10.15.2.7 Example $y^{\prime \prime }-\frac{1}{x^{\frac{3}{2}}}y=0$

Multiplying by $x^{\frac{3}{2}}$

$$x^{\frac{3}{2}}{y}^{\prime \prime }-y=0$$

Multiplying by $x^{\frac{1}{2}}$

$$x^2{y}^{\prime \prime }-x^{\frac{1}{2}}y=0$$

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

$$\begin{array}{rl}(1-2\alpha )& =0\\ {\beta }^2{\gamma }^2{x}^{2\gamma }& =-x^{\frac{1}{2}}\\ (n^2{\gamma }^2-{\alpha }^2)& =0\end{array}$$

Which implies $\alpha =\frac{1}{2},2\gamma =\frac{1}{2},{\beta }^2{\gamma }^2=-1$. Hence $\gamma =\frac{1}{4}$ and ${\beta }^2=-16$ or $\beta =\pm 4i$. Last equation now says $(n^2\frac{1}{16}-\frac{1}{4})=0$ or $n=2$. Hence the solution (C1) is

$$\begin{array}{rl}y\left(x\right)& =x^{\alpha }(c_1{J}_n\left(\beta x^{\gamma }\right)+c_2{Y}_n\left(\beta x^{\gamma }\right))\\ & =\sqrt{x}(c_1{J}_2\left(4i{x}^{\frac{1}{4}}\right)+c_2{Y}_2\left(4i{x}^{\frac{1}{4}}\right))\end{array}$$

By properties of Bessel functions, where $J_n\left(ai\sqrt{x}\right)=i^n{I}_n\left(a\sqrt{x}\right)$, then the above becomes

$$y\left(x\right)=\sqrt{x}(-c_1{I}_2\left(4x^{\frac{1}{4}}\right)+c_2{Y}_2\left(4i{x}^{\frac{1}{4}}\right))$$