Ordinary Diff Eq Exam Review Solutions 108

# Ordinary Diff Eq Exam Review Solutions 108 - x j 2 ± ± =...

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110 1 Solutions Graphs for problems 1 through 8 0 1 0 1 2 3 4 5 6 t (a) 0 1 2 0 1 2 3 4 5 6 t (b) 0 1 ± 1 1 2 3 4 5 6 t (c) 0 2 4 0 1 2 3 4 5 6 t (d) 0 1 0 1 2 3 4 5 6 t (e) 0 1 0 1 2 3 4 5 6 t (f) 0 1 0 1 2 3 4 5 6 t (g) 0 1 ± 1 1 2 3 4 5 6 t (h) 9. R 5 0 f ( t ) dt = R 2 0 ( t 2 ± 4) dt + R 3 2 0 dt + R 5 3 ( ± t + 3) dt = ± t 3 = 3 ± 4 t ²³ ³ 2 0 + 0 + ± ± t 2 = 2 + 3 t ²³ ³ 5 3 = (8 = 3 ± 8) + ( ± 25 = 2 + 15) ± ( ± 9 = 2 + 9) = ± 22 = 3 . 10. R 2 0 f ( u ) du = R 1 0 (2 ± u ) du + R 2 1 u 3 du = (2 u ± u 2 = 2) ³ ³ 1 0 + u 4 = 4 ³ ³ 2 1 = 3 = 2 + 4 ± 1 = 4 = 21 = 4 . 11. R 2 ± 0 j sin x j dx = R ± 0 sin x dx + R 2 ± ± ± sin x dx = ± cos x j ± 0 + cos
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Unformatted text preview: x j 2 ± ± = 4 : 12. R 3 f ( w ) dw = R 1 w dw + R 2 1 1 w dw + R 3 2 1 2 dw = 1 = 2 + ln 2 + 1 = 2 = 1 + ln 2...
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## This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.

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