09SpringHwk02_Soln

09SpringHwk02_Soln - Math 3301 Homework Set 2 Solution 10...

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: Math 3301 Homework Set 2 Solution 10 Points 1. (3 pts) ( ) ( ) ( ) ( ) ( ) 2 2 2 5 2 2 5 2 2 2 2 5 2 8 3 2 5 4 3 2 4 1 3 1 7 3 9 4 4 7 2 3 121 3 2 40 4 3 7 2 8 8 y d y x d x y y x x c c x x x x y y x x y + =- + =- + + = =- + +-- +- +- +- = = = & & Reapplying the initial condition shows that + is the correct sign. The solution is then, ( ) 2 3 121 3 2 40 8 x x y x- + +- = For the interval of validity we need : 2 4 3 2 12 1 0 x x- + . The quadratic is zero at -1.3847 and 2.1847. So, the possible interval of validity is, 1.384 7 .1384 7 2.184 7 2.1847 x x x- < -- < The interval of validity is .1384 7 2.1847 x- because it contains the initial condition and the quadratic is positive in this range. 2. (3 pts) ( ) ( ) 1 1 2 2 1 2 3 4 2 7 4 2 4 2 4 2 7 7 8 8 7 8 2 4 2 8 2 8 2 8 y d y x x d x y x x c c y x x y x x x- = + = + + = = + + = + + & & In the right (implicit) solution the polynomial is always positive and so we wont have any issues with the...
View Full Document

This note was uploaded on 03/13/2009 for the course MATH 2171 taught by Professor Deng during the Spring '08 term at UNC Charlotte.

Page1 / 2

09SpringHwk02_Soln - Math 3301 Homework Set 2 Solution 10...

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