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(a) Verify that y = tan(x + c) is a one-parameter family of solutions of the differential equation y' = 1 + y2. Differentiating y = tan(x + c) we get y' = 1 + tan?(x + c) or y' = 1 + y². Differentiating y = tan(x + c) we get y' = 1 + sec?(x + c) or y' = 1 + y2. Differentiating y = tan(x + c) we get y' = csc(x + c) or y' = 1 + y2. Differentiating y = tan(x + c) we get y' = tan2(x + c) or y' = 1 + y². Differentiating y = tan(x + c) we get y' = sec(x + c) or y' = 1 + y2. (b) Since f(x, y) = 1 + y2 and af/ây = 2y are continuous everywhere, the region R in Theorem 1.2.1 can be taken to be the entire xy-plane. Use the family of solutions in part (a) to find an explicit solution of the first-order initial-value problem y' = 1 + y², y(0) = 0. y = Even though x. = 0 is in the interval (-2, 2), explain why the solution is not defined on this interval. Since tan(x) is discontinuous at x = + the solution is not defined on (-2, 2). (c) Determine the largest interval I of definition for the solution of the initial-value problem in part (b). (Enter your answer using interval notation.)

(a) Verify that y = tan(x + c) is a one-parameter family of solutions of the differential equation y' = 1 + y2. Differentiating y = tan(x + c) we get y' = 1 + tan?(x + c) or y' = 1 + y². Differentiating y = tan(x + c) we get y' = 1 + sec?(x + c) or y' = 1 + y2. Differentiating y = tan(x + c) we get y' = csc(x + c) or y' = 1 + y2. Differentiating y = tan(x + c) we get y' = tan2(x + c) or y' = 1 + y². Differentiating y = tan(x + c) we get y' = sec(x + c) or y' = 1 + y2. (b) Since f(x, y) = 1 + y2 and af/ây = 2y are continuous everywhere, the region R in Theorem 1.2.1 can be taken to be the entire xy-plane. Use the family of solutions in part (a) to find an explicit solution of the first-order initial-value problem y' = 1 + y², y(0) = 0. y = Even though x. = 0 is in the interval (-2, 2), explain why the solution is not defined on this interval. Since tan(x) is discontinuous at x = + the solution is not defined on (-2, 2). (c) Determine the largest interval I of definition for the solution of the initial-value problem in part (b). (Enter your answer using interval notation.)

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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(a) Verify that y = tan(x + c) is a one-parameter family of solutions of the differential equation y' = 1 + y2. Differentiating y = tan(x + c) we get y' = 1 + tan2(x + c) or y' = 1 + y². O Differentiating y = tan(x + c) we get y' = 1 + sec2(x + c) or y' = 1 + y2. O Differentiating y = tan(x + c) we get y' = csc(x + c) or y' = 1 + y2. Differentiating y = tan(x + c) we get y' = tan2(x + c) or y' = 1 + y2. Differentiating y = tan(x + c) we get y' = sec(x + c) or y' = 1 + y2. (b) Since f(x, у) = 1 + y2 and af/ðy = 2y are continuous everywhere, the region R in Theorem 1.2.1 can be taken to be the entire xy-plane. Use the family of solutions in part (a) to find an explicit solution of the first-order initial-value problem y' = 1 + y2, y(0) = 0. y = Even though X = 0 is in the interval (-2, 2), explain why the solution is not defined on this interval. Since tan(x) is discontinuous at x = + the solution is not defined on (-2, 2). (c) Determine the largest interval I of definition for the solution of the initial-value problem in part (b). (Enter your answer using interval notation.)
Transcribed Image Text:(a) Verify that y = tan(x + c) is a one-parameter family of solutions of the differential equation y' = 1 + y2. Differentiating y = tan(x + c) we get y' = 1 + tan2(x + c) or y' = 1 + y². O Differentiating y = tan(x + c) we get y' = 1 + sec2(x + c) or y' = 1 + y2. O Differentiating y = tan(x + c) we get y' = csc(x + c) or y' = 1 + y2. Differentiating y = tan(x + c) we get y' = tan2(x + c) or y' = 1 + y2. Differentiating y = tan(x + c) we get y' = sec(x + c) or y' = 1 + y2. (b) Since f(x, у) = 1 + y2 and af/ðy = 2y are continuous everywhere, the region R in Theorem 1.2.1 can be taken to be the entire xy-plane. Use the family of solutions in part (a) to find an explicit solution of the first-order initial-value problem y' = 1 + y2, y(0) = 0. y = Even though X = 0 is in the interval (-2, 2), explain why the solution is not defined on this interval. Since tan(x) is discontinuous at x = + the solution is not defined on (-2, 2). (c) Determine the largest interval I of definition for the solution of the initial-value problem in part (b). (Enter your answer using interval notation.)
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