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Solving differential equations.

  1. Oct 6, 2012 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    I came across two, not so obvious DEs that have stumped me abit.

    (1) x2y' = xy + x2ey/x
    (2) x2y' + 2xy = 5y3

    2. Relevant equations

    I know these are not separable, and more than likely require an integrating factor to put them into exact form so I can integrate them that way.

    3. The attempt at a solution

    I'm sort of stuck on both of them.

    For (1) I divided through by x2 to get y' = y/x + ey/x which further yields y' - y/x = ey/x which is sadly implicit in nature and not solvable by means of a regular integrating factor.

    Its the same for (2), once again I divide through by x2 to attain y' + (2/x)y = 5y3/x2 which is once again implicit and not solvable by regular means.

    I know I probably need to put these into exact form somehow, but I'm having trouble putting them into the form :

    [itex]M(x, y) + N(x, y) \frac{dy}{dx} = 0[/itex] So I can solve for an integrating factor [itex]\mu (x)[/itex] which satisfies :

    [tex]\frac{\frac{∂M(x, y)}{∂y} - \frac{∂N(x, y)}{∂x}}{N(x, y)} = \frac{d\mu}{dx}[/tex]
     
  2. jcsd
  3. Oct 7, 2012 #2

    rock.freak667

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    For the first one, try a substitution of y=Vx.
     
  4. Oct 7, 2012 #3
    For the second one, observe x^2y' + 2xy = (x^2y)'.
     
  5. Oct 7, 2012 #4

    Zondrina

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    I didn't know what substitution was, but I learned it quickly and I managed to get the solution to the first equation to be :

    -e-y/x = ln|x| + c which I decided to obviously leave implicit as no real solutions of y would exist.

    As for the second equation, voko you mentioned I could re-write this as :

    d/dx [x2y] = 5y3 I'm not sure how this would help though?
     
  6. Oct 7, 2012 #5
    What is in square brackets is another function u(x). Can you express y via x and u?

    Speaking of the first one, consider the case when c is a large negative constant.
     
  7. Oct 7, 2012 #6

    Zondrina

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    So for the first one I COULD re-write it as :

    y = xln|d-ln|x|| where d is some arbitrary constant ( I get d after some arithmetic ).

    As for the second one, are you hinting that a substitution v(x) = x2y which simplifies to x-2v(x) = y is what I need here?
     
  8. Oct 7, 2012 #7
    For the first one, I think it should be y = -x ln...

    The second one, try :)
     
  9. Oct 7, 2012 #8

    Zondrina

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    For the first one when I integrate and sub back m obvious choice of v = y/x I get :

    -e^-y/x = ln|x| + c
    e^-y/x = d - ln|x|
    -y/x = ln|d - ln|x||
    y = -xln|d-ln|x||

    I c wut u did thur.

    As for the second one I get the answer as : y = ±[itex]\sqrt{\frac{x}{2} + c}[/itex]

    I think that's good?
     
  10. Oct 7, 2012 #9
    I do not think the second result is correct. Please show your steps.
     
  11. Oct 7, 2012 #10

    Zondrina

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    x2y' + 2xy = 5y3
    d/dx [x2y] = 5y3

    Substituting v = x2y yields :

    dv/dx = 5y3

    Also notice that y = vx-2 so :

    dv/dx = 5(vx-2)3
    dv/dx = 5v3x-6

    Separating variables and integrating gives us :

    -(1/2v2) = -x-5 + c

    Subbing back v = x2y and simplifying :

    -(1/2(x2y)2 = -x-5 + c
    2x4y2 = x5+c
    y2 = x/2 + c/x4
    y = ±[itex]\sqrt{\frac{x}{2} + \frac{c}{x^4}}[/itex]

    I really should write my steps out haha.
     
  12. Oct 7, 2012 #11
    Good so far.

    Wrong, both the LHS and the RHS.
     
  13. Oct 7, 2012 #12

    Zondrina

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    Oh snap, silly mistake. I should also stop rushing when showing my work.

    -(1/2(x2y)2) = -x-5 + c
    1/(2x4y2) = x-5 - c
    2x4y2 = 1/(x-5-c)
    y2 = 1/(2x4(x-5-c))

    y = ±[itex]\sqrt{\frac{1}{2x^4(x^{-5}-c)}}[/itex]
     
  14. Oct 7, 2012 #13
    That's better :) To be completely sure, you could try and plug that into the original equation.
     
  15. Oct 7, 2012 #14

    Zondrina

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    Yup it works perfectly. Although I DO have a technicality question out of my own curiosity.

    Say I was given an initial condition y(x0) = y0.

    If y0 > 0, then the solution corresponds to the positive portion of y.

    If y0 < 0, then the solution corresponds to the negative portion of y.

    Now if y0 = 0, I'm not really sure what happens here? Which solution would I pick or would it never be the case that y0 = 0 because y is discontinuous there regardless of my choice. Seems logical, but some confirmation would be nice.

    Thanks for all your help btw :)
     
  16. Oct 7, 2012 #15
    y = 0 is a singular solution in this equation. It corresponds to an infinite c. Note the regular solutions approach it as x goes to zero.
     
  17. Oct 7, 2012 #16

    Zondrina

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    Ahh I see. That makes sense now, thank you.
     
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