09SpringHwk02_Soln

# 09SpringHwk02_Soln - Math 3301 1(3 pts Homework Set 2...

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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 32 40 4 3 7 2 0 8 8 y dy x dx 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 32 40 8 x x y x - + + - = For the interval of validity we need : 2 40 32 121 0 x x - + . The quadratic is zero at -1.3847 and 2.1847. So, the possible interval of validity is, 1.3847 .13847 2.1847 2.1847 x x x -¥ < £ - - £ £ £ < ¥ The interval of validity is .13847 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 dy x x dx 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 won’t have any issues with the root being negative out of it. Also, the explicit solution is a polynomial and so will exist for for all x ’s.

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• Spring '08
• Deng

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