homework set 1-Solutions - B ˙ ILKENT UNIVERSITY...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

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.

Page1 / 2

homework set 1-Solutions - B ˙ ILKENT UNIVERSITY...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online