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First order O.D.E. with no analytic solution?

  1. Nov 10, 2008 #1
    1. The problem statement, all variables and given/known data

    Does the following differential equation have a simple analytic solution?

    [tex]\frac{dy}{dx}=y-\frac{2x}{y}[/tex]

    2. Relevant equations

    Don't know.

    3. The attempt at a solution

    I would have said that there is no simple solution to this equation, I tried to find the integrating factor, which would just be [tex]-x[/tex] in this case, but i don't know what to do about the [tex]-\frac{2x}{y}[/tex] part? The standard formula's I have don't work.

    Any hints and pointers would be much appreciated!
     
    Last edited: Nov 10, 2008
  2. jcsd
  3. Nov 10, 2008 #2

    Office_Shredder

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    y-dy/dx = -2x/y multiply both sides by y

    y2 - y*dy/dx = -2x

    Try substituting v = y2
     
  4. Nov 10, 2008 #3
    Okay,

    [tex]y^{2}=v[/tex] gives,

    [tex]y*\frac{dy}{dx}=2x+v[/tex] rearranging,

    [tex]\int y dy=\int 2x dx +\int v dx[/tex] integrating,

    [tex]\frac{y^{2}}{2}=x^{2}+vx+k[/tex]

    Is that the right thing to do?

    Final answer being,

    [tex]\frac{y^{2}}{2}=x^{2}+y^{2}x+k[/tex]
     
  5. Nov 10, 2008 #4

    HallsofIvy

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    No, you can't do that! v is a function of y which is itself a function of x: the integral of "v dx" is NOT vx because v is not a constant. If you are going to set v= y2 then completely replace y with v.

    If v= y2 then dv/dx= 2y dy/dx so y dy/dx= (1/2)v dv/dx and the equation becomes (1/2)v dv/dx- v= 2x.

    But, frankly, I don't see how that helps. Office Shredder, do you have more to say?
     
  6. Nov 10, 2008 #5

    Mark44

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    What Office_Shredder meant was to substitue v = y^2 everywhere that y occurs, which means that you will also need to replace dy with some expression involving dv. The resulting DE should have only x, dx, v, and dv in it.
     
  7. Nov 10, 2008 #6
    Thanks guys, I didn't think you could do that with a differential equation, but it was the only way that the substitution could have helped. I think me it has no analytic soution.

    Thanks again!
     
  8. Nov 10, 2008 #7

    Office_Shredder

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    [tex] \frac{dy}{dx} = y - \frac{2x}{y}[/tex]

    So multiplying through by y and re-arranging

    [tex]y^2- y\frac{dy}{dx} = 2x[/tex]

    If [tex]v=y^2[/tex]

    Then [tex]dv/dx = 2y\frac{dy}{dx}[/tex]
    We get

    [tex]v -\frac{1}{2}\frac{dv}{dx} = 2x[/tex]

    And this is good because when you have a polynomial of degree n on the RHS, you just guess v(x) = a generic polynomial of degree n (in this case 1).

    [tex]v(x) = ax+b[/tex] Then

    [tex]ax+b - \frac{1}{2}a = 2x[/tex]

    Then a=2, b=1 by simple algebra

    So [tex]y = \sqrt{2x+1}[/tex] (plus or minus). I plugged this in and got

    [tex]y\frac{dy}{dx} = \sqrt{2x+1}\frac{2}{2\sqrt{2x+1}} = 1[/tex]

    and

    [tex]\sqrt{2x+1}^2 - 2x = 2x+1 -2x = 1[/tex]

    So in fact the solution works
     
  9. Nov 10, 2008 #8

    HallsofIvy

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    Ouch! I can't believe I made that silly mistake!
     
  10. Nov 10, 2008 #9

    Office_Shredder

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  11. Nov 10, 2008 #10
    Ahh, I see. Now that makes a lot of sense. Sorry about that, I need to go back to my text book. :redface:

    In my worksheet it gave me four first order ode's and said one had no simple analytical solution, and that was the only one I couldn't see how to solve, so now I'm concerned. Does the following look okay to you?

    [tex]y*\frac{dy}{dx}-y^{2}+2=0[/tex]

    Using the substitution [tex]v=y^{2}[/tex] again gives,

    [tex]\int\frac{1}{2v-4} dv=\int dx[/tex]

    [tex]\ln{(2v-4)^{1/2}}=x+k[/tex]

    [tex]\ln{(2y^{2}-4)}^{1/2}=x+k[/tex]

    [tex]y=(\frac{k*e^{2x}+4}{2})^{1/2}[/tex]

    The other two are trivial,

    [tex]y\frac{dy}{dx}=y^{2}-2y[/tex] which gives,

    [tex]y=2+Ce^{x}[/tex]

    and,

    [tex]y\frac{dy}{dx}=y^{2}-2x*y[/tex] which gives,

    [tex]y=2*x+2+ke^{x}[/tex]
    What do you think?
     
  12. Nov 10, 2008 #11
    Also I should note that the general solution was required if it can be found. Sorry I should have said that sooner,
     
  13. Nov 10, 2008 #12

    Office_Shredder

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    If you can find a general solution, just find the complementary solution of the homogeneous equation (which is easy) and add it to any particular solution. Contrary to whatever anyone may say about how it's more elegant to find the full set of solutions in one swoop through a clever integration, they're lying. Always
     
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