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Homogeneous Differential Equation

  1. Sep 8, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the general solution of the differential equation:

    (x+y)y'=x-y


    2. Relevant equations

    I want to solve this as a homogeneous differential equation, so our equations are:
    v=[itex]\frac{y}{x}[/itex], y=vx, [itex]\frac{dy}{dx}[/itex]=v+x[itex]\frac{dv}{dx}[/itex]


    3. The attempt at a solution

    I need to get this into the form [itex]\frac{dy}{dx}[/itex]=F([itex]\frac{y}{x}[/itex]), so I rewrite it as y'=[itex]\frac{x-y}{x+y}[/itex]. Dividing by x I get, y'=[itex]\frac{1-\frac{y}{x}}{1+\frac{y}{x}}[/itex]. From here, I substitute to get v+x[itex]\frac{dv}{dx}[/itex]=[itex]\frac{1-v}{1+v}[/itex]. When looking at the solution manual, however, it says that it should be in the form x(v+1)v'=-(v2+2v-1) before I integrate. I don't see how i can get it into this form. Also, it gives the answer as y2+2xy-x2=C. I can correctly solve this differential equation using different methods, but I would really like to know how to solve this using the homogeneous method. I would really appreciate any help with this. Thanks!
     
  2. jcsd
  3. Sep 8, 2011 #2

    hunt_mat

    User Avatar
    Homework Helper

    You are almost there...

    You have to compute:
    [tex]
    \frac{1-v}{1+v}-v=x\frac{dv}{dx}
    [/tex]
    This will be a separable equation which is solved in the usual fashion.
     
  4. Sep 8, 2011 #3
    you just have to take subtract v on both sides of the equation you got man......(xdv/dx=(1-v)/(1+v) - v..
     
  5. Sep 8, 2011 #4
    Ok, so I rewrite it as [itex]\frac{1}{x}[/itex]dx=[itex]\frac{1+v}{1-v}[/itex]-[itex]\frac{1}{v}[/itex]dv. Do I need to do an additional substitution on [itex]\frac{1+v}{1-v}[/itex]?
     
  6. Sep 8, 2011 #5
    you solved it wrong on right side... you will get dv/{(1-v)/(1+V) - v}....which comes out to be (1+v)/(1-v^2-2v)...just substitute 1-v^2-2v as t....
     
  7. Sep 8, 2011 #6
    I'm having trouble seeing how it comes out to (1+v)/(1-v2-2v)
     
  8. Sep 8, 2011 #7

    hunt_mat

    User Avatar
    Homework Helper

    Do the addition as I said to get:
    [tex]
    \frac{1-v}{1+v}-v=\frac{1-v-v(1+v)}{1+v}=\frac{1-v-v-v^{2}}{1+v}=\frac{1-2v-v^{2}}{1+v}
    [/tex]
     
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