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ps2sol - 18.03 PS2 Solutions-Spring 09 Part B[1 8pts = 2 1...

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18.03 PS2 Solutions—Spring 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 0 = 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 ′′ > 0, so the approximation is smaller than the exact value. 1c. The table shows that E 0 . 24 h when h 1 / 4. Thus E h as expected. h y est (1) E E/h E/h 2 0 - 0 . 67 0 0 . 125 - 0 . 64 0 . 03 0 . 24 1 . 92 0 . 25 - 0 . 61 0 . 06 0 . 24 0 . 96 0 . 5 - 0 . 52 0 . 15 0 . 30 0 . 60 1 . 0 - 0 . 24 0 . 43 0 . 43 0 . 43 1d. See Figures 1 and 2. 2a. In a small amount of time Δ t , measured in hours, the fraction of air in the building that’s heated by the furnace is Δ t/ 2. Before accounting for the loss of heat to the outside, the new average temperature in the building is Δ t 2 · 90 + parenleftbigg 1 - Δ t 2 parenrightbigg · T = T + Δ t 2 (90 - T ) , so Δ T = Δ t 2 (90 - T ) and Δ T Δ t = 1 2 (90 - T ) . Thus, the furnace increases the temperature at a rate of (1 / 2)(90 - T ) F/hr. From PS1, the outside air increases the temperature at k 1 (30 - T ) F/hr. (Note, if T > 30 this rate is negative, as expected.) The overall rate is
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