Chapter 13: Q64E (page 610)
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
The obtained integral is \(J_0^'(x) = - {J_1}(x) = {J_{ - 1}}(x)\).
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Chapter 13: Q64E (page 610)
In Problems 29 and 30 use (22) or (23) to obtain the given result.
\({J_0}(x) = {J_{ - 1}}(x) = {J_1}(x)\)
The obtained integral is \(J_0^'(x) = - {J_1}(x) = {J_{ - 1}}(x)\).
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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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