Solve this equation to show that the stationary path isa sin wa262y=a sinh (@(z − 1)) +ßsinh(@(2 − 2))-whereα ==1sinhw1sinh1+a sin 2w26²b cosh w2wb cosh 2wa sin w262W]+bx cosh wx2w Let a,isb and w be constants, with w‡0.theEuler-Lagrange equation for the functional2S[y] = [² da (y² + w²y² + 2y(a sin wa + b sinhwa)),1y(1) = 0, y(2) = 1,y" - w²y = a sin wx+bsinh wx, y(1) = 0, y(2) = 1.2

Let be constants .Given that : Then, we can write…

Answered: Solve this equation to show that the…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Related questions

Question

Solve this equation to show that the stationary path is
a sin wa
262
y=a sinh (@(z − 1)) +ßsinh(@(2 − 2))
-
where
α =
=
1
sinhw
1
sinh
1+
a sin 2w
26²
b cosh w
2w
b cosh 2w
a sin w
262
W
]
+
bx cosh wx
2w

Let a,
is
b and w be constants, with w‡0.
the
Euler-Lagrange equation for the functional
2
S[y] = [² da (y² + w²y² + 2y(a sin wa + b sinhwa)),
1
y(1) = 0, y(2) = 1,
y" - w²y = a sin wx+bsinh wx, y(1) = 0, y(2) = 1.
2

Expert Solution

Step 1: ''Introduction to the solution''

Let $a comma b comma omega not equal to 0$ be constants .

Given that : $S open square brackets y close square brackets equals integral subscript 1 superscript 2 space d x space open parentheses y apostrophe squared plus omega squared y squared plus 2 y open parentheses a sin omega x plus b sin h omega x close parentheses close parentheses comma$

$y left parenthesis 1 right parenthesis equals 0 comma y left parenthesis 2 right parenthesis equals 1$

Then, we can write $S left square bracket y right square bracket equals integral subscript 1 superscript 2 F left parenthesis x comma y comma y apostrophe right parenthesis d x......... left parenthesis 1 right parenthesis$

where $F left parenthesis x comma y comma y apostrophe right parenthesis equals open parentheses y apostrophe squared plus omega squared y squared plus 2 y open parentheses a sin omega x plus b sin h omega x close parentheses close parentheses space left parenthesis s a y right parenthesis$

Then, the Euler-Lagrange's equation is given by $fraction numerator partial differential F over denominator partial differential y end fraction minus fraction numerator d over denominator d x end fraction open parentheses fraction numerator partial differential F over denominator partial differential y apostrophe end fraction close parentheses equals 0............ left parenthesis 2 right parenthesis$

$rightwards double arrow 2 omega squared y plus 2 open parentheses a sin omega x plus b sin h omega x close parentheses equals 2 y apostrophe apostrophe$

$rightwards double arrow y apostrophe apostrophe minus omega squared y equals open parentheses a sin omega x plus b sin h omega x close parentheses........... left parenthesis 3 right parenthesis$

where $y left parenthesis 1 right parenthesis equals 0 comma y left parenthesis 2 right parenthesis equals 1.......... left parenthesis 4 right parenthesis$