Question

(a) Use (10) of Section 6.4 and Problem 32 of Exercises 6.4 to show that

$y_{3/2}\left(x\right)=-\sqrt{\frac{2}{\mathrm{\pi x}}}\left(\frac{\cos x}{x}+\sin x\right)$

(b) Use the definition ${|}_v\left(x\right)=i^{-v}{J}_v\left(ix\right)$to show that

localid="1664351596582" $I_{1/2}\left(x\right)=\sqrt{\frac{2}{\mathrm{\pi x}}}\sin hxand{I}_{-1/2}\left(x\right)=\sqrt{\frac{2}{\mathrm{\pi x}}}\cos hx.$

(c) Use (16) of Section 6.4 and part (b) to show that

$K_{12}\left(x\right)=\sqrt{\frac{\mathrm{\pi }}{2x}}{e}^{-x}$

Short answer

Hence proved a, b and c.

Step-by-step solution

Definition of Power series solutions

If X = X₀is an ordinary point of the differential equation , we can always find two linearly independent solutions in the form of a power series centered at x₀, 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 fromto the closest singular point.

Differentiation

Let us show that

$y_{3/2}\left(x\right)=[-\sqrt{\frac{2}{\mathrm{\pi x}}}\left(\frac{\cos x}{x}+\sin x\right)$

The equation (10) of Section 6.4 is:

$Y_v\left(x\right)=\frac{\cos v{\mathrm{\pi J}}_{\mathrm{v}}\left(\mathrm{x}\right)-{\mathrm{J}}_{-\mathrm{v}}\left(\mathrm{x}\right)}{\sin v\mathrm{\pi }}$

$Y_v\left(x\right)=\frac{\mathrm{cos\pi }{J}_v\left(x\right)-J_{-v}\left(x\right)}{\sin v\mathrm{\pi }}$

The Problem 32 of Section 6.4 gives us:

role="math" localid="1664353870179" $J_{3/2}\left(x\right)=-\sqrt{\frac{2}{{\mathrm{\pi x}}^3}}\cos x-\sqrt{\frac{2}{\mathrm{\pi x}}}\sin x.$

and

$J_{-3/2}\left(x\right)=-\sqrt{\frac{2}{{\mathrm{\pi x}}^3}}\cos x-\sqrt{\frac{2}{\mathrm{\pi x}}}\sin x.$

Substituting $\frac{3}{2}$for in (10), we obtain:

$Y_{3/2}\left(x\right)=\frac{\cos {\displaystyle \frac{3\mathrm{\pi }}{2}}{J}_{3/2}\left(x\right)-J_{3/2}\left(x\right)}{\sin {\displaystyle \frac{3\mathrm{\pi }}{2}}}$

$$\begin{gathered} \frac{0.J_{3/2}\left(x\right)-J_{-3/2}\left(x\right)}{-1}=J_{-3/2}\left(x\right) \\ \stackrel{3}{=}-\sqrt{\frac{2}{{\mathrm{\pi x}}^3}}\cos x-\sqrt{\frac{2}{\mathrm{\pi x}}}\sin x \end{gathered}$$

Notice that this is equivalent to (1).

Substitution

(b)

Let us show that

$I_{1/2}\left(x\right)=\sqrt{\frac{2}{\mathrm{\pi x}}}\sin hxand{I}_{-1/2}\left(x\right)=\sqrt{\frac{2}{\mathrm{\pi x}}}\cos hx.$

Equations (15) of Section 6.4 is given by:

$I_v=i^{-v}{J}_v\left(ix\right)$

From this, we have:

$I_{1/2}\left(x\right)=1^{-1/2}{J}_{1/2}\left(ix\right)andI1/2\left(x\right)=i^{1/2}J1/2\left(ix\right)$

In Section 6.4 it is shown that

role="math" localid="1664357914317" $J_{1/2}\left(x\right)=\sqrt{\frac{2}{\mathrm{\pi x}}}\sum _{n=0}^{\infty }\frac{{(-1)}^n}{(2n+1)!}{x}^{2n}$

and

role="math" localid="1664357994797" $J_{-1/2}\left(x\right)=\sqrt{\frac{2}{\mathrm{\pi x}}}\sum _{n=0}^{\infty }\frac{{(-1)}^n}{\left(2n\right)!}{x}^{2n}$

Substituting ( 7 ) and (8) into (6), we find:

role="math" localid="1664358201560" $I_{1/2}\left(x\right)=\sqrt{\frac{2}{\mathrm{\pi x}}}\sum _{n=0}^{\infty }\frac{{(-1)}^n}{(2n+1)!}{x}^{2n-1}=\sqrt{\frac{2}{\mathrm{\pi x}}}\sin hx$

and

$I_{-1/2}\left(x\right)=\sqrt{\frac{2}{\mathrm{\pi x}}}\sum _{n=0}^{\infty }\frac{1}{\left(2n\right)!}{x}^{2n}=\sqrt{\frac{2}{\mathrm{\pi x}}}\cos hx$

Find the value

(c)

Let us show that

$K_{1/2}\left(x\right)=\sqrt{\frac{\mathrm{\pi }}{2x}}{e}^{-x}$

The equation (16) of Section 6.4 is given by:

$K_v=\frac{\mathrm{\pi }}{2}\frac{I_{-v}-I_v\left(x\right)}{\sin v\mathrm{\pi }}$

Substituting $\frac{1}{2}$for v in (16) and using part (b), we obtain:

$K_{1/2}=\frac{\mathrm{\pi }}{2}\frac{I_{-1/2}-I_{1/2}\left(x\right)}{\sin {\displaystyle \frac{\mathrm{\pi }}{2}}}$

$$\begin{gathered} \frac{\mathrm{\pi }}{2}\left[\begin{array}{c}\sqrt{\frac{2}{\mathrm{\pi x}}}\cos hx-\sqrt{\frac{2}{\mathrm{\pi x}}}\sin hx\end{array}\right] \\ \sqrt{\frac{\mathrm{\pi }}{2x}}\left[\frac{e^x+e^{-x}}{2}-\frac{e^x-e^{-x}}{2}\right] \\ \sqrt{\frac{\mathrm{\pi }}{2x}}{e}^{-x} \end{gathered}$$