Which of the following statements are true for this initial value problem? Select all that apply. dy (y - 4) = x - 6 with y (6) = 4 dx O y = x - 2 is the only solution of this initial-value problem. O y = x - 2 and y = 10 - x are both solutions of this initial value problem. This initial-value problem cannot have a solution because the conditions of the existence and uniqueness theorem for first-order linear equations are not satisfied. □ A locally unique solution is not guaranteed to exist by the local existence and uniqueness theorem for first-order x-6 differential equations because is not continuous at the point (6, 4). y-4 O The existence and uniqueness theorem for first-order linear equations ensures the existence of a unique local solution of this initial value problem because x - 6 is continuous at the point (6,4).
Which of the following statements are true for this initial value problem? Select all that apply. dy (y - 4) = x - 6 with y (6) = 4 dx O y = x - 2 is the only solution of this initial-value problem. O y = x - 2 and y = 10 - x are both solutions of this initial value problem. This initial-value problem cannot have a solution because the conditions of the existence and uniqueness theorem for first-order linear equations are not satisfied. □ A locally unique solution is not guaranteed to exist by the local existence and uniqueness theorem for first-order x-6 differential equations because is not continuous at the point (6, 4). y-4 O The existence and uniqueness theorem for first-order linear equations ensures the existence of a unique local solution of this initial value problem because x - 6 is continuous at the point (6,4).
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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Transcribed Image Text:Which of the following statements are true for this initial value problem? Select all that apply.
dy
(y-4) = x 6 with y (6) = 4
dx
y = x 2 is the only solution of this initial-value problem.
y = x - 2 and y = 10 x are both solutions of this initial value problem.
O This initial-value problem cannot have a solution because the conditions of the existence and uniqueness theorem for
first-order linear equations are not satisfied.
A locally unique solution is not guaranteed to exist by the local existence and uniqueness theorem for first-order
x-6
is not continuous at the point (6, 4).
y - 4
differential equations because
O The existence and uniqueness theorem for first-order linear equations ensures the existence of a unique local
solution of this initial value problem because x - 6 is continuous at the point (6, 4).
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