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Strange ODE from my final today?

  1. Dec 6, 2011 #1
    1. The problem statement, all variables and given/known data

    So I had my final exam today in ODE and I had an equation which appeared to be exact, but was not. I also tried to find a special integrating factor to make it exact, but no success. I then attempted to manipulate it into a linear eq, tried separable variables, even tried to get it into a Bernoulli form, all of which no luck. This was the extra credit question so I know it must be difficult, but I cannot find an example similar to it in our text, "Fundamentals of Differential Equations" by Nagle. Here is the equation and I was told I could use any method I like to solve it.

    x(x-y-2)dx+y(y-x+4)=0

    2. Relevant equations

    It is not an exact differential, cannot obtain a special integrating factor. Thanks again for any assistance, this problem has been bugging me all day!

    3. The attempt at a solution
     
  2. jcsd
  3. Dec 7, 2011 #2

    Mark44

    Staff: Mentor

    There should be a dy somewhere in this problem.
     
  4. Dec 7, 2011 #3
    Crap you're correct Mark, the equation actually read as x(x-y-2)dx+y(y-x+4)dy=0. I have determined that I need to do some sort of substitution, using v=[itex]y/x[/itex], which ought to transform it into some sort of separable variables. However, I am still running into trouble attempting to separate into dv=dx. Here is where I get stuck:

    [itex]\frac{dy}{dx}[/itex]=-[itex]\frac{x(x-y-2)}{y(y-x+4)}[/itex]

    dividing by x

    =-[itex]\frac{(1-y/x-2/x)}{y/x(y/x-1+4/x)}[/itex]=-[itex]\frac{(x-y-2)}{(y^{2}/x-y+4y/x)}[/itex]

    substituting [itex]v[/itex]=[itex]y/x[/itex], [itex]y=vx[/itex], [itex]\frac{dy}{dx}[/itex]=[itex]\frac{dv}{dx}x[/itex]+[itex]v[/itex] and some rearranging and I get stuck...

    [itex]\Rightarrow[/itex] [itex]\frac{dv}{dx}[/itex] = -[itex]\frac{(x-vx-2)}{(v^{2}x-vx^{2}+4vx)}[/itex]-[itex]\frac{v}{x}[/itex]

    Do I now need to come up with another substitution? Thanks for you help.

    Joe
     
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