Solving another bessel equation by substitution

Question

For a problem solving class I need to find the general solution of ODE $x^2y''+(\frac{3}{16}+x)y=0$ in terms of $J_{\nu}$ and $J_{-\nu}$, if possible. $\nu$ represents the Bessel parameter.

A hint is given, namely that useful substitutions would be $y=2u \sqrt{x}$ and $\sqrt{x}=z$; this should lead to the ODE being reduced to a Bessel equation. Substituting this value and applying the chain rule leads to the ODE $z^2u''+zu'+(4z^2-(\frac{1}{4})^2)u=0$, which almost has the form of a Bessel equation with $\nu=\frac{1}{4}$, apart from the coefficient 4 in front of the $z^2$ in the $u$ coefficient.

What is the proper way of solving this ODE by reduction to a Bessel equation?
Can this be done?

Answer 1

What happens is that you will get $J_{1/4}(2z)$ and $Y_{1/4}(2z)$. To see this, try to put $s=2z$ and $v(s)=u(z)$. The terms with derivatives will behave very nicely: $$ zu'(z)=sv'(s)\quad\text{and}\quad z^2u''(z)=s^2v''(s). $$ Are you sure of the original differential equation? It yields simpler solutions, as Jack mentions.

Update

Your original differential equation transforms into $$ z^2u''(z)+zu'(z)+(4z^2-(1/2)^2)u(z)=0. $$ Note the $(1/2)^2$, and not $(1/4)^2$.

The same idea as before gives $J_{1/2}(2z)$ and $Y_{1/2}(2z)$. Now, these functions can be simplified into cosines and sines with some square roots of $x$ and so on.

Answer 2

By setting $$ y(x) = x^{1/4}\, f(\sqrt{x}) $$ the original ODE boils down to the ordinary differential equation: $$ f''(x) + 4\,f(x) = 0. $$

Answer 3

The previous remark answers your question, keeping track of the expansions \begin{gather*} \cos z = J_0(z)-2J_2(z)+2J_4(z)-\cdots;\\ \sin z= 2J_1(z)-2J_3(z)+\cdots. \end{gather*}