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Unformatted text preview: homework 2 October 29, 2007 1. The general solution is x = c 1 e t + c 2 e t , and the initial conditions are satisfied if c 1 = 1 4 , c 2 = 1. Thus x = 1 4 e t + e t ≥ 2 r 1 4 e t e t. Then the minimum value of the solution is x min = x ( ln 2) = 1. 2. The characteristic equation is λ 2 λ 2 = 0 , Thus the values of λ are λ = 2 and λ = 1; the general solution is x = c 1 e 2 t + c 2 e t, and the initial conditions are satisfied if c 1 = α +2 3 and c 2 = 2 α 2 3 ; Then when c 1 = 0, that is α = 2, the solution will approaches zero as t → ∞ . 3. The general solution is x = c 1 e ( α 1) t + c 2 e αt , So, when α 1 < 0 and α < 0, that is, α < 0, x will tend to zero as t → ∞ ; when α 1 > 0 and α > 0, that is, α > 1, x will become unbounded as t → ∞ . 4. The characteristic equation is r 2 r + 1 4 = 0 , so the roots are r 1 = r 2 = 1 2 . Thus the general solution of the differential equation is x = c 1 e t/ 2 + c 2 te t/ 2 ....
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This note was uploaded on 01/28/2011 for the course MATH 150 taught by Professor T.qian during the Spring '09 term at HKUST.
 Spring '09
 T.Qian

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