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Differential Equation with Bessel Function

  1. Nov 23, 2015 #1
    <<Moderator note: Missing template due to move from other forum.>>

    Good afternoon. I'm trying to solve a differential equation with bessel function solutions. I am trying to solve

    \begin{equation*}
    y''(x)+e^{2x}y(x)=0
    \end{equation*}

    using the substitution ##z=e^x##. The textbook this problem is from (Mathematical Methods in the Physical Sciences, Third Edition) also includes that a differential equation in the form

    \begin{equation}
    y''+\frac{1-2a}{x}y'+\Big[(bcx^{c-1})^2+ \frac{a^2-p^2c^2}{x^2}\Big]y=0
    \end{equation}

    has a the solution

    \begin{equation}
    y=x^aZ_p(bx^c)
    \end{equation}

    I believe that I need to use this to solve the problem.

    Using this substitution suggested in the problem I then have ##x=ln(z)## and I solve the derivatives of the functions of ##y(x)## in terms of a new function ##y(x)=w(ln(z))##.

    \begin{equation*}
    y(x)=w(ln(z)) \\
    y'(x)=\frac{w'(ln(z))}{z} \\
    y''(x)=\frac{w''(ln(z))-w'(ln(z))}{z^2}
    \end{equation*}

    Doing the substitution I get

    \begin{equation*}
    \frac{w''}{z^2}-\frac{w'}{z^2}+z^2w=0
    \end{equation*}

    multiplying through by ##z^2## to get rid of the denominators gives me

    \begin{equation*}
    w''-w'+z^4w=0
    \end{equation*}

    I then compare this to the differential equation of form

    \begin{equation}
    w''+\frac{1-2a}{z}w'+\Big[(bcz^{c-1})^2+ \frac{a^2-p^2c^2}{z^2}\Big]w=0
    \end{equation}

    and by inspection get the values for ##a##, ##b##, ##c## and ##p##

    \begin{equation*}
    1-2a=-1 \rightarrow a=1 \\
    (bc)^2=1 \rightarrow b= \frac{1}{c} = \frac{1}{3} \\
    2c-2 = 4 \rightarrow c=3 \\
    a^2-p^2c^2=0 \rightarrow p= \frac{1}{3}
    \end{equation*}

    I then go ahead and put these values into the solution ##w=z^aZ_p(bz^c)## and then substitute ##z=e^x## back in and yet a solution of

    \begin{equation*}
    y=e^xZ_{1/3}\Big(\frac{1}{3}e^{3x}\Big)=e^x\Big(AJ_{1/3}\big(\frac{1}{3}e^{3x}\big)+BN_{1/3}\big(\frac{1}{3}e^{3x}\big)\Big)
    \end{equation*}

    where ##A## and ##B## are arbitrary constants.

    If however I use a computer system to solve the D.E it gives a solution of

    \begin{equation*}
    y(x)=A J_0\left(\sqrt{e^{2 x}}\right)+B N_0\left(\sqrt{e^{2 x}}\right)
    \end{equation*}

    I do not believe these are equal and am unsure how to check. Also I am unsure if this is even the correct method for solving D.E with Bessel function solutions as I haven't seen an example using substitution and am having a hard time finding other resources on the method given in the text. Is there another method for solving D.Es with Bessel function solutions? Also is there a name for the method used above so I can try to find more resources on it?

    Thanks for any help you can offer, I do appreciate it!

    Edit: Also I just realized this is a homework question and is posted in the wrong forum. Miscellaneous question is how can I get this post switched to the correct form? Sorry about this!
     
    Last edited by a moderator: Nov 24, 2015
  2. jcsd
  3. Nov 24, 2015 #2

    Orodruin

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    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Everywhere in your equations, you have ##w## and its derivatives as a function of ##\ln x##. This means it does not satisfy the equation you found as ##w'(f(x)) \neq w'(x)##. You need to have ##w## as a function of the same argument variable as you have in your equation, i.e., ##w(z) = y(\ln z)## or, equivalently, ##w(e^x) = y(x)##.
     
    Last edited: Nov 24, 2015
  4. Nov 26, 2015 #3
    Thank you, my prof mentioned the same when I handed it in. I'll be redoing it with the correct variables to make sure I understand it.
     
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