Exercises 1-15: (a) Find the general solution of the differential equation. (b) Impose the initial conditions to obtain the unique solution of the initial value prob- lem. (c) Describe the behavior of the solution y(t) as t→-∞o and as t→ ∞o. Does y(t) ap- proach -00, +00, or a finite limit? 1. y"+y'- 2y = 0, y(0) = 3, y'(0) = -3 2. y" - y = 0, y(2) = 1, y'(2) = 0 "1
Exercises 1-15: (a) Find the general solution of the differential equation. (b) Impose the initial conditions to obtain the unique solution of the initial value prob- lem. (c) Describe the behavior of the solution y(t) as t→-∞o and as t→ ∞o. Does y(t) ap- proach -00, +00, or a finite limit? 1. y"+y'- 2y = 0, y(0) = 3, y'(0) = -3 2. y" - y = 0, y(2) = 1, y'(2) = 0 "1
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:SES
y(t) = (30=%) e-² + (2x + %) ².
e-2t
e
Since lim
problem will tend to zero as t increases if the coefficient of e' in the solution is
= 0 and lime = +00, the solution of the initial value
zero. Therefore, lim, y(t) = 0 if yo = -2yo.
Exercises 1-15:
(a) Find the general solution of the differential equation.
(b) Impose the initial conditions to obtain the unique solution of the initial value prob-
lem.
(c) Describe the behavior of the solution y(t) as t→-oo and as too. Does y(t) ap-
proach -∞, +∞o, or a finite limit?
1. y"+y' - 2y = 0, y(0) = 3, y'(0) = -3
= 1, y'(2) = 0
2. y" - y = 0, y(2)
3. y" - 4y' + 3y = 0,
4. 2y" - 5y' + 2y = 0,
y(0) = -1, y'(0) = 1
5. y" - y = 0, y(0) = 1,
6. y" + 2y = 0, y(-1) = 0,
7. y" + 5y' + 6y = 0,
8. y" - 5y' + 6y = 0,
9. y" - 4y = 0, y(3)
10. 8y" - 6y' + y = 0, y(1) = 4,
11. 2y" - 3y = 0,
y(-2) = 3,
y(0) = -1,
y'(0) = -1
y'(0) = -5
y'(-1) = 2
y(0) = 1,
y(0) = 1,
= 0, y'(3) = 0
y(1) = 2,
y(0) = 0,
12. y" - 6y' + 8y = 0,
13. y" + 4y + 2y = 0,
14. y" - 4y'- y = 0,
y(0) = 1,
15. 2y" - y = 0, y(0) = -2,
y'(0) = -1
y'(0) = -1
y'(1) =
y'(-2) = 0
y'(1) = -8
y'(0) = 4
y'(0) = 2 + √5
y'(0) = √2
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