91Ó°ÊÓ

Let the Bessel equation be\({x^2}y'' + xy' + \left( {{x^2} - {n^2}} \right)y = 0\). This equation hastwo linearly independent solutionsfor a fixed value of\(n\).A Bessel equation of the first kind,indicated by\({J_n}(x)\), is one of these solutions that may be derived usingFrobinous approach.

\(\begin{array}{l}{y_1} = {x^a}{J_p}\left( {b{x^c}} \right)\\{y_2} = {x^a}{J_{ - p}}\left( {b{x^c}} \right)\end{array}\)

At\(x = 0\), this solution is regular. The second solution, which is singular at\(x = 0\), is represented by\({Y_n}(x)\)and is calleda Bessel function of the second kind.

\({y_3} = {x^a}\left( {\frac{{cosp\pi {J_p}\left( {b{x^c}} \right) - {J_{ - p}}\left( {b{x^c}} \right)}}{{sinp\pi }}} \right)\)

Let the given function be\(mx'' + k{e^{ - \alpha t}}x = 0\)… (1), and the change of variables be\(s = \frac{2}{\alpha }\sqrt {\frac{k}{m}} {e^{ - \alpha t/2}}\).

By using the chain rule, the derivatives are,

\(\begin{array}{c}\frac{{dx}}{{dt}} = \frac{{dx}}{{ds}} \times \frac{{ds}}{{dt}}\\\frac{{{d^2}x}}{{d{t^2}}} = \frac{d}{{dt}}\left( {\frac{{dx}}{{ds}}} \right)\frac{{ds}}{{dt}} + \frac{{dx}}{{ds}}\frac{d}{{dt}}\left( {\frac{{ds}}{{dt}}} \right)\\ = \frac{{ds}}{{dt}}\frac{d}{{ds}}\left( {\frac{{dx}}{{ds}}} \right)\frac{{ds}}{{dt}} + \frac{{dx}}{{ds}}\left( {\frac{{{d^2}s}}{{d{t^2}}}} \right)\end{array}\)

\(\frac{{{d^2}x}}{{d{t^2}}} = \frac{{{d^2}x}}{{d{s^2}}}{\left( {\frac{{ds}}{{dt}}} \right)^2} + \frac{{dx}}{{ds}}\frac{{{d^2}s}}{{dt}}\)… (2)

And also,

\(\begin{array}{c}\frac{{ds}}{{dt}} = \frac{d}{{dt}}\left( {\frac{2}{\alpha }\sqrt {\frac{k}{m}} {e^{ - \alpha t/2}}} \right)\\ = \frac{{ - \alpha }}{2}\left( {\frac{2}{\alpha }\sqrt {\frac{k}{m}} {e^{ - \alpha t/2}}} \right)\end{array}\)

\(\frac{{ds}}{{dt}} = - \sqrt {\frac{k}{m}} {e^{ - \alpha t/2}}\)… (3)

\(\begin{array}{c}\frac{{{d^2}s}}{{d{t^2}}} = \frac{d}{{dt}}\left( { - \sqrt {\frac{k}{m}} {e^{ - \alpha t/2}}} \right)\\ = \frac{{ - \alpha }}{2}\left( { - \sqrt {\frac{k}{m}} {e^{ - \alpha t/2}}} \right)\end{array}\)

\(\frac{{{d^2}s}}{{d{t^2}}} = \frac{\alpha }{2}\sqrt {\frac{k}{m}} {e^{ - \alpha t/2}}\) … (4)

Let substitute the equation (3) and (4) into (2) that yields,

\(\frac{{{d^2}x}}{{d{t^2}}} = \frac{{{d^2}x}}{{d{s^2}}}{\left( { - \sqrt {\frac{k}{m}} {e^{ - \alpha t/2}}} \right)^2} + \frac{{dx}}{{ds}} \times \frac{\alpha }{2}\sqrt {\frac{k}{m}} {e^{ - \alpha t/2}}\)

\(\frac{{{d^2}x}}{{d{t^2}}} = \frac{k}{m}{e^{ - \alpha t}}\frac{{{d^2}x}}{{d{s^2}}} + \frac{\alpha }{2}\sqrt {\frac{k}{m}} {e^{ - \alpha t/2}}\frac{{dx}}{{ds}}\) … (5)

Let substitute the equation (5) into (1) that yields,

\(\begin{array}{c}m\left( {\frac{k}{m}{e^{ - \alpha t}}\frac{{{d^2}x}}{{d{s^2}}} + \frac{\alpha }{2}\sqrt {\frac{k}{m}} {e^{ - \alpha t/2}}\frac{{dx}}{{ds}}} \right) + k{e^{ - \alpha t}}x = 0\\k{e^{ - \alpha t}}\frac{{{d^2}x}}{{d{s^2}}} + \frac{{\alpha m}}{2}\sqrt {\frac{k}{m}} {e^{ - \alpha t/2}}\frac{{dx}}{{ds}} + k{e^{ - \alpha t}}x = 0\\\frac{4}{{{\alpha ^2}m}} \times k{e^{ - \alpha t}}\frac{{{d^2}x}}{{d{s^2}}} + \frac{4}{{{\alpha ^2}m}} \times \frac{{\alpha m}}{2}\sqrt {\frac{k}{m}} {e^{ - \alpha t/2}}\frac{{dx}}{{ds}} + \frac{4}{{{\alpha ^2}m}} \times k{e^{ - \alpha t}}x = 0\\\frac{4}{{{\alpha ^2}}} \times \frac{k}{m}{e^{ - \alpha t}}\frac{{{d^2}x}}{{d{s^2}}} + \frac{2}{\alpha }\sqrt {\frac{k}{m}} {e^{ - \alpha t/2}}\frac{{dx}}{{ds}} + \frac{4}{{{\alpha ^2}}} \times \frac{k}{m}{e^{ - \alpha t}}x = 0\\{\left( {\frac{2}{\alpha }\sqrt {\frac{k}{m}} {e^{ - \alpha t/2}}} \right)^2}\frac{{{d^2}x}}{{d{s^2}}} + \left( {\frac{2}{\alpha }\sqrt {\frac{k}{m}} {e^{ - \alpha t/2}}} \right)\frac{{dx}}{{ds}} + {\left( {\frac{2}{\alpha }\sqrt {\frac{k}{m}} {e^{ - \alpha t/2}}} \right)^2}x = 0\\{s^2}\frac{{{d^2}x}}{{d{s^2}}} + s\frac{{dx}}{{ds}} + {s^2}x = 0\end{array}\)