(a) Here we have F = y² + w²y² + 2y (a sin wx+bsinh wx), so F₁ = 2y' and Fy = 2w²y + 2(a sin wx + b sinh wx), and by (4.2) in the course notes the Euler-Lagrange equation is Solve y" - w2y = a sin wx+bsinh wx, y(1) = 0, y(2) = 1. (b) Because of the boundary conditions, it is convenient to write the general solution of the homogeneous equation y" — w²y = 0 in the form - c = a sinh(c(2 − 1))+Bsinh(c(2 −2)), - where a and ẞ are constants. A particular solution of the inhomogeneous equation y" — w²y = a sin wx + b sinh wx is y = p cos wx + q sin wx + rx coshwx + sx sinhwx, and substituting this into the equation gives p = 0, q = −a/(2w²), r = b/(2w), s = 0. Hence the general solution is I how come?? - y = a sinh(@(2 − 1)) +ßsinh(c( −2)) - a sin wx 2w2 bx cosh wx + 2w
(a) Here we have F = y² + w²y² + 2y (a sin wx+bsinh wx), so F₁ = 2y' and Fy = 2w²y + 2(a sin wx + b sinh wx), and by (4.2) in the course notes the Euler-Lagrange equation is Solve y" - w2y = a sin wx+bsinh wx, y(1) = 0, y(2) = 1. (b) Because of the boundary conditions, it is convenient to write the general solution of the homogeneous equation y" — w²y = 0 in the form - c = a sinh(c(2 − 1))+Bsinh(c(2 −2)), - where a and ẞ are constants. A particular solution of the inhomogeneous equation y" — w²y = a sin wx + b sinh wx is y = p cos wx + q sin wx + rx coshwx + sx sinhwx, and substituting this into the equation gives p = 0, q = −a/(2w²), r = b/(2w), s = 0. Hence the general solution is I how come?? - y = a sinh(@(2 − 1)) +ßsinh(c( −2)) - a sin wx 2w2 bx cosh wx + 2w
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 92E
Related questions
Question
![(a) Here we have F = y² + w²y² + 2y (a sin wx+bsinh wx), so F₁ = 2y' and
Fy = 2w²y + 2(a sin wx + b sinh wx), and by (4.2) in the course notes the
Euler-Lagrange equation is
Solve
y" - w2y = a sin wx+bsinh wx,
y(1) = 0, y(2) = 1.
(b) Because of the boundary conditions, it is convenient to write the general
solution of the homogeneous equation y" — w²y = 0 in the form
-
c = a sinh(c(2 − 1))+Bsinh(c(2 −2)),
-
where a and ẞ are constants. A particular solution of the inhomogeneous
equation y" — w²y = a sin wx + b sinh wx is
y = p cos wx + q sin wx + rx coshwx + sx sinhwx, and substituting this into the
equation gives p = 0, q = −a/(2w²), r = b/(2w), s = 0. Hence the general
solution is
I how come??
-
y = a sinh(@(2 − 1)) +ßsinh(c( −2))
-
a sin wx
2w2
bx cosh wx
+
2w](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa6c8ed7d-75cc-4e27-869e-3ad6a1efc0b4%2Fa0c2c72d-42ca-45ae-b5b5-7239c45f527c%2Fucoq2di_processed.png&w=3840&q=75)
Transcribed Image Text:(a) Here we have F = y² + w²y² + 2y (a sin wx+bsinh wx), so F₁ = 2y' and
Fy = 2w²y + 2(a sin wx + b sinh wx), and by (4.2) in the course notes the
Euler-Lagrange equation is
Solve
y" - w2y = a sin wx+bsinh wx,
y(1) = 0, y(2) = 1.
(b) Because of the boundary conditions, it is convenient to write the general
solution of the homogeneous equation y" — w²y = 0 in the form
-
c = a sinh(c(2 − 1))+Bsinh(c(2 −2)),
-
where a and ẞ are constants. A particular solution of the inhomogeneous
equation y" — w²y = a sin wx + b sinh wx is
y = p cos wx + q sin wx + rx coshwx + sx sinhwx, and substituting this into the
equation gives p = 0, q = −a/(2w²), r = b/(2w), s = 0. Hence the general
solution is
I how come??
-
y = a sinh(@(2 − 1)) +ßsinh(c( −2))
-
a sin wx
2w2
bx cosh wx
+
2w
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