Chapter 13: Q45E (page 591)
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: Q45E (page 591)
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 formula obtained in Example \(6\) along with part (a) of Problem \(27\) to derive the recurrence relation \(2v{J_v}(x) = x{J_{v + 1}}(x) + x{J_{v - 1}}(x)\).
In Problems 29 and 30 use (22) or (23) to obtain the given result.
\(\int_0^x r {J_0}(r)dr = x \times {J_1}(x)\)
In Problems 29 and 30 use (22) or (23) to obtain the given result.
\(\int_0^x r {J_0}(r)dr = x \times {J_1}(x)\)
In Problems 29 and 30 use (22) or (23) to obtain the given result.
\({J_0}(x) = {J_{ - 1}}(x) = {J_1}(x)\)
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.
\(16{x^2}y'' + 32xy' + \left( {{x^4} - 12} \right)y = 0\)
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