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In Problems 1 through 16, a homogeneous second-order lin- ear differential equation, two functions y1 and y2, and a pair of initial conditions are given. First verify that y1 and y2 are solutions of the differential equation. Then find a particular solution of the form y = c1y1 + c2y2 that satisfies the given initial conditions. Primes denote derivatives with respect to x. 1. y" – y = 0; yı = e*, y2 = e=*; y(0) = 0, y'(0) = 5 7. y" + y' = 0; yı = 1, y2 = e-*; y(0) = -2, y'(0) = 8 8. y" – 3y' = 0; yı1 = 1, y2 = e3x; y(0) = 4, y'(0) = -2 9. y" + 2y' + y = 0; y1 = e-x, y2 = xe-*; y(0) = 2, y'(0) = -1 10. y" – 10y' + 25y = 0; yı = e5x, y2 = xe5x; y(0) = 3, y'(0) = 13 11. y" – 2y' +2y = 0; y1 = e* cos x, y2 = e* sin x; y(0) = 0, y'(0) = 5 12. y" + 6y' + 13y = 0; yı = e3x cos 2x, y2 = e-3x sin 2x; y (0) = 2, y'(0) = 0 13. x?y" – 2xy' + 2y = 0; y1 = x, y2 = x²; y(1) = 3, y'(1) = 1 14. x?y" + 2xy' – 6y = 0; y1 = x², y2 = x-3; y(2) = 10, y'(2) = 15 15. ху" — ху" + у%3D 0;B у1 %3D х, у2 %3D х Inx; у(1) %3D 7, y'(1) = 2 16. x2y" + xy' + y = 0; yı = y(1) = 2, y'(1) = 3 %3D %3D Зх. %3D %3D %3D = cos(In x), y2 = sin(In x);

In Problems 1 through 16, a homogeneous second-order lin- ear differential equation, two functions y1 and y2, and a pair of initial conditions are given. First verify that y1 and y2 are solutions of the differential equation. Then find a particular solution of the form y = c1y1 + c2y2 that satisfies the given initial conditions. Primes denote derivatives with respect to x. 1. y" – y = 0; yı = e*, y2 = e=*; y(0) = 0, y'(0) = 5 7. y" + y' = 0; yı = 1, y2 = e-*; y(0) = -2, y'(0) = 8 8. y" – 3y' = 0; yı1 = 1, y2 = e3x; y(0) = 4, y'(0) = -2 9. y" + 2y' + y = 0; y1 = e-x, y2 = xe-*; y(0) = 2, y'(0) = -1 10. y" – 10y' + 25y = 0; yı = e5x, y2 = xe5x; y(0) = 3, y'(0) = 13 11. y" – 2y' +2y = 0; y1 = e* cos x, y2 = e* sin x; y(0) = 0, y'(0) = 5 12. y" + 6y' + 13y = 0; yı = e3x cos 2x, y2 = e-3x sin 2x; y (0) = 2, y'(0) = 0 13. x?y" – 2xy' + 2y = 0; y1 = x, y2 = x²; y(1) = 3, y'(1) = 1 14. x?y" + 2xy' – 6y = 0; y1 = x², y2 = x-3; y(2) = 10, y'(2) = 15 15. ху" — ху" + у%3D 0;B у1 %3D х, у2 %3D х Inx; у(1) %3D 7, y'(1) = 2 16. x2y" + xy' + y = 0; yı = y(1) = 2, y'(1) = 3 %3D %3D Зх. %3D %3D %3D = cos(In x), y2 = sin(In x);

Calculus: Early Transcendentals
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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In Problems 1 through 16, a homogeneous second-order lin- ear differential equation, two functions y1 and y2, and a pair of initial conditions are given. First verify that y1 and y2 are solutions of the differential equation. Then find a particular solution of the form y = c1y1 + c2y2 that satisfies the given initial conditions. Primes denote derivatives with respect to x. 1. y" – y = 0; yı = e*, y2 = e=*; y(0) = 0, y'(0) = 5 7. y" + y' = 0; yı = 1, y2 = e-*; y(0) = -2, y'(0) = 8 8. y" – 3y' = 0; yı1 = 1, y2 = e3x; y(0) = 4, y'(0) = -2 9. y" + 2y' + y = 0; y1 = e-x, y2 = xe-*; y(0) = 2, y'(0) = -1 10. y" – 10y' + 25y = 0; yı = e5x, y2 = xe5x; y(0) = 3, y'(0) = 13 11. y" – 2y' +2y = 0; y1 = e* cos x, y2 = e* sin x; y(0) = 0, y'(0) = 5 12. y" + 6y' + 13y = 0; yı = e3x cos 2x, y2 = e-3x sin 2x; y (0) = 2, y'(0) = 0 13. x?y" – 2xy' + 2y = 0; y1 = x, y2 = x²; y(1) = 3, y'(1) = 1 14. x?y" + 2xy' – 6y = 0; y1 = x², y2 = x-3; y(2) = 10, y'(2) = 15 15. ху" — ху" + у%3D 0;B у1 %3D х, у2 %3D х Inx; у(1) %3D 7, y'(1) = 2 16. x2y" + xy' + y = 0; yı = y(1) = 2, y'(1) = 3 %3D %3D Зх. %3D %3D %3D = cos(In x), y2 = sin(In x);
Transcribed Image Text:In Problems 1 through 16, a homogeneous second-order lin- ear differential equation, two functions y1 and y2, and a pair of initial conditions are given. First verify that y1 and y2 are solutions of the differential equation. Then find a particular solution of the form y = c1y1 + c2y2 that satisfies the given initial conditions. Primes denote derivatives with respect to x. 1. y" – y = 0; yı = e*, y2 = e=*; y(0) = 0, y'(0) = 5 7. y" + y' = 0; yı = 1, y2 = e-*; y(0) = -2, y'(0) = 8 8. y" – 3y' = 0; yı1 = 1, y2 = e3x; y(0) = 4, y'(0) = -2 9. y" + 2y' + y = 0; y1 = e-x, y2 = xe-*; y(0) = 2, y'(0) = -1 10. y" – 10y' + 25y = 0; yı = e5x, y2 = xe5x; y(0) = 3, y'(0) = 13 11. y" – 2y' +2y = 0; y1 = e* cos x, y2 = e* sin x; y(0) = 0, y'(0) = 5 12. y" + 6y' + 13y = 0; yı = e3x cos 2x, y2 = e-3x sin 2x; y (0) = 2, y'(0) = 0 13. x?y" – 2xy' + 2y = 0; y1 = x, y2 = x²; y(1) = 3, y'(1) = 1 14. x?y" + 2xy' – 6y = 0; y1 = x², y2 = x-3; y(2) = 10, y'(2) = 15 15. ху" — ху" + у%3D 0;B у1 %3D х, у2 %3D х Inx; у(1) %3D 7, y'(1) = 2 16. x2y" + xy' + y = 0; yı = y(1) = 2, y'(1) = 3 %3D %3D Зх. %3D %3D %3D = cos(In x), y2 = sin(In x);
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