Question

Assume that b in equation (20) can be pure imaginary, that is,. Use this assumption to express thegeneral solution of the given differential equation in terms ofthe modified Bessel functions In and Kn.

(a) y0 2 x2y 5 0

(b) xy0 1 y9 2 7x3

Short answer

The general solutions of the given differential equation are

$$\begin{gathered} y=x^{-1}\left(C_1{J}_{1/2}\left(\frac{1}{2}{x}^2\right)+C_2{J}_{-1/2}\left(\frac{1}{2}{x}^2\right)\right) \\ y=x^{-1}\left(C_1{J}_{1/2}\left(\frac{1}{2}{x}^2\right)+C_2{Y}_{1/2}\left(\frac{1}{2}{x}^2\right)\right) \end{gathered}$$

Step-by-step solution

Define Bessel’s equation.

Let the Bessel equation be $x^{2}y\text{'}\text{'}+xy\text{'}+\left(x^2-n^2\right)y=0$ This equation has two linearly independent solutions for a fixed value of . A Bessel equation of the first kind, indicated by J_n(X) , is one of these solutions that may be derived using Frobinous approach. $$\begin{gathered} y_1=x^a{J}_p\left(b{x}^c\right) \\ {y}_2=x^a{J}_{-p}\left(b{x}^c\right) \end{gathered}$$

At ,X =0 this solution is regular. The second solution, which is singular aX=0 t , is represented by Y_n(X) and is called a Bessel function of the second kind.

$y_3=x^a\left[\frac{cosp\pi J_p\left(b{x}^c\right)-J_{-p}\left(b{x}^c\right)}{sinp\pi }\right]$

Determine the general form of the Bessel’s equation.

(a)

Let the given differential equation be $\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{x}}^2+{\mathrm{y}}^2$… (1), and the substitution be

$y=-\frac{1}{u}\frac{du}{dx}$…(2)

Then, the first derivative is,

$\frac{dy}{dx}=-\frac{1}{u}\frac{d^{2}u}{d{x}^2}+-\frac{1}{u^2}{\left(\frac{du}{dx}\right)}^2$… (3)

Substitute (3) and (2) into the equation (1) yields,

$-\frac{1}{u}\frac{d^{2}u}{d{x}^2}+\frac{1}{u^2}{\left(\frac{du}{dx}\right)}^2=x^2+\frac{1}{u^2}\left(\frac{du}{dx}\right)$

From this, it follows:

$\frac{d^{2}u}{d{x}^2}+x^{2}u=0$… (4)

Find the general solution and the value of constants.

(b)

The equation becomes in the following form:

$\mathrm{y}\text{'}\text{'}+\frac{1-2\mathrm{a}}{\mathrm{x}}\mathrm{y}\text{'}+\left({\mathrm{b}}^2{\mathrm{c}}^2{\mathrm{x}}^{2\mathrm{c}-2}+\frac{{\mathrm{a}}^2-{\mathrm{p}}^2{\mathrm{c}}^2}{{\mathrm{x}}^2}\right)\mathrm{y}=0$… (5)

That yields,

$\mathrm{y}={\mathrm{x}}^{\mathrm{a}}[{\mathrm{c}}_1{\mathrm{J}}_{\mathrm{p}}\left({\mathrm{bx}}^{\mathrm{c}}\right)+{\mathrm{c}}_2{\mathrm{Y}}_{\mathrm{p}}\left({\mathrm{bx}}^{\mathrm{c}}\right)]$… (6)

The equation (5) is equivalent to (6), if

$\begin{array}{ccc}1-2\mathrm{a}=0& \iff & \mathrm{a}=\frac{1}{2}\\ 2\mathrm{c}-2=2& \iff & \mathrm{c}=2\\ {\mathrm{b}}^2{\mathrm{c}}^2=1& \iff & \mathrm{b}=\frac{1}{2}\\ {\mathrm{a}}^2-{\mathrm{p}}^2{\mathrm{c}}^2=0& \iff & \mathrm{p}=\frac{1}{4}\end{array}$

Substitute the values in the equation (6).

$\mathrm{u}\left(\mathrm{x}\right)={\mathrm{x}}^{1/2}\left[{\mathrm{c}}_1{\mathrm{J}}_{1/4}\left(\frac{1}{2}{\mathrm{x}}^2\right)+{\mathrm{c}}_2{\mathrm{J}}_{-1/4}\left(\frac{1}{2}{\mathrm{x}}^2\right)\right]$ … (7)

Obtain the general solution.

Let $u\left(x\right)=x^{1/2}w\left(t\right)$, where $t=\frac{1}{2}{x}^2$ and $w\left(t\right)=c_1{J}_{1/4}\left(t\right)+c_2{J}_{-1/4}\left(t\right)$. So, the equation (2) becomes,

$\begin{array}{rcl}y& =& -\frac{1}{u}\frac{du}{dx}\\ & =& -\frac{1}{x^{1/2}w\left(t\right)}\frac{d}{dx}{x}^{1/2}w\left(t\right)\\ & =& -\frac{1}{x^{1/2}w\left(t\right)}\left[x^{1/2}\frac{dw}{dt}\frac{dt}{dw}+\frac{1}{2}{x}^{-1/2}w\right]\\ & =& -\frac{1}{x^{1/2}w\left(t\right)}\left[x^{3/2}\frac{dw}{dt}+\frac{1}{2x^{1/2}}w\right]\\ & =& -\frac{1}{2xw}\left[2x^2\frac{dw}{dt}+w\right]\\ & =& -\frac{1}{2xw}\left[4t\frac{dw}{dt}+w\right]\\ & & \end{array}$

The general solution by using superposition principle is,

$\begin{array}{rcl}y& =& C_1{x}^{-1}{J}_{1/2}\left(\frac{1}{2}{x}^2\right)+C_2{x}^{-1}{J}_{-1/2}\left(\frac{1}{2}{x}^2\right)\\ & =& x^{-1}\left(C_1{J}_{1/2}\left(\frac{1}{2}{x}^2\right)+C_2{J}_{-1/2}\left(\frac{1}{2}{x}^2\right)\right)\\ & & \end{array}$

There is another general solution obtained from the Bessel’s equation of second order. (i.e.)$y=x^{-1}\left(C_1{J}_{1/2}\left(\frac{1}{2}{x}^2\right)+C_2{Y}_{1/2}\left(\frac{1}{2}{x}^2\right)\right)$.

Hence, the general solutions are $y=x^{-1}\left(C_1{J}_{1/2}\left(\frac{1}{2}{x}^2\right)+C_2{J}_{-1/2}\left(\frac{1}{2}{x}^2\right)\right)$and $y=x^{-1}\left(C_1{J}_{1/2}\left(\frac{1}{2}{x}^2\right)+C_2{Y}_{1/2}\left(\frac{1}{2}{x}^2\right)\right)$.