Section 1.3 Notes - dy dx = f x,y y x = y has a unique...

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Section 1.3: Initial-Value Problems, Boundary-Value Problems, and Existence of Solutions Deep The solutions for initial-value problems are not necesarily unique. Theorem 1 (Basic Existence and Uniqueness Theorem) Given the differential equation dy dx = f ( x, y ) If the fuction f is a continuous function of x and y in D of xy plane, and ∂f ∂y is also a continuous function of x and y in D and let ( x 0 , y 0 ) be a point, then the initial value problem
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Unformatted text preview: dy dx = f ( x,y ) y ( x ) = y has a unique solution which is valid in some interval about the initial point x . Example 1 dy dx = x 2 + y 2 y (1) = 3 To apply Theorem 1, we check the conditions. f ( x,y ) = x 2 + y 2 ∂f ( x,y ) ∂y = 2 y 1 Both f and ∂f ∂y are continuous. So, since the conditions are met, The differential equation dy dx = x 2 + y 2 has a unique solution. 2...
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