285hw2 - HW 2 1. Sec. 1.3: 13. dy p = 3 y ; y (0) = 1: dx...

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HW 2 1. Sec. 1.3: 13. dy dx = 3 p y ; y (0) = 1 : The rate function is given by f ( x; y ) = 3 p y: It is known from calculus that f ( x; y ) = 3 p y is a continuous function for all x and y , in particular in a neighborhood of the point x 0 = 0 and y 0 = 1 . So, by the existence theorem, solution exists close to the point (0 ; 1) . @ @y 3 p y = 1 3 y 2 3 is continuous in a neighborhood of the point (0 ; 1) . Indeed, 1 is a constant, thereby a continuous function for all x and y ; 3 y 2 3 is a continuous function for all x and y . Notice that 3 y 2 3 does not vanish in a neighborhood of the point of (1 ; 0) . Hence, the ratio of continuous function 1 and 3 y 2 3 is continuous because the denominator is not zero around (0 ; 1) . So, by the uniqueness theorem, we see that there is a unique solution of the foregoing initial value problem. Answer: solution exists and is unique. (students do not have to include the sketch). 1
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2. Sec. 1.3: 14. dy dx = 3 p y ; y (0) = 0 : The rate function is given by f ( x; y ) = 3 p y: It is known from calculus that f ( x; y ) = 3 p y is a continuous function for all x and y , in particular in a neighborhood of the point x 0 = 0 and y 0 = 0 . So, by the existence theorem, solution exists close to the point (0 ; 0) . If y 6 = 0 , then @ @y 3 p y = 1 3 y 2 3 : It is readily seen that lim y ! 0 @ 3 p y = lim y ! 0 1 3 y 2 3 = 1 ; hence, the partial derivative f y cannot be continuous at (0 ; 0) . The condi- tion for uniqueness fails. So, the uniqueness theorem does not guarantee the uniqueness of solution.
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This note was uploaded on 01/26/2012 for the course MATH 285 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.

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285hw2 - HW 2 1. Sec. 1.3: 13. dy p = 3 y ; y (0) = 1: dx...

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