Question

Find the solutions of the following differential equations in terms of Bessel functions by using equations (16.1) and (16.2).

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{(}\mathbf{4}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{)}\mathbf{y}\mathbf{=}\mathbf{0}$

Short answer

The solution of the differential equation $\mathrm{y}\text{'}\text{'}-\frac{1}{\mathrm{x}}\mathrm{y}\text{'}+(4+\frac{1}{{\mathrm{x}}^2})\mathrm{y}=0$is

role="math" localid="1659265236089" $\mathrm{y}=\mathrm{x}\left[{\mathrm{AJ}}_0\left(2\mathrm{x}\right)+{\mathrm{BN}}_0\left(2\mathrm{x}\right)\right]$.

Step-by-step solution

Concept of Bessel functions by using equations (16.1) and (16.2):

Consider the standard form as,

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

Solution to the above mentioned standard equation is,

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

Find the solutions of y''-1xy'+(4+1x2)y=0differential equations:

Consider the given equation as follow.

$\mathrm{y}\text{'}\text{'}-\frac{1}{\mathrm{x}}\mathrm{y}\text{'}+(4+\frac{1}{{\mathrm{x}}^2})\mathrm{y}=0$

Compare the given equation with standard differential equation as follows:

$$\begin{gathered} 1-2\mathrm{a}=-1 \\ {\left(\mathrm{bc}\right)}^2=4 \\ 2(\mathrm{c}-1)=0 \\ {\mathrm{a}}^2-{\mathrm{p}}^2{\mathrm{c}}^2=1 \end{gathered}$$

Therefore, the given values are as follows:

$$\begin{gathered} \mathrm{a}=\frac{1}{6} \\ \mathrm{c}=\frac{14}{2} \\ \mathrm{p}=\frac{1}{3} \\ \mathrm{b}=4 \end{gathered}$$

Hence, the solution is $y=x{Z}_0\left(2x\right)$.

Therefore, $\mathrm{y}=\mathrm{x}\left[{\mathrm{AJ}}_0\left(2\mathrm{x}\right)+{\mathrm{BN}}_0\left(2\mathrm{x}\right)\right]$.