In Exercises 1-4, we refer to a function f, but we do not provide its formula. However, we do assume that f satisfies the hypotheses of the Uniqueness Theorem in the entire ty-plane, and we do provide various solutions to the given differential equation. Finally, we specify an initial condition. Using the Uniqueness Theorem, what can you conclude about the solution to the equation with the given initial condition? dy 1. = f(t, y) dt yı(t) = 3 for all t is a solution, initial condition y(0) = 1 3. dy dt yı(t) = 1 + 2 for all t is a solution, y2(t) = -1² for all t is a solution, initial condition y(0) = 1 = f(t, y) 2. 4. dy dt yı (t) = 4 for all t is a solution, y2 (t) = 2 for all t is a solution, y3 (t) = 0 for all t is a solution, initial condition y(0) = 1 = f(y) dy dt yı (t) = -1 for all t is a solution, y2 (t) = 1 + 1² for all t is a solution, initial condition y(0) = 0 = f(t, y)

In Exercises 1-4, we refer to a function f, but we do not provide its formula. However, we do assume that f satisfies the hypotheses of the Uniqueness Theorem in the entire ty-plane, and we do provide various solutions to the given differential equation. Finally, we specify an initial condition. Using the Uniqueness Theorem, what can you conclude about the solution to the equation with the given initial condition? dy 1. = f(t, y) dt yı(t) = 3 for all t is a solution, initial condition y(0) = 1 3. dy dt yı(t) = 1 + 2 for all t is a solution, y2(t) = -1² for all t is a solution, initial condition y(0) = 1 = f(t, y) 2. 4. dy dt yı (t) = 4 for all t is a solution, y2 (t) = 2 for all t is a solution, y3 (t) = 0 for all t is a solution, initial condition y(0) = 1 = f(y) dy dt yı (t) = -1 for all t is a solution, y2 (t) = 1 + 1² for all t is a solution, initial condition y(0) = 0 = f(t, y)

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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Question 3.)

What is the question asking for? my understanding is that uniquness theorem is true when f(x,y) and partial derivative are continuous then there should only be one solution, but why are we given various solutions?

In Exercises 1-4, we refer to a function f, but we do not provide its formula. However,
we do assume that f satisfies the hypotheses of the Uniqueness Theorem in the entire
ty-plane, and we do provide various solutions to the given differential equation. Finally,
we specify an initial condition. Using the Uniqueness Theorem, what can you conclude
about the solution to the equation with the given initial condition?
dy
1. = f(t, y)
dt
yı(t) = 3 for all t is a solution,
initial condition y(0) = 1
dy
3. = f(t, y)
dt
yı(t) = 1 + 2 for all t is a solution,
y2(t) = -1² for all t is a solution,
initial condition y(0) = 1
2.
4.
dy
dt
yı (t) = 4 for all t is a solution,
y2 (t) = 2 for all t is a solution,
y3 (t) = 0 for all t is a solution,
initial condition y(0) = 1
= f(y)
dy
dt
yı (t) = -1 for all t is a solution,
y2 (t) = 1 + 1² for all t is a solution,
initial condition y(0) = 0
= f(t, y)

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