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Nonseparable Differential Equation

  1. Feb 29, 2008 #1
    I know that this differential equation is not separable, but is there a way to solve it?

    dy/dx=y+x

    I've tried a substitution of y=vx:

    (dv/dx)x+v=x+vx
    (dv/dx)x=x+vx-v
    dv/dx=1+v-(v/x)

    I'm stuck trying to rewrite that as a product of v and x.

    Any help is appreciated!
     
  2. jcsd
  3. Feb 29, 2008 #2

    arildno

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    Set u=y+x,

    then you get:
    du/dx=dy/dx+1,

    whereby your diff.eq reads:

    du/dx=1+u, which IS separable.

    Alternatively, find an integrating factor to your diff.eq.
     
  4. Feb 29, 2008 #3
    By using that method, my answer was: eˣ-x-1.

    Correct?

    Thanks!
     
  5. Feb 29, 2008 #4
    Of course there are many methods to solve the above equations as they are a system of linear differential equations. The methods solved above are great for their simplicity but not so great in terms of generality.
     
  6. Feb 29, 2008 #5

    arildno

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    Why not check it out?

    We have: y=e^x-x-1,

    whereby:

    dy/dx=e^x-1=(e^x-x-1)+x=y+x

    so that is indeed A solution.

    You still lack the general solution.
     
  7. Mar 2, 2008 #6
    isn't the general solution, the same thing but with the constant not being defined?

    if not then please clarify it for me.
     
  8. Mar 2, 2008 #7

    rock.freak667

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    Yes it is...but your solution was [itex]y=e^x-x-1[/itex] which is a particular solution. the general sol'n would be [itex]y=e^x-x-1+C[/itex]....C= a constant. Never forget the constant of integration.
     
  9. Mar 2, 2008 #8
    The general solution is:

    [tex]y(x)= A\cdot e^x -x- 1[/tex]

    If you set A=1 then you get the particular solution of altcmdesc. However, the general solution is also obtained via the method of Arildno.

    Nothing to do with adding a constant just like that, rock.freak667, you have to add it at the right place.
     
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