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Unformatted text preview: Problem 26 Page 143
I suppose you ﬁnished part (a) and part (b). Then from part (b), you should get ym
ym = 4(6 + β )3
27(4 + β )2 (1) For part (c), we need to determine the smallest value of β for which ym ≥ 4, hence
√
4(6 + β )3
≥ 4 ⇐⇒ β ≥ 6 + 6 3
27(4 + β )2
√
Hence the smallest value for β is 6 + 6 3.
ym ≥ 4 ⇐⇒ (2) Here is the details: First dividing 4 on both sides and moving 27(4 + β )2 on the right hand side:
4(6 + β )3
≥ 4 ⇐⇒ (6 + β )3 ≥ 27(4 + β )2
2
27(4 + β ) (3) Then we may do a sub u = 4 + β , then we get
(6 + β )3 ≥ 27(4 + β )2 ⇐⇒ (u + 2)3 ≥ 27u2 (4) then expand (u + 2)3 = u3 + 6u2 + 12u + 8, then we get
(u + 2)3 ≥ u2 ⇐⇒ u3 + 6u2 + 12u + 8 ≥ 27u2 (5) Then move 27u2 on the left hand side
u3 + 3u2 + 12u + 8 ≥ 27u2 ⇐⇒ u3 − 21u2 + 12u + 8 ≥ 0 (6) ...
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This note was uploaded on 02/14/2012 for the course MATH 334 taught by Professor Dallon during the Fall '08 term at BYU.
 Fall '08
 DALLON
 Differential Equations, Equations

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