In applications, most initial value problems will have a unique solution. In fact, the existence of unique solutions is so important that there is a theorem about the existence and uniqueness of a solution. Consider the initial value problemafare continuous functions in some rectangle R = {(x,y):a0, satisfies the initial value problem =y with y(0) = 0.dydx(c) Now show that the constant function y = 0 also satisfies the initial value problem given in part (b). Hence, this initial value problem does not have a unique solution.=dydx13dydx(d) Finally, show that the conditions for the theorem are not satisfied for the initial value problem in part (b). Begin by identifying f(x,y) in the initial value problem in (b).f(x,y)=afду=Answer parts (a) through (d).for x ≥ 0., and yC113dyThus, it holds that the constant function y = 0 satisfies the initial value problem -= y³ with y(0) = 0.dx=f(x,y). y (xo) yo. Iff anddx

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In applications, most initial value problems will have a unique solution. In fact, the existence of unique solutions is so important that there is a theorem about the existence and uniqueness of a solution. Consider the initial value problem =f(x,y). y(x) =yo. Iff and dx af dy are continuous functions in some rectangle R={(x,y):a

In applications, most initial value problems will have a unique solution. In fact, the existence of unique solutions is so important that there is a theorem about the existence and uniqueness of a solution. Consider the initial value problem =f(x,y). y(x) =yo. Iff and dx af dy are continuous functions in some rectangle R={(x,y):a

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter7: Analytic Trigonometry

Section7.6: The Inverse Trigonometric Functions

Problem 91E

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