This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
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...
View Full Document
- Spring '11
- Continuous function, Lipschitz continuity