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Tutorial5 - Thus by (3) the general solution of the ode is...

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Unformatted text preview: Thus by (3) the general solution of the ode is x ( t ) = e- 2 t ( A cos t + B sin t ) . The derivative is ˙ x ( t ) =- 2 e- 2 t ( A cos t + B sin t ) + e- 2 t (- A sin t + B cos t ) . Applying the initial conditions x (0) = 1 = A, ˙ x (0) = 0 =- 2 A + B, which give us A = 1 , B = 2 . Therefore, x ( t ) = e- 2 t (cos t + 2sin t ) . Clearly when t increases, e- 2 t decays to zero and cos t + 2sin t oscillates. Thus x ( t ) has decaying oscillation for large t. 2 Remark . In general note that μ 6 = 0, it is easy to see that for the solu- tion (3) when t increases there are three cases: (i) If λ > 0 then x ( t ) has growing oscillation; (ii) If λ = 0 then x ( t ) has steady oscillation; (iii) If λ < 0 then x ( t ) has decaying oscillation. 2 Case 3. When Δ = 0 , (2) has only one real root given by r =- b 2 a . Then the solution of (1) can be written as x ( t ) = ( c 1 + c 2 t )exp- b 2 a t ¶ , (4) with c 1 , c 2 constants. Example 2. Consider the initial value problem ¨ x + ˙ x + 0 . 25 x = 0 , x (0) = 2 , ˙ x (0) = b. Find the solution as a function of b and then determine the critical value of b that separates solutions that grow positively from those that eventually grow negatively . 2 The characteristic equation is r 2 + r + 0 . 25 = ( r + 1 / 2) 2 = 0 , with a repeated root r =- 1 / 2 . Therefore the general solution is x ( t ) = ( c 1 + c 2 t ) e- t/ 2 with derivative ˙ x ( t ) = ‡- c 1 2 + c 2- c 2 2 t · e- 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.

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Tutorial5 - Thus by (3) the general solution of the ode is...

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