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Use the Fundamental Existence and Uniqueness theorem to determine if the IVP will have a inique solution on the given intervals. If the theorem fails, make sure to state which part of he theorem failed. See page 2 of this document for the theorem statement. ODE: (1 + y³) ay = x² dx nitial value: y(−1) = −1 nterval: (-2,2)

Use the Fundamental Existence and Uniqueness theorem to determine if the IVP will have a inique solution on the given intervals. If the theorem fails, make sure to state which part of he theorem failed. See page 2 of this document for the theorem statement. ODE: (1 + y³) ay = x² dx nitial value: y(−1) = −1 nterval: (-2,2)

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter6: The Trigonometric Functions
Section6.3: Trigonometric Functions Of Real Numbers
Problem 77E
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Use the Fundamental Existence and Uniqueness theorem to determine if the IVP will have a unique solution on the given intervals. If the theorem fails, make sure to state which part of the theorem failed. See page 2 of this document for the theorem statement. ODE: (1+y³) y = x² dy Initial value: y(-1) = −1 Interval: (-2,2)
Transcribed Image Text:Use the Fundamental Existence and Uniqueness theorem to determine if the IVP will have a unique solution on the given intervals. If the theorem fails, make sure to state which part of the theorem failed. See page 2 of this document for the theorem statement. ODE: (1+y³) y = x² dy Initial value: y(-1) = −1 Interval: (-2,2)
The Fundamental Existence-Uniqueness Theorem for a first-order IVP If F(x, y) and y' = F(x, y), y(x) = yo OF (x,y) are real and continuous functions in a rectangle ду R = {(x, y): a < x <b,c<y<d} which contains the point (xo, Yo), then the IVP has a unique solution y = f(x) in some subinterval (xo - h, xo + h) of (a, b).
Transcribed Image Text:The Fundamental Existence-Uniqueness Theorem for a first-order IVP If F(x, y) and y' = F(x, y), y(x) = yo OF (x,y) are real and continuous functions in a rectangle ду R = {(x, y): a < x <b,c<y<d} which contains the point (xo, Yo), then the IVP has a unique solution y = f(x) in some subinterval (xo - h, xo + h) of (a, b).
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