3.15.2.8 Example \(x^{2}y^{\prime \prime }-xy+\left ( x^{2}+1\right ) y=0\)

\[ x^{2}y^{\prime \prime }-xy+\left ( x^{2}+1\right ) y=0 \]

Comparing the above to (C) \(x^{2}y^{\prime \prime }+\left ( 1-2\alpha \right ) xy^{\prime }+\left ( \beta ^{2}\gamma ^{2}x^{2\gamma }-\left ( n^{2}\gamma ^{2}-\alpha ^{2}\right ) \right ) y=0\) shows that

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

Which implies \(\alpha =1\) and \(\gamma =1\) and \(\beta ^{2}\gamma ^{2}=1\) or \(\beta =1\). Last equation now becomes \(-\left ( n^{2}-1\right ) =1\) or \(n^{2}=0\) or \(n=0\). Hence the solution (C1) becomes

\begin{align*} y\left ( x\right ) & =x^{\alpha }\left ( c_{1}J_{n}\left ( \beta x^{\gamma }\right ) +c_{2}Y_{n}\left ( \beta x^{\gamma }\right ) \right ) \\ & =x\left ( c_{1}J_{0}\left ( x\right ) +c_{2}Y_{0}\left ( x\right ) \right ) \end{align*}