Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

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\({s^2}\frac{{{d^2}x}}{{d{s^2}}} + s\frac{{dx}}{{ds}} + {s^2}x = 0\).

Hence showed that the differential equation of the aging spring becomes \({s^2}\frac{{{d^2}x}}{{d{s^2}}} + s\frac{{dx}}{{ds}} + {s^2}x = 0\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1:Define Bessel’s equation.

Let the Bessel equation be \({x^2}y'' + xy' + \left( {{x^2} - {n^2}} \right)y = 0\). This equation has two linearly independent solutions for 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 using Frobinous approach.

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

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

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

Step 2: Determine thederivatives.

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,

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

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

And also,

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

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

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

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

Step 3: Find the value of derivatives.

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)

Step 4: Proof for reducing the given DE.

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

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

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

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\({s^2}\frac{{{d^2}x}}{{d{s^2}}} + s\frac{{dx}}{{ds}} + {s^2}x = 0\).

Step by Step Solution

Step 1:Define Bessel’s equation.

Step 2: Determine thederivatives.

Step 3: Find the value of derivatives.

Step 4: Proof for reducing the given DE.

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

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\({s^2}\frac{{{d^2}x}}{{d{s^2}}} + s\frac{{dx}}{{ds}} + {s^2}x = 0\).

Step by Step Solution

Step 1:Define Bessel’s equation.

Step 2: Determine thederivatives.

Step 3: Find the value of derivatives.

Step 4: Proof for reducing the given DE.

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