= 1 + y², - 1 ≤t≤ 1, y(0) = 0. unique solution, find the solution. dy dt 1 + y² doesn't satisfy Lipschitz condition on D = {(t, y), 0 ≤ t ≤ posed (from part (a)), but f doesn't satisfies Lipschitz condition (f mple does not contradict the theorem on Lecture Notes Page 6: (1 hitz condition in the variable y on the set D, then the initial-value p
= 1 + y², - 1 ≤t≤ 1, y(0) = 0. unique solution, find the solution. dy dt 1 + y² doesn't satisfy Lipschitz condition on D = {(t, y), 0 ≤ t ≤ posed (from part (a)), but f doesn't satisfies Lipschitz condition (f mple does not contradict the theorem on Lecture Notes Page 6: (1 hitz condition in the variable y on the set D, then the initial-value p
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![Consider the initial-value problem
dy
dt
(a) This problem has a unique solution, find the solution.
(b) Show that f(t, y) = 1 + y² doesn't satisfy Lipschitz condition on D = {(t, y)], 0 ≤ t ≤ 1, -∞ < y < ∞}
in the variable y.
(c) This problem is well posed (from part (a)), but ƒ doesn't satisfies Lipschitz condition (from part (b)).
Explain why this example does not contradict the theorem on Lecture Notes Page 6: (If ƒ is continuous
and satisfies a Lipschitz condition in the variable y on the set D, then the initial-value problem is well
posed.)
=1+y², −1≤ t ≤ 1, y(0) = 0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe2f2c4bd-bf4c-4a3b-a0a2-6333c3306a45%2F685026fa-7654-438d-99b0-ee717c9f90f8%2Ftml8uvd_processed.png&w=3840&q=75)
Transcribed Image Text:Consider the initial-value problem
dy
dt
(a) This problem has a unique solution, find the solution.
(b) Show that f(t, y) = 1 + y² doesn't satisfy Lipschitz condition on D = {(t, y)], 0 ≤ t ≤ 1, -∞ < y < ∞}
in the variable y.
(c) This problem is well posed (from part (a)), but ƒ doesn't satisfies Lipschitz condition (from part (b)).
Explain why this example does not contradict the theorem on Lecture Notes Page 6: (If ƒ is continuous
and satisfies a Lipschitz condition in the variable y on the set D, then the initial-value problem is well
posed.)
=1+y², −1≤ t ≤ 1, y(0) = 0.
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