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Unformatted text preview: AMATH 351 Homework 3 Keys Section3.1 16,21 Section3.2 3,18,22,24 Section3.3 9,21 Section 3.1 16. 4 y 00 y = 0 , y ( 2) = 1 , y ( 2) = 1 Characteristic equation 4 λ 2 1 = 0 ⇒ λ = ± 1 2 So the general solution is y ( t ) = C 1 e t 2 + C 2 e t 2 Apply given conditions ( y ( 2) = C 1 e 1 + C 2 e = 1 y ( 2) = C 1 2 e 1 C 2 2 e = 1 = ⇒ ( C 1 = e 2 C 2 = 3 2 e So the speci c solution is y ( t ) = 1 2 e t 2 +1 + 3 2 e t 2 1 → ∞ as t → + ∞ 21. y 00 y 2 y = 0 , y (0) = α, y (0) = 2 Characteristic equation λ 2 λ 2 = 0 ⇒ ( λ 2)( λ + 1) = 0 ⇒ λ 1 = 2 , λ 2 = 1 So the general solution is y ( t ) = C 1 e 2 t + C 2 e t 1 Apply initial conditions: ( y (0) = C 1 + C 2 = α y (0) = 2 C 1 C 2 = 2 ⇒ ( C 1 = α +2 3 C 2 = 2 α 2 3 So the speci c solution is y ( t ) = α + 2 3 e 2 t + 2 α 2 3 e t If we want y ( t ) → as t → + ∞ , the coe cient for e 2 t must be , i.e....
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 Spring '08
 WAlker
 Math, Calculus, Linear Algebra, Elementary algebra, Complex number, 1 g, Abel

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