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Second order diff.eq. help?

  1. Feb 14, 2008 #1
    Second order diff.eq. help??

    well i am trying to find a solution to this diff. eq, but i get stuck somewhere.
    [tex]\ 4x^{2}y''+y=0[/tex]

    I first took this substitution

    y'=p, y"=p' so the diff. eq becomes of this form

    [tex]\ 4x^{2}p'+p=0[/tex]
    i think this can be done with the separable of variables thing. so
    [tex]\ 4x^{2}\frac{dp}{dx}=-p[/tex]

    [tex]\frac{dp}{p}=-\frac{dx}{4x^{2}}[/tex] now i integrate bot sides
    [tex] \int\frac{dp}{p}[/tex] = [tex] -\int\frac{dx}{4x^{2}}[/tex], after some calculations i get

    [tex] p=A e^{\frac{1}{4x}[/tex], where A is a constant that we get because of integrating bot siedes. NOw i go back to p=y'

    [tex] y'= A e^{\frac{1}{4x}[/tex], now i think all i need to do is integrate, but i get stuck integrating the right part of this eq. I do not think it has any closed form, right?

    but i know that the answer to this equation is in a closed form, that is, it is an elementary function, so where am i going wrong?
    I also tried to consider the eq [tex]\ 4x^{2}p'+p=0[/tex] as a linear one, but also when i tried to derive an integrating factor i ended up with a similar expression as [tex] y'= A e^{\frac{1}{4x}[/tex],
    so, any hints on how to solve this?
  2. jcsd
  3. Feb 14, 2008 #2
    You can not apply the substitution [itex]y'(x)=p(x),\,y''=p'(x)[/itex] in your equation since it does not contains [itex]y'(x)[/itex]
    Try the substitution [itex]y(x)=\sqrt{x}\,p(x)[/itex] to arrive to [itex]x\,p''(x)+p'(x)=0[/itex] and then try your substitution. i.e. [itex]p'(x)=w(x),\,p''(x)=w'(x)[/itex]
  4. Feb 14, 2008 #3
    How does one know that such a substitution will work?? Is there a theorem that provides us with these kind of substitutions or???
  5. Feb 14, 2008 #4
    I'm not sure how the above works. Why not substitute into the equation y=x^n and then solve for n. I say this because if you differentiate twice you lose two powers of x.
  6. Mar 10, 2008 #5
    Why can't you use the Power Series Method where you let the solution y=[tex]\sum_{0}a_n{}x^{n}[/tex] where a is from to infinite. The you will replace y along with its derivatives in that equation
  7. Mar 10, 2008 #6
    I guess i have to wait for another month then, untill we get to that chapter! We haven't yet done this method of inifinite taylor series, or whatever!.
  8. Mar 10, 2008 #7
    DEs with this form are called Euler differential equations. A substuition of the form:

    z = ln(x)

    with [tex]\frac{dy}{dx} = \frac{dy}{dz}\frac{dz}{dx}[/tex], can be used to convert this to a constant coefficient second order equation.
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