Question

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

$\mathbf{xy}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{+}\mathbf{xy}\mathbf{=}\mathbf{0}$

Short answer

The solution of the differential equation $\mathrm{xy}\text{'}\text{'}+5\mathrm{y}+\mathrm{xy}=0\mathrm{is}\mathrm{y}={\mathrm{x}}^{-2}\left[{\mathrm{AJ}}_2\left(\mathrm{x}\right)+{\mathrm{BN}}_2\left(\mathrm{x}\right)\right]$ .

Step-by-step solution

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

Consider the standard form is,

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

Solution to the above mentioned standard equation is,

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

Find the solutions of xy''+5y'+xy=0differential equations:

Consider the given equation as follow.

$\mathrm{xy}\text{'}\text{'}+5\mathrm{y}\text{'}+\mathrm{xy}=0$

Rearrange the above equation as follow.

$\mathrm{y}\text{'}\text{'}+\frac{5\mathrm{y}\text{'}}{\mathrm{x}}+\mathrm{y}=0$

Compare the given equation with standard differential equation as follows:

$$\begin{gathered} 1-2\mathrm{a}=5....\left(1\right) \\ {\left(\mathrm{bc}\right)}^2=1....\left(2\right) \\ 2(\mathrm{c}-1)=0.....\left(3\right) \\ {\mathrm{a}}^2-{\mathrm{p}}^2{\mathrm{c}}^2=0......\left(4\right) \end{gathered}$$

Therefore, the values are as follows:

$$\begin{gathered} -2\mathrm{a}=5-1 \\ \mathrm{a}=-\frac{4}{2} \\ \mathrm{a}=-2 \end{gathered}$$

By solving equation (1), you have

$$\begin{gathered} \left(\mathrm{c}-1\right)=0 \\ \mathrm{c}=1 \end{gathered}$$

Substitute -2 for a and 1 for c into equation (4).

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

And putting 1 for c into equation (2) you get,

b = 1

Hence, the solution is

$\mathrm{y}={\mathrm{x}}^{-2}{\mathrm{Z}}_2\left(\mathrm{x}\right)$

Therefore, $\mathrm{y}={\mathrm{x}}^{-2}\left[{\mathrm{AJ}}_2\left(\mathrm{x}\right)+{\mathrm{BN}}_2\left(\mathrm{x}\right)\right]$.