Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Homogeneous DE Problem

  1. Jan 12, 2007 #1
    1. The problem statement, all variables and given/known data

    I'm told that this is homogenous
    [tex](x^2-xy)y'+y^2 = 0[/tex]

    2. The attempt at a solution

    This is going to be very painful for me to type out...

    [tex](x^2-xy)\frac{dy}{dx} = -y^2[/tex]

    [tex]\frac{dy}{dx} = \frac{-y^2}{(x^2-xy)}[/tex]

    [tex]\frac{dy}{dx} = \frac{-y^2}{(x^2-xy)}[/tex]

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

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

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

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

    [tex]\frac{dv}{-v^2} = \frac {dx}{x}[/tex]

    [tex]\int-v^-^2dv = \int\frac {dx}{x}[/tex]

    [tex]v^-^1 =ln|x|+C[/tex]

    [tex]\frac{y^-^1}{x^-^1} = ln|x|+C[/tex]

    [tex]y^-^1 = x ln|x|+C[/tex]

    [tex]\frac{1}{y} = x ln|x|+C[/tex]

    [tex]\frac{1}{y} = \frac {ln|x|+C}{x}[/tex]

    [tex] 0 = \frac {y ln|x|+C}{x}[/tex]

    [tex] -Cx = y ln|x|[/tex]

    But the answer is...

    [tex] xln|y| - y = Cx[/tex]
     
    Last edited: Jan 12, 2007
  2. jcsd
  3. Jan 12, 2007 #2

    cristo

    User Avatar
    Staff Emeritus
    Science Advisor

    Try going from here again. Add another line in if you need to. Put all the v's over a common denominator before you multiply by dx.
     
  4. Jan 12, 2007 #3

    HallsofIvy

    User Avatar
    Science Advisor

    [tex]\frac{-v^2}{1-v}-v= \frac{-v^2-v+v^2}{1-v}= \frac{-v}{1-v}[/tex]
    The left hand side in your next line is wrong.

     
  5. Jan 12, 2007 #4
    Oh I understand the error, thanks, I'll work it out from there
     
  6. Jan 12, 2007 #5
    I've messed up again...

    [tex]v+x\frac{dv}{dx} = \frac{-v^2}{(1-v)}[/tex]
    [tex]\frac{v(1-v)}{-v^2}dv = \frac{dx}{(x)}[/tex]
    [tex]\frac{v-v^2}{-v^2}dv = \frac{dx}{(x)}[/tex]
    [tex]\int v^-^1-1 dv = \int \frac{dx}{(x)}[/tex]
    [tex]ln|v|-x = ln|x|+C[/tex]
    [tex]ln|\frac {y}{x}|-x = ln|x|+C[/tex]

    and no need to go any farter.. I know its wrong...
     
  7. Jan 12, 2007 #6
    is it cus this line

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

    should look like this?

    [tex]\frac{v(1-v)}{-v^2} + (1-v)dv = \frac{dx}{(x)}[/tex]
     
  8. Jan 12, 2007 #7

    cristo

    User Avatar
    Staff Emeritus
    Science Advisor

    Try going from this line. Simplify before you multiply by dx.

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

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

    Simplify from here, and then carry on!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook