# Differential Equation with Bessel Function

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1. Nov 23, 2015

### PatsyTy

<<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

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

has a the solution

y=x^aZ_p(bx^c)

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

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

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?

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. Nov 24, 2015

### Orodruin

Staff Emeritus
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
3. Nov 26, 2015

### PatsyTy

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.