Section 1.5 In Exercises 1 and 2, we refer to a function ƒ, but we do not provide its formula. However, we do assume that f satisfies th hypothesis of the Uniqueness Theorem in the entire ty-plane,...

Only do number 2 pleaseSection 1.5<br>In Exercises 1 and 2, we refer to a function ƒ, but we do not provide its formula. However, we do assume that f satisfies th<br>hypothesis of the Uniqueness Theorem in the entire ty-plane, and we do provide various solutions to the given differentia<br>equation. Finally, we specify an initial condition. Using the Uniqueness Theorem, what can you conclude about th<br>solution to the equation with the given initial condition?<br>dy<br>1.<br>= f(t, y)<br>dt<br>dy<br>2.<br>= f(t, y)<br>dt<br>yı(t) = 4 for all t is a solution,<br>Y2(t) = 2 for all t is a solution,<br>Y3(t) = 0 for all t is a solution,<br>initial condition y(0) = 1.<br>y1 (t) = -1 for all t is a solution,<br>y2 (t) = 1+ t2 for all t is a solution,<br>initial condition y(0) = 0.<br>

Extracted text: Section 1.5 In Exercises 1 and 2, we refer to a function ƒ, but we do not provide its formula. However, we do assume that f satisfies th hypothesis of the Uniqueness Theorem in the entire ty-plane, and we do provide various solutions to the given differentia equation. Finally, we specify an initial condition. Using the Uniqueness Theorem, what can you conclude about th solution to the equation with the given initial condition? dy 1. = f(t, y) dt dy 2. = f(t, y) 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. y1 (t) = -1 for all t is a solution, y2 (t) = 1+ t2 for all t is a solution, initial condition y(0) = 0.

Jun 05, 2022
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