In Exercises 53-58, verify that the given function is a particular solu- tion to the specified nonhomogeneous equation. Find the general solu- tion and evaluate its arbitrary constants to find the unique solution sat- isfying the equation and the given initial conditions. x" 53. y" + y' = x, yp x, y(0) = 0, y'(0) = 0 2 54. y" + y = x, yp = 2 sin xx, y(0) = 0, y'(0) = 0 55. - ½y" + y' + y = 4e³ (cos x − sinx), = Ур 2ex cos x, - 56. y" — y' — 2y = - y(0) = 0, y'(0) = 1 - - 1 − 2x, yp = x − 1, y(0) = 0, y'(0) = 1 57. y″ − 2y' + y = 2e*, yp = x²e*, y(0) = 1, y'(0) = 0 - 58. y” − 2y' + y = x¯¹e³, x > 0, = Ур xeln x, y(1) = e, y'(1) = 0
In Exercises 53-58, verify that the given function is a particular solu- tion to the specified nonhomogeneous equation. Find the general solu- tion and evaluate its arbitrary constants to find the unique solution sat- isfying the equation and the given initial conditions. x" 53. y" + y' = x, yp x, y(0) = 0, y'(0) = 0 2 54. y" + y = x, yp = 2 sin xx, y(0) = 0, y'(0) = 0 55. - ½y" + y' + y = 4e³ (cos x − sinx), = Ур 2ex cos x, - 56. y" — y' — 2y = - y(0) = 0, y'(0) = 1 - - 1 − 2x, yp = x − 1, y(0) = 0, y'(0) = 1 57. y″ − 2y' + y = 2e*, yp = x²e*, y(0) = 1, y'(0) = 0 - 58. y” − 2y' + y = x¯¹e³, x > 0, = Ур xeln x, y(1) = e, y'(1) = 0
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Transcribed Image Text:In Exercises 53-58, verify that the given function is a particular solu-
tion to the specified nonhomogeneous equation. Find the general solu-
tion and evaluate its arbitrary constants to find the unique solution sat-
isfying the equation and the given initial conditions.
x"
53. y" + y' = x, yp
x, y(0) = 0, y'(0) = 0
2
54. y" + y = x, yp
=
2 sin xx, y(0) = 0, y'(0) = 0
55.
-
½y" + y' + y = 4e³ (cos x − sinx),
=
Ур 2ex cos x,
-
56. y" — y' — 2y =
-
y(0) = 0, y'(0) = 1
-
-
1 − 2x, yp = x − 1, y(0) = 0, y'(0) = 1
57. y″ − 2y' + y = 2e*, yp = x²e*, y(0) = 1, y'(0) = 0
-
58. y” − 2y' + y = x¯¹e³, x > 0,
=
Ур xeln x, y(1) = e, y'(1) = 0
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