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Unformatted text preview: B ˙ ILKENT UNIVERSITY Department of Mathematics MATH 225, LINEAR ALGEBRA and DIFFERENTIAL EQUATIONS, Solution of Homework set 1 # 1 U. Mu˘gan June 13, 2008 Homework problems from the 2 nd Edition, SECTION 1.3 12 (22) ) 2 Since, f ( x,y ) = x ln y , and ∂f ∂y = x y are both continues in a neighborhood of (1 , 1), by the theorem of existence and uniqueness the solution exists and unique in some neighborhood of (1 , 1). 14 (24) ) Similar to previous problem, f ( x,y ) = y 1 / 3 is continuous in a neighborhood of (0 , 0), but ∂f/∂y = (1 / 3) y 2 / 3 is not, so the theorem guarantees existence but not uniqueness in some neighborhood of x = 0. 15 (25) ) f ( x,y ) = ( x y ) 1 / 2 is not continuous at (2 , 2) because it is not even defined if y > x . Hence the theorem guarantees neither existence nor uniqueness in any neighborhood of the point x = 2. 17 (27) ) Since f ( x,y ) = x 1 y and ∂f ∂y = x 1 y 2 are both continuous near the point (0 , 1), so the theorem guarantees both existence and uniqueness of a solution in some neighborhood of x = 0....
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This note was uploaded on 06/18/2010 for the course COMPUTER S Math 225 taught by Professor Yosumhoca during the Spring '10 term at Bilkent University.
 Spring '10
 YosumHoca

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