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# f08236ps3 - 92.236 Engineering Dierential Equations...

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92.236 Engineering Diferential Equations Practice Exam # 3 Solutions Fall 2008 Problem 1. (15 pts.) Solve the d.e. y (3) + 4 y pp + 4 y p = 0 The characteristic equation is r 3 + 4 r 2 + 4 r = 0 r ( r 2 + 4 r + 4 ) = 0 r ( r + 2) 2 = 0, which gives r = 0 and r = - 2 (double root). 7 pts. Therefore, y = c 1 e 0 x + c 2 e - 2 x + c 3 xe - 2 x , or y = c 1 + c 2 e - 2 x + c 3 xe - 2 x . 8 pts. Problem 2. (10 pts.) Solve the d.e. y pp - 4 y p + 5 y = 0 The characteristic equation is r 2 - 4 r + 5 = 0, which gives r = - ( - 4) ± r ( - 4) 2 - 4(1)(5) 2(1) = 4 ± - 4 2 = 4 ± 2 i 2 = 2 ± 1 i . 4 pts. Therefore, y = c 1 e 2 x cos(1 x ) + c 2 e 2 x sin(1 x ) or y = c 1 e 2 x cos( x ) + c 2 e 2 x sin( x ) . 6 pts. Problem 3. (15 points) Solve the d.e. y pp + 4 y = 8 Step 1. Find y c by solving the homogeneous d.e. y pp + 4 y = 0. Characteristic equation: r 2 + 4 = 0 r 2 = - 4 r = ± - 4 = ± 2 i . 2 pts. Therefore, y c = c 1 e 0 x cos(2 x ) + c 2 e 0 x sin(2 x ) = c 1 cos(2 x ) + c 2 sin(2 x ). 2 pts. Step 2. Find y p . Method 1. Undetermined Coe±cients. The nonhomogeneous term 8 in the given d.e. is a degree zero polynomial, so we should guess that y p is a degree zero polynomial: y p = A. 4 pts. This guess does not duplicate any term in y c , so there is no need to modify the guess. y = A y p = 0 y pp = 0 . Therefore, the left side of the d.e. is y pp + 4 y = 0 + 4 A = 4 A. We want this to equal the nonhomogeneous term 8: 4 A = 8 A = 2 Therefore, y p = 2. 6 pts.

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f08236ps3 - 92.236 Engineering Dierential Equations...

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