The solution of problem as spherical Bessel function using definition of ${\mathbf{j}}_{\mathbf{n}}\left(x\right)$ and ${\mathit{y}}_{\mathbf{n}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ in terms of ${\mathbf{j}}_{\left(2n+1\right)\mathbf{/}\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ and ${\mathbf{y}}_{\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{/}\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$. Also obtain solutions in terms of $\mathbf{sin}\mathbf{}\mathbf{x}$and $\mathbf{}\mathbf{cos}\mathbf{}\mathbf{x}$. Compare the answers.

The differential equation,

${\mathbf{Y}}^{\mathbf{"}}\mathbf{+}\frac{\mathbf{1}\mathbf{-}\mathbf{2}\mathbf{a}}{\mathbf{x}}{\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{+}\left[{\left({\mathrm{bcx}}^{c-1}\right)}^2+\frac{a^2-p^2{c}^2}{x^2}\right]\mathbf{y}\mathbf{=}\mathbf{0}$ ….. (1)

Having solution,

$\mathbf{y}\mathbf{=}{\mathbf{x}}^{\mathbf{a}}{\mathbf{Z}}_{\mathbf{p}}\left({\mathrm{bx}}^c\right)$ …… (2)

Here, $\mathbf{Z}$stands for $\mathbf{J}$ or $\mathbf{N}$ or any linear combination of them and $\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{,}\mathbf{c}\mathbf{,}\mathbf{p}$ are constants.

${\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{"}}\mathbf{+}\mathbf{4}{\mathbf{xy}}^{\mathbf{\text{'}}}\mathbf{+}\left(x^2+2\right)\mathbf{y}\mathbf{=}\mathbf{0}$

Now, Divide both side by ${\mathbf{x}}^{\mathbf{2}}$:

$\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{y}}^{\mathbf{"}}\mathbf{+}\frac{\mathbf{4}\mathbf{x}}{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{+}\left(\frac{x^2+2}{x^2}\right)\mathbf{y}\mathbf{=}\mathbf{0}$ …… (3)

Now obtain the equations as follows:

$1-2\mathrm{a}=4$ …(4)

$$\begin{gathered} {\left(bc\right)}^2=1\dots \dots \left(5\right) \\ 2\left(c-1\right)=0\dots \dots \left(6\right) \\ {a}^2-p^2{c}^2=2\dots \dots \left(7\right) \end{gathered}$$

From these equations, you obtain:

$\mathrm{a}=-\frac{3}{2},\mathrm{b}=1,\mathrm{c}=1,\mathrm{p}=\frac{1}{2}$

Using (1), (2), and (4) in (3) as follows:

$$\begin{gathered} y=x^{-3/2}{Z}_{1/2}\left(1x^1\right) \\ =x^{-3/2}{Z}_{1/2}\left(x^1\right) \end{gathered}$$

So, the general solution of (1) is as follows:

$\mathrm{y}={\mathrm{x}}^{-3/2}\left[{\mathrm{AJ}}_{1/2}\left(\mathrm{x}\right)+{\mathrm{BY}}_{1/2}\left(\mathrm{x}\right)\right]$

Here, and are arbitrary constants.

Also:

localid="1659327482226" $$\begin{gathered} j_n\left(x\right)=\sqrt{\frac{\pi }{2x}}{J}_{\left(2n+1\right)/2}\left(x\right)\dots \dots \left(8\right) \\ {j}_n\left(x\right)=x^n{\left(-\frac{1}{x}\frac{d}{dx}\right)}^n\left(\frac{\sin x}{x}\right)\dots \dots \left(9\right) \\ {y}_n\left(x\right)=\sqrt{\frac{\pi }{2x}}{y}_{\left(2n+1\right)/2}\left(x\right)\dots \dots \left(10\right) \\ {y}_n\left(x\right)=x^n{\left(-\frac{1}{x}\frac{d}{dx}\right)}^n\left(\frac{\sin x}{x}\right)\dots \dots \left(11\right) \end{gathered}$$

Now for $n=0$ in (8) as follows:

$j_0\left(x\right)=\sqrt{\frac{\pi }{2x}}{J}_{1/2}\left(x\right)$

And in (10) as follows:

$$\begin{gathered} y_0\left(x\right)=\sqrt{\frac{\pi }{2x}}{y}_{1/2}\left(x\right) \\ {y}_{1/2}\left(x\right)=\sqrt{\frac{2x}{\pi }}{y}_0\left(x\right) \\ y=x^{-3/2}\left[A{J}_{1/2}\left(x\right)+B{Y}_{1/2}\left(x\right)\right] \\ =x^{-3/2}\left[A\sqrt{\frac{2x}{\pi }}{j}_0\left(x\right)+B\sqrt{\frac{2x}{\pi }}{y}_{0\left(x\right)}\right] \end{gathered}$$

Further simplify as follows:

$$\begin{gathered} y=x^{-1}\sqrt{\frac{2}{\pi }}\left[A{j}_0\left(x\right)+B{Y}_0\left(x\right)\right] \\ =\sqrt{\frac{2}{\pi }}\frac{1}{x}\left[A{j}_0\left(x\right)+B{Y}_0\left(x\right)\right] \\ {j}_0\left(x\right)=x^0{\left(-\frac{1}{x}\frac{d}{dx}\right)}^0\left(\frac{\sin x}{x}\right) \end{gathered}$$

For $\mathrm{n}=0$ in (6) and (11) $j_0\left(x\right)=\frac{\sin x}{x}$ as follows:

$$\begin{gathered} y_0\left(x\right)=-x^0{\left(-\frac{1}{x}\frac{d}{dx}\right)}^0\left(\frac{\cos x}{x}\right) \\ =\frac{\cos x}{x} \end{gathered}$$

The general solution is given by putting value in equation (9) as follows:

$$\begin{gathered} Y=x^{3/2}\left[A{J}_{1/2}\left(x\right)+B{Y}_{1/2}\left(x\right)\right] \\ =\sqrt{\frac{2}{\pi }}\left[A{J}_0\frac{\sin x}{x^2}-B\frac{\cos x}{x^2}\right] \end{gathered}$$

Comparing with $y=\frac{a_0\sin x}{x^2}+\frac{a_1\cos x}{x^2}$ the answer differ by constant only.

So here, $a_0=\sqrt{\frac{2}{\pi }}A$ and $a_1=-\sqrt{\frac{2}{\pi }}B$.

Both answers are same.