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Unformatted text preview: Part I: Answer each of the following 15 multiple choice questions. Each correct answer is worth 2 marks. Question 1 2 3 4 5 6 7 8 9 10 Answer (d) (b) (a) (a) (d) (b) (c) (c) (d) (e) Question 11 12 13 14 15 Total Answer (e) (b) (c) (b) (d) 1. Which one of the following equations is solvable by the method of integrating factor? (a) dy dt + ty = y 2 (b) dy dt = 2 ty t 2 + 2 y (c) y = t 2 + 1 y (d) dy dt 2 y + 1 2 t = 2 t + 1 2 y (e) y 00 + 4 y 5 y = 0 Solution : Equations (a), (b), (c) are firstorder nonlinear equations whereas equation (e) is a secondorder equation. Only equation (d) is a firstorder linear equation which can be solved by the method of integrating factor. We rewrite the equation (d) as dy dt 4 t + 1 2 y = t 2 . The integrating factor is given by μ ( t ) = exp ( Z 4 t + 1 2 dt ) = e t 2 t/ 2 . (d) 2. Consider the initial value problem y 00 y + 1 4 y = 0 , y (0) = 1 , y (0) = b. Find the critical value of b that distinguishes y ( t ) grows positively and negatively, as t → ∞ . (a) 1 4 (b) 1 2 (c) 1 (d) 2 (e) 4 Solution : The characteristic equation is λ 2 λ + 1 4 = ( λ 1 2 ) 2 = 0 which gives the repeated root λ = 1 / 2. The differential equation therefore has the general solution y ( t ) = ( c 1 + c 2 t ) e 1 2 t . Initial conditions require that c 1 = 1 and c 2 = b 1 / 2. To find the critical value of b , we need the expression of the first derivative y ( t ). The derivative y ( t ) is given by y ( t ) = h‡ c 1 2 + c 2 · + c 2 2 t i e 1 2 t = • b + b 1 2 ¶ t 2 ‚ e 1 2 t . Since e 1 2 t is always positive, when t is (infinitely) large, the sign of y ( t ) is determined by the sign of ( b 1 / 2). That is, the critical value of b is 1 / 2. (b) 2 3. The velocity v ( t ) of a falling object satisfies the initial value problem dv dt + 0 . 2 v = 9 . 8 , v (0) = 0 . Find the time that must elapse for the object to reach 98% of its limiting velocity. (a) 5ln50 (b) 25ln10 (c) 50ln5 (d) 25 (e) 50 Solution : The differential equation is a firstorder linear equation which can be solved by the method of integrating factor. The general solution of the differential equation is v ( t ) = 49 + ce t/ 5 . The initial condition requires that c = 49. Hence, v ( t ) = 49 ‡ 1 e t/ 5 · . As t → ∞ , v → 49. The limiting velocity is denoted v ∞ = 49. To find the elapsed time we set 49 ( 1 e t/ 5 ) = (0 . 98) · v ∞ , 1 e t/ 5 = 0 . 98 , e t/ 5 = 0 . 02 , t 5 = ln 1 . 02 , t = 5ln50 . (a) 4. Determine the value of b which makes the following differential equation exact. ( x + ye 2 xy ) dx + bxe 2 xy dy = 0 . (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 Solution : The given equation can be written as M ( x,y ) dx + N ( x,y ) dy = 0 , where M ( x,y ) = x + ye 2 xy , N ( x,y ) = bxe 2 xy ....
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This note was uploaded on 09/30/2010 for the course MATH MATH152 taught by Professor Kcc during the Spring '10 term at HKUST.
 Spring '10
 KCC
 Math

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