Section 1.5In Exercises 1 and 2, we refer to a function f, but we do not provide its formula. However, we do assume that f satisfies thehypothesis of the Uniqueness Theorem in the entire ty-plane, and we do provide various solutions to the given differentialequation. Finally, we specify an initial condition. Using the Uniqueness Theorem, what can you conclude about thesolution to the equation with the given initial condition?dy1.f(t, y)dyf(t, y)2.dtdty1(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.y1 (t) = -1 for all t is a solution,y2 (t) = 1+ t2 for all t is a solution,initial condition y(0) = 0.In Exercises 3 and 4, an initial condition for the differential equationdy= y(y – 1)(y – 3)dtis given. What does the Existence and Uniqueness Theorem say about the corresponding solution?3. y(0) = 44. y(0) = –1

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Section 1.5 In Exercises 1 and 2, we refer to a function f, but we do not provide its formula. However, we do assume that f satisfies the hypothesis 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 = f(t, y) ip = f(t, y) 1. 2. 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. dt y1 (t) = -1 for all t is a solution, 2(t) = 1+t2 for all t is a solution, initial condition y(0) = 0. In Exercises 3 and 4, an initial condition for the differential equation dy = y(y – 1)(y – 3) dt is given. What does the Existence and Uniqueness Theorem say about the corresponding solution? 3. y(0) = 4 4. y(0) = -1

Section 1.5 In Exercises 1 and 2, we refer to a function f, but we do not provide its formula. However, we do assume that f satisfies the hypothesis 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 = f(t, y) ip = f(t, y) 1. 2. 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. dt y1 (t) = -1 for all t is a solution, 2(t) = 1+t2 for all t is a solution, initial condition y(0) = 0. In Exercises 3 and 4, an initial condition for the differential equation dy = y(y – 1)(y – 3) dt is given. What does the Existence and Uniqueness Theorem say about the corresponding solution? 3. y(0) = 4 4. y(0) = -1

Calculus For The Life Sciences

2nd Edition

ISBN:9780321964038

Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Chapter11: Differential Equations

Section11.CR: Chapter 11 Review

Problem 1CR

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