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Unformatted text preview: 18.03 PS2 SolutionsSpring 09 Part B: [1. 8pts = 2 + 1 + 3 + 2]; [2. 14pts = 2 + 2 + 2 + 4 + 2 + 2]; [3. 4pts = 2 + 2]; [4. 6pts = 5 + 1] 1a. y = 1, y 1 = 1 + 1 /n , y 2 = (1 + 1 /n ) 2 . For k 0, y k +1 = y k + (1 /n ) y k = (1 + 1 /n ) y k . So y k = (1 + 1 /n ) k and the n th Euler approximation is y n = (1 + 1 /n ) n 1b. y &gt; 0, so the approximation is smaller than the exact value. 1c. The table shows that E . 24 h when h 1 / 4. Thus E h as expected. h y est (1) E E/h E/h 2 . 67 . 125 . 64 . 03 . 24 1 . 92 . 25 . 61 . 06 . 24 . 96 . 5 . 52 . 15 . 30 . 60 1 . . 24 . 43 . 43 . 43 1d. See Figures 1 and 2. 2a. In a small amount of time t , measured in hours, the fraction of air in the building thats heated by the furnace is t/ 2. Before accounting for the loss of heat to the outside, the new average temperature in the building is t 2 90 + parenleftbigg 1 t 2 parenrightbigg T = T + t 2 (90 T ) , so T = t 2 (90 T ) and T t = 1 2 (90 T ) ....
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This note was uploaded on 05/06/2010 for the course 18 18.03 taught by Professor Unknown during the Spring '09 term at MIT.
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