Question

The first $J{\text{'}}_v\left(x\right)=\frac{v}{x}{J}_v\left(x\right)-J_v\left(x\right)-J_{v+1}\left(x\right)and{J}_v\left(x\right)=-\frac{v}{x}{J}_v\left(x\right)+J_{v-1}\left(x\right)$-order differential equation$dy/dx=x^2+y^2$cannot be solved in terms of elementary functions. However, a solution can be expressed in terms of Bessel functions.

(a) Show that the substitution $y=-\frac{1}{u}\frac{du}{dx}$leads to the equation$u"+x^{2}u=\text{'}0$

(b) Use (20) in Section 6.4 to find the general solution of$u^n+x^{2}u=0$

(c) Use (22) and (23) in Section 6.4 in the forms

as an aid to show that a one-parameter family of solutions of$dy/dx=x^2+y^2$is given by

$y=x\frac{J_{3/4}\left({\displaystyle \frac{1}{2}}{x}^2\right)-C{J}_{-3/4}\left({\displaystyle \frac{1}{2}}{x}^2\right)}{C{J}_{1/4}\left({\displaystyle \frac{1}{2}}{x}^2\right)+J_{1/4}\left({\displaystyle \frac{1}{2}}{x}^2\right)}$

Short answer

(a) Find the derivation of the given y

$\frac{d^{2}u}{d{x}^2}+x^{2}u=0$

(b) role="math" localid="1664877495208" $u\left(x\right)=x^{1/2}\left[C_1{J}_{1/4}\left(\frac{1}{2}{x}^2\right)+C_2{J}_{1/4}\left(\frac{1}{2}{x}^2\right)\right]$

(c) Rewrite $u\left(x\right)$ as u(x)$=x^{1/2}w\left(t\right)$

Step-by-step solution

Definition of Power series solutions

If $x=x_0$is an ordinary point of the differential equation , we can always find two linearly independent solutions in the form of a power series centered at$x_0$, that is,

$y=\sum _{n=0}^x{C}_n{(x-x_0)}^n$

A power series solution converges at least on some interval defined by $\left|x-x_0\right|<R$, where R is the distance from$x_0$to the closest singular point.

Differentiation

We were given the next equation

$\frac{dy}{dx}=x^2+y^2$ (1)

and the substitution

role="math" localid="1664875399677" $y=-\frac{1}{u}\frac{du}{dx}$ (2)

From (2), we find:

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

Substituting (2) and (3) into (1), we obtain:

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

From this, it follows:

$\frac{d^{2}u}{d{x}^2}+x^{2}u=0$

The general solution of the next equation:

$y"+\frac{1-2a}{x}y\text{'}+\left(b^2{c}^2{x}^{2c-2}+\frac{a^2-p^2{c}^2}{x2}\right)y=0,,$(4)

is given by:

$y=x^a\left[c_1{J}_p\left(b{x}^c\right)+c_2{Y}_p\left(b{x}^0\right)\right]$ (5)

The equation (4) is equivalent to (5) iff

$$\begin{gathered} 1-2a=0\iff a=\frac{1}{2} \\ 2c-2=2\iff c=2 \\ {b}^2{c}^2=1\iff b=\frac{1}{2} \\ {a}^2-p^2{c}^2=0\iff p-\frac{1}{4} \end{gathered}$$

Substitution

Substituting this into (6), we obtain;

$u\left(x\right)=x^{1/2}\left[C_1{J}_{1/4}\left(\frac{1}{2}{x}^2\right)+C_2{J}_{-1/4}\left(\frac{1}{2}{x}^2\right)\right]$

Notice that we can write u(x) in the next form:$u\left(x\right)=x^{1/2}$w(t) , where$t=\frac{1}{2}{x}^2$and .role="math" localid="1664881695366" $w\left(t\right)=c_1{J}_{1/4}\left(t\right)+c_2{J}_{-1/4}\left(t\right)$ Let us rewrite (2):

$y=-\frac{1}{u}\frac{du}{dx}=-\frac{1}{x^{1/2}w\left(t\right)}\frac{d}{dx}{x}^{1/2}w\left(t\right)$

role="math" localid="1664881105723" $$\begin{gathered} =-\frac{1}{x^{1/2}w\left(t\right)}\left[x^{1/2}\frac{dw}{dt}\frac{dt}{dw}+\frac{1}{2}{x}^{-1/2}w\right] \\ =-\frac{1}{x^{1/2}w\left(L\right)}\left[x^{3/2}\frac{dw}{dt}+\frac{1}{2x^{1/2}}w\right] \\ =-\frac{1}{2xw}\left[2x^2\frac{dw}{dt}+w\right] \end{gathered}$$

$=-\frac{1}{2xw}\left[4t\frac{dw}{dt}+w\right]$

Substitution and Derivations

Now, we will use the given equations:

$J_v^{\text{'}}\left(x\right)=\frac{v}{x}{J}_v\left(x\right)-J_{v+1}\left(x\right)$

and

$J_v^{\text{'}}\left(x\right)=-\frac{v}{x}{J}_v\left(x\right)+J_{v-1}\left(x\right).$

Let us now rewrite one part of the (8):

$4t\frac{dw}{dt}+w=4t\frac{d}{dt}\left[C_1{J}_{1/4}\left(t\right)+C_{2J}-1/4\left(t\right)\right]+C_1{J}_{1/4}\left(t\right)+C_{2}J-1/4\left(t\right)$

$$\begin{gathered} 4t\frac{dw}{dt}+w=4t\frac{d}{dt}\left[C_1{J}_{1/4}\left(t\right)+C_{2J}-1/4\left(t\right)\right]+C_1{J}_{1/4}\left(t\right)+C_{2}J-1/4\left(t\right) \\ \stackrel{\left(9\right),\left(10\right)}{=}4t\left[c_1\left(J_{3/4}\left(t\right)-\frac{1}{4t}{J}_{1/4}\left(t\right)\right)+C_2\left(-\frac{1}{4t}{J}_{1/4}\left(t\right)-J_{3/4}\left(t\right)\right)\right]+C_1{J}_{(1/4)}\left(t\right)+C_{2}J=1/4\left(t\right) \\ =4C_{1}t{J}_{-3/4}\left(t\right)-4c_{2}t{J}_{3/4}\left(t\right) \\ =2C_1{x}^2{J}_{3/4}\left(\frac{1}{2}{x}^2\right)-2C_2{x}^2{J}_{3/4}\left(\frac{1}{2}{x}^2\right) \end{gathered}$$

Substitute this and the definition of into (8) :

$$\begin{gathered} y=\frac{2c_1{x}^2{J}_{3/4}\left({\displaystyle \frac{1}{2}}{x}^2\right)-2C_2{x}^2{J}_{3/4}\left({\displaystyle \frac{1}{2}}{x}^2\right)}{2x\left[c_1{J}_{1/4}\left({\displaystyle \frac{1}{2}}{x}^2\right)+C_2{J}_{1/4}\left({\displaystyle \frac{1}{2}}{x}^2\right)\right]} \\ =x\frac{-c_1{J}_{3/4}\left({\displaystyle \frac{1}{2}}{x}^2\right)+2C_2{J}_{3/4}\left({\displaystyle \frac{1}{2}}{x}^2\right)}{c_1{J}_{1/4}\left({\displaystyle \frac{1}{2}}{x}^2\right)+C_2{J}_{1/4}\left({\displaystyle \frac{1}{2}}{x}^2\right)} \end{gathered}$$

Substitute $c;=\frac{c_1}{c_2}$into this to obtain: $y=x\frac{J_{3/4}\left({\displaystyle \frac{1}{2}{x}^2}\right)-c{J}_{3/4}\left({\displaystyle \frac{1}{2}{x}^2}\right)}{c{J}_{1/4}\left({\displaystyle \frac{1}{2}{x}^2}\right)+J_{1/4}\left({\displaystyle \frac{1}{2}{x}^2}\right)}.$.

(a) Find the derivation of the given y.

$\frac{d^{2}u}{d{x}^2}+x^{2}u=0.$

(b) $u\left(x\right)=x^{1/2}\left[C_1{J}_{1/4}\left(\frac{1}{2}{x}^2\right)+c_2{J}_{1/4}\left(\frac{1}{2}{x}^2\right)\right]$

(c) Rewrite u(x)as$u\left(x\right)=x^{1/2}w\left(t\right)$ .