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First order DE

  1. Sep 13, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Calculate the general solution of the following DE reducible to a homogeneous one: x+y-2+(x-y+4)y'=0.


    2. Relevant equations
    Not sure.


    3. The attempt at a solution
    My idea is first to write the DE into the homogeneous form and then solve it via any method that work.
    I've read on the Internet that a homogeneous DE is of the form y'(x)=ay(x) where a is a constant.
    What I've done is [itex]\frac{dy}{dx}=\frac{2-x-y}{x-y+4}[/itex] for x-y different from 4. So it seems I could write [itex]\frac{2-x-y}{x-y+4} =\alpha y(x)[/itex] but I do not see how it's possible.
    I don't really know how to go further.
     
  2. jcsd
  3. Sep 13, 2011 #2

    HallsofIvy

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    There are different meaning for "homogeous" for d.e.s of first order and higher order. A first order equation, of the form dy/dx= f(x,y)/g(x,y) is "homogeneous" if replacing both x and in f(x,y)/g(x,y) by ax and ay respectively, gives the same as f(x,y)/g(x,y)- that is, the "a" cancels out. Here, f(x,y)= 2- x- y and g(x,y)= x-y+ 4 and f(ax,ay)= 2- ax- ay and g(ax, ay)= ax- ay+ 4 so that f(x,y)/g(x,y)= (2- ax- ay)/(ax- ay+ 4) which does not equal f(x,y)/g(x,y). This equation is not "homogeneous" in that sense.

    For higher order, linear equations, a differential equation is "homogeneous" if there are NO terms that does not involve y or a derivative of y. This d.e. is not a linear equation to begin with so it is NOT homogeneous in that sense.
     
  4. Sep 13, 2011 #3

    fluidistic

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    Thank you for the information.
    Hmm so the problem is wrongly stated?
    I am not sure how to solve the DE. I had read the integrating factor method can solve any first order DE but I think it requires the DE to be linear which isn't the case here.
    What other method(s) could I use here?
     
  5. Sep 14, 2011 #4

    ehild

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    What about substitution:


    z=x-y+4?

    ehild
     
  6. Sep 14, 2011 #5

    ehild

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    The other method can be to check if the ODE is exact, that is
    (x+y-2)dx+(x-y+4)dy =d(F(x,y).

    ehild
     
  7. Sep 15, 2011 #6

    fluidistic

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    Thanks for your help!
    Hmm, I didn't reach anything useful.
    Just checked out, it is exact. I reach [itex]F(x,y)=\frac{x^2}{2}-\frac{y^2}{2}+xy-2x+4y[/itex] and according to Boa's mathematical methods for physics, "the solution is F(x,y)=C". But obviously it isn't the solution to the DE since I should get y(x)...
     
  8. Sep 15, 2011 #7
    I hope I helped...

    Write it as

    [tex](x + y)dx - (2x - 2y + 4)dy = 0[/tex]

    Don't check because it isn't exact, so you need an integrating factor.
     
  9. Sep 15, 2011 #8

    Char. Limit

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    Let F(x,y)=C. Then [itex]x^2 - y^2 + 2xy + 4x + 8y = 2C_1 = C_2[/itex]. That's an implicit function (or rather, collection of functions) of x in terms of y. It can be solved using the quadratic formula, arranging the terms like so:

    [tex](-1) y^2 + (8+2x) y + (x^2 + 4x - C_2) = 0[/tex]
     
  10. Sep 15, 2011 #9

    fluidistic

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    Thanks for helping too. I don't really understand why I can write the DE under this form.
    The one of ehild was understandable to me and looked exact too. But from my last step in my last post, I reached y(x) but it didn't satisfy the original DE so I think I made at least 1 mistake somewhere.
    I will come back tomorrow on this problem, must sleep now.

    Edit: Just saw your post char.limit. That is exactly what I've done on my draft but made a mistake I guess. Will come back tomorrow on it.
    By the way, could someone explain me the post of flyingpig? Looks an interesting approach I'm missing.
     
  11. Sep 15, 2011 #10

    Char. Limit

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    As I said above, you didn't make a mistake. You just neglected to go all the way with it.
     
  12. Sep 15, 2011 #11

    LCKurtz

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    :tongue2:
    That is the solution:
    [tex]\frac{x^2}{2}-\frac{y^2}{2}+xy-2x+4y = C[/tex]

    It defines y implicitly. If you must have an explicit formula, solve for y using the quadratic formula. But it will be ugly.

    [Edit] Never mind, I see someone else types faster :tongue2:
     
  13. Sep 16, 2011 #12

    ehild

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    Is that useful?
    z=x-y+4
    y=x-z+4
    y'=1-z'

    Original equation: x+y-2+(x-y+4)y'

    Substituting z:

    x+(x-z+4)-2+(x-(x-z+4)+4)(1-z')=0

    2x-z+2+z(1-z')=0

    2x-z+2+z-zz'=0

    2x+2-zz'=0

    2x+2=zz'
    can you find y(x)? Is the solution the same as yours? :smile:
     
    Last edited: Sep 16, 2011
  14. Sep 16, 2011 #13

    fluidistic

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    Unfortunately I reach different answers. I had solved the quadratic yesterday and I just redid the arithmetics, fixing some error. I reach [itex]y(x)=x+4 \pm \sqrt {2x^2+4x+C}[/itex].
    Using ehild's z-substitution, I reach [itex]y(x)=x+4 \pm \sqrt {2x^2+16+C}[/itex].
    I can't seem to find any error in either methods; must be another arithmetics errors...
     
  15. Sep 16, 2011 #14
    I need to make a correction. I read the question wrong. Don't even take my post seriously ...
     
  16. Sep 16, 2011 #15

    fluidistic

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    I see. No problem :)
    Thanks for your will of helping me though.
     
  17. Sep 17, 2011 #16

    ehild

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    I just wonder how could you get 16 instead of 4x. Next time show your derivations in detail, to spare extra work for everybody.


    [itex]2x+2=zz'[/itex]


    [itex]\int{(2x+2)dx}=\int{zdz}[/itex]

    [itex]x^2+2x+C'=\frac{1}{2}z^2[/itex]

    [itex]\mp \sqrt{2x^2+4x+C}=z[/itex]

    [itex]y=x-z+4[/itex]

    [itex]y=x+4\pm \sqrt{2x^2+4x+C}[/itex]

    ehild
     
  18. Sep 17, 2011 #17
    Oh wait, this is exact...Sorry lol
     
  19. Sep 17, 2011 #18

    fluidistic

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    My bad, this is exactly what I've found using your method. I meant that using your method I reach [itex]y=x+4\pm \sqrt{2x^2+4x+C}[/itex] while using the other method, I find a different answer.
    I start from the general solution [itex]\frac{x^2}{2}-\frac{y^2}{2}+xy-2x+4y=C[/itex]. Multiplying everything by -2, I get [itex]y^2+y(-2x-8)-x^2+4x+C=0[/itex]. Solving the quadratic for y(x) gives me a "+16" rather than a "+4x" compared to your result.
    [itex]y(x)= \frac{2x+8 \pm \sqrt {4x^2+16x +64 -4(-x^2+4x+C)} }{2}=x+4 \pm \frac{ \sqrt {8x^2+64+C}}{2}=x+4 \pm \sqrt {2x^2+16+C}[/itex].
    I don't see a single mistake in any step of either derivations, yet I reach a different result.
     
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