3.15.2.1 Example \(x^{2}y^{\prime \prime }+xy^{\prime }+\left ( ax^{2}-n^{2}\right ) y=0\)

\begin{equation} x^{2}y^{\prime \prime }+xy^{\prime }+\left ( ax^{2}-n^{2}\right ) y=0 \tag {1}\end{equation}

Comparing (1) to (C) shows that

\begin{align*} \left ( 1-2\alpha \right ) & =1\\ 2\gamma & =2\\ a & =\beta ^{2}\gamma ^{2}\\ \gamma ^{2} & =1\\ \alpha & =0 \end{align*}

Solving shows that \(\gamma =1,\beta =\sqrt {a}\). Hence the solution from (C1) can now be written directly as

\[ y\left ( x\right ) =c_{1}J_{n}\left ( \sqrt {a}x\right ) +c_{2}Y_{n}\left ( \sqrt {a}x\right ) \]

Another way to obtain this solution is to use the transformation

\[ x=\frac {1}{\sqrt {a}}z \]

Which converts (1) to

\begin{equation} z^{2}y^{\prime \prime }+zy^{\prime }+\left ( x^{2}-v^{2}\right ) y=0 \tag {2}\end{equation}

This is now in standard form (A) which has solution

\[ y\left ( z\right ) =c_{1}J_{v}\left ( z\right ) +c_{2}Y_{v}\left ( z\right ) \]

Replacing back \(z=\sqrt {a}x\) in the above gives

\[ y\left ( x\right ) =c_{1}J_{v}\left ( \sqrt {a}x\right ) +c_{2}Y_{v}\left ( \sqrt {a}x\right ) \]

So the rule is that, the term is \(\left ( ax^{2}-n^{2}\right ) y\,\) then we can just replace \(J_{n}\left ( x\right ) \) and \(Y_{n}\left ( x\right ) \) in the standard solution with \(J_{n}\left ( \sqrt {a}x\right ) \) and \(Y_{n}\left ( \sqrt {a}x\right ) \). For example \(x^{2}y^{\prime \prime }+xy^{\prime }+\left ( 4x^{2}-9\right ) y=0\) will have the solution \(y\left ( x\right ) =c_{1}J_{3}\left ( 2x\right ) +c_{2}Y_{3}\left ( 2x\right ) \).