2360_2010Fall_hw1

2360_2010Fall_hw1 - Homework#1 Section 1.2 8 y y = e t sin...

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Unformatted text preview: Homework #1 Section 1.2 8. y- y =- e t sin( t ) , y (0) =- 1 y = e t cos( t ) + ce t y = e t cos( t )- e t sin( t ) + ce t y- y = e t cos( t )- e t sin( t ) + c e t- ( e t cos( t ) + c e t ) =- e t sin( t ) y (0) =- 1 y (0) = e cos(0) + c e = 1 + c =- 1 c =- 2 14. y = y ( y + 1)-2-1.5-1-0.5 0.5 1 1.5 2-2-1.5-1-0.5 0.5 1 1.5 2 y ( y + 1) = 0 y = 0, which is an unstable equilibrium solution; y =- 1, which is stable. 16. y = 1 (C). Constant slopes of 1. 17. y = y (D). Slopes are equal along each y-value. 18. y = y t (F) Slopes are zero along y = 0, slopes are vertical when t = 0. 19. y = t 2 (B) Slopes are zero when t = 0. 20. y = t 2 + y 2 (E) Larger slopes than y = t 2 . 1 21. y = 1 t (A) Negative slopes when t < 0, positive slopes when t > 0, same slopes along each t value. 22. y = y 2- 4 y 00 = 2 yy = 2 y ( y 2- 4) = 2 y ( y- 2)( y + 2) = 0 y = 0 , y =- 2 , y = 2 Minus 4 Minus 2 2 4 Minus 4 Minus 2 2 4 Solutions are concave up for y > 2 and y (- 2 , 0). Solutions are concave down for y <- 2 and y (0 , 2). 2 Section 1.3 1. y = 1 + y Separable dy 1 + y = dt with constant solution y =- 1. 2. y = y- y 3 Separable dy y- y 3 = dt with constant solutions y = 1 , 0. 3. y = sin( t + y ) Not separable, no constant solutions....
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This note was uploaded on 10/24/2010 for the course APPM 2360 at Colorado.

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2360_2010Fall_hw1 - Homework#1 Section 1.2 8 y y = e t sin...

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