Apr 16, 2008 #2

    benorin

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    * Use long division: [tex]\frac{1-3v}{1+3v}=-1+\frac{2}{1+3v}[/tex] then

    [tex]\int \left(\frac{1-3v}{1+3v}-\frac{1}{v}\right)\, dv = \int \left(-1+\frac{2}{1+3v}-\frac{1}{v}\right)\, dv = -v+{\textstyle\frac{2}{3}} \ln |1+3v| -\ln |v|+C[/tex]​

    A direction field (a.k.a http://mathworld.wolfram.com/SlopeField.html" [Broken], see link for a better definition) is a plot of unit vectors with slope determined by the DE (e.g., on our direction field at (5,1) we plot a unit vector with a slope of [itex]y^{\prime}=\frac{5+3}{5-1}= 2[/itex]).

    An http://mathworld.wolfram.com/IntegralCurve.html" [Broken] is a particular solution to a differential equation corresponding to a specific value of the equation's free parameters.
     
    Last edited by a moderator: May 3, 2017
  4. Apr 16, 2008 #3

    tiny-tim

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    Hey guys! :smile:

    Isn't it (1 - v)dv/(1 + v)² ? :confused:
     
  5. Apr 16, 2008 #4

    benorin

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    Thanks tiny-tim.
    Oops :blushing:...

    [tex]x\frac{dv}{dx}=\frac{1+3v}{1-v}-v[/tex]
    [tex] \frac{1-v}{(1+v)^2}dv = \frac{dx}{x}[/tex]
     
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