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Second order non homogeneous DE

  1. Jan 30, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Solve [itex]x^2y''-3y=x^3[/itex].
    Show that there are many solutions [itex]\phi[/itex] such that [itex]\phi (0)= \phi '(0)=0[/itex].


    2. Relevant equations
    Not sure.


    3. The attempt at a solution

    It's a Cauchy-Euler equation so that I made the ansatz [itex]\phi (x)=x^\alpha[/itex]. I reached that [itex]x^\alpha [\alpha (\alpha -1 )-3]=x^3[/itex]. For this to be true for all x, I think that [itex]\alpha (\alpha -1 )-3[/itex] must equal [itex]0[/itex].
    If this is right, I reach that [itex]\alpha = \frac{1 \pm \sqrt { 13}}{2}[/itex]. Because these 2 roots are real and distinct, I have that [itex]y(x)=c_1 x ^{ \frac{1+\sqrt 13}{2} }+c_2 x ^{ \frac{1-\sqrt 13}{2} }[/itex]. However this does not agree with Wolfram alpha: http://www.wolframalpha.com/input/?i=x^2y''-3y=x^3.
    I don't know what I'm missing/did wrong. Any help is appreciated!
     
  2. jcsd
  3. Jan 30, 2012 #2

    HallsofIvy

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    You are correct that the standard method of solving such an equation gives that "general solution". However, the leading coefficient, [itex]x^2[/itex] is 0 at x= 0. The "existance and uniqueness" theorem does not apply here so that is NOT the only possible solution.
     
  4. Jan 30, 2012 #3

    fluidistic

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    Wow, I didn't pay any attention to this.
    So my solution is right for all x different from 0 if I understood well, correct?
    For x=0 the DE becomes [itex]-3y(x)=0\Rightarrow y(x)=0[/itex]. So there exist a solution for x=0 and it's the trivial one... is this correct?
     
  5. Feb 1, 2012 #4

    fluidistic

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    Hmm this can't be right. I have no idea how to get the solution when x=0. I don't understand what's wrong in my reasoning in the last post.
    How does one get the "x³/3" extra term that wolfram alpha have?
     
  6. Feb 1, 2012 #5
    I know this isn't the best way to learn, but click on "Show Steps" at WolframAlpha and it will instruct you to use variation of parameters to find the particular solution.
     
  7. Feb 1, 2012 #6

    Dick

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    x^3/3 is an inhomogeneous solution. Just try it. It works. And your homogeneous solutions look like the same ones WA gives if you simplify the radicals. y(x)=0 doesn't not work. Setting x=0 is not likely to give a solution to an ODE.
     
    Last edited: Feb 1, 2012
  8. Feb 2, 2012 #7

    fluidistic

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    Oh I see guys, I didn't realize that the Cauchy-Euler equation was necessarily homogeneous! So I followed brainlessly wikipedia's article to solve the homogeneous equation and never the inhomogeneous!
    Also I didn't know wolfram alpha could give such a detailled answer, I'm speechless.
    I have however a few questions:
    1)The general solution (mine+ the x³/3 term) seems to have 3 linearly independent terms, and since it's a second order DE, I'd think that the general solution would have at most 2 linearly independent terms. What's going on here?
    2)According to my friend, the general solution close to x=0 is [itex]\phi (x)=ax^{(1+\sqrt {13})/2 }+x^3/3[/itex]. So he dropped the term of the general solution anywhere else that contains a negative exponent. Is it only because otherwise [itex]\phi[/itex] would diverge in x=0?
    I had set x=0 into the DE, because I wanted to know if there was a solution around x=0 (as HallsofIvy said, we cannot instantly say if there exist a solution there, and if there exist a solution we cannot say it's unique, there might be an infinity of them, so x=0 is a special case to take care of). So I wonder how to analyse the DE around x=0. I just take the general solution of the DE anywhere else x=0 and modify it so that it fits in x=0? That looks so artificial to me... That way I could never get an infinity of solutions in x=0, while in fact it would be possible. I'm missing something here.
     
  9. Feb 2, 2012 #8

    Dick

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    Yes, you drop the negative exponent solution so you can get the boundary conditions at x=0 to work. Try to determine the constant 'a' in your solution from the boundary conditions at x=0.
     
  10. Feb 2, 2012 #9

    fluidistic

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    [itex]\phi (x)=c_1x^{(1+\sqrt 13 )/2}+\frac{x^3}{3}[/itex].
    [itex]\phi ' (x)= \left ( \frac{1+\sqrt 13}{2} \right )c_1 x ^{\frac{\sqrt 13 -1 }{2}} +x^2[/itex].
    Boundary conditions: [itex]\phi(0)=\phi ' (0)=0[/itex]. But both [itex]\phi (0)[/itex] and [itex]\phi' (0)[/itex] are worth 0 when x=0, no matter what [itex]c_1[/itex] is worth. So [itex]c_1\in \mathbb{R}[/itex] or even [itex]\mathbb {C}[/itex]. So that to answer the original question... this show that there are an infinity of solutions satisfying the given boundary conditions?! Wow that's really impressive.

    Also I'm astonished the general solution when x is different from 0 contains 3 linearly independent terms while the DE is of order 2. I didn't know that was possible.
     
  11. Feb 2, 2012 #10

    Dick

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    y''=1 has the general solution c1+c2*x+x^2/2. The two linearly independent solutions are 1 and x. The x^2/2 is the inhomogeneous solution. It doesn't count as a linearly independent solution of the ODE. There's no arbitrary constant in front of it. Same here.
     
    Last edited: Feb 2, 2012
  12. Feb 2, 2012 #11

    fluidistic

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    Ah I see, you are right. Thanks a lot for all.
     
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