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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...
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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.
 Spring '08
 Deng
 Math, Differential Equations, Equations

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