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dy = 6x² + 2x − 3 and y(0) = 4. - dx What is y(2)? Suppose Oy(2) = 14 Oy(2) = 16 Oy(2)= 18 Oy(2) = 22

dy = 6x² + 2x − 3 and y(0) = 4. - dx What is y(2)? Suppose Oy(2) = 14 Oy(2) = 16 Oy(2)= 18 Oy(2) = 22

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
### Differential Equation Problem Suppose \(\frac{dy}{dx} = 6x^2 + 2x - 3\) and \(y(0) = 4\). #### What is \(y(2)\)? - ⭕ \(y(2) = 14\) - ⭕ \(y(2) = 16\) - ⭕ \(y(2) = 18\) - ⭕ \(y(2) = 22\) To determine the correct answer, integrate the given differential equation \(\frac{dy}{dx} = 6x^2 + 2x - 3\) and apply the initial condition \(y(0) = 4\).
Transcribed Image Text:### Differential Equation Problem Suppose \(\frac{dy}{dx} = 6x^2 + 2x - 3\) and \(y(0) = 4\). #### What is \(y(2)\)? - ⭕ \(y(2) = 14\) - ⭕ \(y(2) = 16\) - ⭕ \(y(2) = 18\) - ⭕ \(y(2) = 22\) To determine the correct answer, integrate the given differential equation \(\frac{dy}{dx} = 6x^2 + 2x - 3\) and apply the initial condition \(y(0) = 4\).
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