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Differential Equation - should be simple, but?

  1. Mar 7, 2007 #1
    1. The problem statement, all variables and given/known data

    I am studying for the exams, and not terribly practiced on DE's yet. From a mock-test I found this DE - it be solved? A complete solution would be appreciated so I can understand the answer and compare it to my notes please...


    dy/dx - 2xy = ( y ln (y) ) / x

    3. The attempt at a solution

    I have attempted so far:

    a) Separate variables.
    I cannot, I get stuck at something like dy + ( -2XY -(Y ln(y)) ) dx = 0, which I tried to make dy*x/y + (-2x^2 -lny )dx = 0 and cannot find a way to separate further

    b) Find a multiplicator (is it how is it called? sorry I am not a native English speaker and not yet very familiar with English mathematics terminology), anyway, something to multiply it with so that it can become directly differentable. I don't know how to go about it at the moment.

    c) Find a suitable transformation. No luck...

    Thanks for reading, thanks for the solution if you provide it...
     
    Last edited: Mar 7, 2007
  2. jcsd
  3. Mar 7, 2007 #2

    Dick

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    Write it as:

    (x/y)*dy-(2*x^2+ln(y))*dx=0 and think about it again. "Multiplicator"="Integrating factor".
     
  4. Mar 7, 2007 #3
    Thanks for the answer. I actually reached that point, but unless I am a complete dummy, it still is not an exact DE.

    ....M(x,y).............N(x,y)
    (2x^2 + lny)dx + (-x/y)dy = 0

    dM/dy = 1/y != dN/dx = -1/y so it is not exact.

    .So how do I tackle the minus? Is it still a matter of finding a correct integrating factor? Please take it a step further for me, this is actually the point I am stuck at, it is so frustrating I cannot describe it
     
    Last edited: Mar 7, 2007
  5. Mar 7, 2007 #4

    Dick

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    You are right. I missed the sign. Back to the drawing board.
     
  6. Mar 7, 2007 #5
    i hope it's not bad form to re-use my post from a similar type of question:


    from this point:


    yes, you can convert it to exact by trying:


     
    Last edited: Mar 7, 2007
  7. Mar 8, 2007 #6

    dextercioby

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    SO you have the equation

    [tex] \frac{dy}{dx}=2xy+\frac{y}{x}\ln y [/tex]

    Make the substitution
    [tex] y(x)=e^{p(x)} [/tex]

    The resulting ODE is

    [tex] \frac{dp}{dx}=\frac{p}{x}+2x [/tex]

    For [itex] x\neq 0 [/itex], we get

    [tex] \frac{1}{x}\frac{dp}{dx}-\frac{p}{x^2}=2 [/tex]

    which is the same as

    [tex] \frac{d}{dx}\left(\frac{p(x)}{x}\right)=2 [/tex]

    The rest is trivial.
     
  8. Mar 8, 2007 #7
    Thank you very much dick, mybsaccownt and dextercioby for the help, I have succeded in solving it both ways proposed (without looking at your solutions while I was doing it ofc :) )

    My solution was: -2x + y + (lny)/x = c

    Thanks again very much for your help.

    For dextercioby's solution, what hinted us to this transformation pls? Was it the presence of lny in the equation?
     
    Last edited: Mar 8, 2007
  9. Mar 9, 2007 #8

    dextercioby

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    Yes, of course, that way, i got rid of a nasty function of "y".
     
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