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(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

Advanced Engineering Mathematics
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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(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
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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