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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. O Differentiating y = tan(x + c) we get y'=csc(x + c) or 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'= tan²(x + c) or y'= 1 + y². O Differentiating y = tan(x + c) we get y' = sec(x + c) or y'= 1 + y². (b) Since f(x, y) = 1 + y² and affay = 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=tan(x) 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 = 2 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.) (-∞, -2) U (-2,2) U (∞0,2) X

(a) Verify that y = tan(x + c) is a one-parameter family of solutions of the differential equation 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'= 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'= tan²(x + c) or y'= 1 + y². O Differentiating y = tan(x + c) we get y' = sec(x + c) or y'= 1 + y². (b) Since f(x, y) = 1 + y² and affay = 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=tan(x) 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 = 2 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.) (-∞, -2) U (-2,2) U (∞0,2) X

Advanced Engineering Mathematics
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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(a) Verify that y =tan(x + c) is a one-parameter family of solutions of the differential equation 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'= 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' = tan²(x + c) or y' = 1 + y². O Differentiating y = tan(x + c) we get y' = sec(x + c) or y' = 1 + y². (b) Since f(x, y) = 1 + y² and af/ay = 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 = tan(x) Even though x = 0 is in the interval (-2, 2), explain why the solution is not defined on this interval. the solution is not defined on (-2, 2). Since tan(x) is discontinuous at x = 2 X (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.) (-∞, -2) U (-2,2) U (0,2) X
Transcribed Image Text:(a) Verify that y =tan(x + c) is a one-parameter family of solutions of the differential equation 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'= 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' = tan²(x + c) or y' = 1 + y². O Differentiating y = tan(x + c) we get y' = sec(x + c) or y' = 1 + y². (b) Since f(x, y) = 1 + y² and af/ay = 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 = tan(x) Even though x = 0 is in the interval (-2, 2), explain why the solution is not defined on this interval. the solution is not defined on (-2, 2). Since tan(x) is discontinuous at x = 2 X (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.) (-∞, -2) U (-2,2) U (0,2) X
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