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In Problems 29 and 30 use (22) or (23) to obtain the given result.

\({J_0}(x) = {J_{ - 1}}(x) = {J_1}(x)\)

Short Answer

Expert verified

The obtained integral is \(J_0^'(x) = - {J_1}(x) = {J_{ - 1}}(x)\).

Step by step solution

01

Define differential recurrence relation.

Recurrence formulas that relate Bessel functions of different orders are important in theory and in applications.

\(\frac{d}{{dx}}\left( {{x^{ - v}}{J_v}(x)} \right) = - {x^{ - v}}{J_{v + 1}}(x)\)… (1)

\(\frac{d}{{dx}}\left( {{x^v}{J_v}(x)} \right) = {x^v}{J_{v - 1}}(x)\) … (2)

02

Obtain the integration.

Substitute the value\(v = 0\)in the equation (1).

\(\begin{array}{c}\frac{d}{{dx}}\left( {{x^0}{J_0}(x)} \right) = - {x^0}{J_{0 + 1}}(x)\\\frac{d}{{dx}}\left( {{J_0}(x)} \right) = {J_1}(x)\end{array}\)

\(J_0^'(x) = - {J_1}(x)\)… (3)

Substitute the value\(v = 0\)in the equation (2).

\(\begin{array}{c}\frac{d}{{dx}}\left( {{x^0}{J_0}(x)} \right) = {x^0}{J_{0 - 1}}(x)\\\frac{d}{{dx}}\left( {{J_0}(x)} \right) = {J_{ - 1}}(x)\end{array}\)

\(J_0^'(x) = {J_{ - 1}}(x)\)… (4)

From (3) and (4) yields the result,

\(J_0^'(x) = - {J_1}(x) = {J_{ - 1}}(x)\)

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Most popular questions from this chapter

Use the change of variables \(s = \frac{2}{\alpha }\sqrt {\frac{k}{m}} {e^{ - \alpha t/2}}\) to show that the differential equation of the aging spring \(mx'' + k{e^{ - \alpha t}}x = 0\),\(\alpha > 0\) becomes \({s^2}\frac{{{d^2}x}}{{d{s^2}}} + s\frac{{dx}}{{ds}} + {s^2}x = 0\).

In Problems \(23 - 26\) first use \((20)\) to express the general solution of the given differential equation in terms of Bessel functions. Then use \((26)\) and \((27)\) to express the general solution in terms of elementary functions.

\(4{x^2}y'' - 4xy' + \left( {16{x^2} + 3} \right)y = 0\)

Proceed as on page \(269\) to derive the elementary form of \({J_{ - 1/2}}(x)\) given in \((27)\).

Show that \(y = {x^{1/2}}w\left( {\frac{2}{3}\alpha {x^{3/2}}} \right)\) is a solution of the given form of Airy’s differential equation whenever w is a solution of the indicated Bessel’s equation. (Hint: After differentiating, substituting, and simplifying, then let \(t = \frac{2}{3}\alpha {x^{3/2}}\))

(a) \(y'' + {\alpha ^2}xy = 0,x > 0;{t^2}w'' + tw' + \left( {{t^2} - \frac{1}{9}w} \right) = 0,t > 0\)

(b) \(y'' - {\alpha ^2}xy = 0,x > 0;{t^2}w'' + tw' - \left( {{t^2} - \frac{1}{9}w} \right) = 0,t > 0\)

Show that \(y = {x^{1/2}}w\left( {\frac{2}{3}\alpha {x^{3/2}}} \right)\) is a solution of the given form of Airy’s differential equation whenever w is a solution of the indicated Bessel’s equation. (Hint: After differentiating, substituting, and simplifying, then let \(t = \frac{2}{3}\alpha {x^{3/2}}\))

(a) \(y'' + {\alpha ^2}xy = 0,x > 0;{t^2}w'' + tw' + \left( {{t^2} - \frac{1}{9}w} \right) = 0,t > 0\)

(b) \(y'' - {\alpha ^2}xy = 0,x > 0;{t^2}w'' + tw' - \left( {{t^2} - \frac{1}{9}w} \right) = 0,t > 0\)

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