Solving another bessel equation by substitution 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? 

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*}
