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Unformatted text preview: Math 33b final exam June 13, 2007 Name: UCLA ID: Section (circle one): 1A Kree ColeMcLaughlin Tuesday 1B Kree ColeMcLaughlin Thursday 1C Will Conley Tuesday 1D Will Conley Thursday Directions: Fill in your name, UCLA ID and section above. Do not turn the page until instructed to do so. You have 3 hours to complete the exam. You may use only your brain and a pen or pencil. There are 7 problems, of varying weights. Extra scratch paper is included; if you need more, raise your hand. If your work on a problem appears on a different page, indicate clearly where it may be found. Show all the necessary steps involved in finding your solutions, unless otherwise instructed. In the interest of us not losing pages of your exam, please refrain from detaching pages from the exam (except for the table of integrals at the back). Good luck. Problem Maximum Score Problem Maximum Score 1 10 5 16 2 12 6 14 3 12 7 14 4 12 Total 90 1 1. (10 pts total) Consider the firstorder differential equation (2 y cos x xy sin x ) dx + (3 x cos x ) dy = 0 . a. (5 pts) This equation is not exact. Find integers a and b such that multiplying the equation by the integrating factor x a y b makes it exact. b. (5 pts) Find the unique solution to the differential equation satisfying the initial condition y ( π ) = 1. Solution. Set P ( x, y ) = (2 y cos x xy sin x )( x a y b ) = (2 x a y b +1 cos x x a +1 y b +1 sin x ), and set Q ( x, y ) = (3 x cos x )( x a y b ) = 3 x a +1 y b cos x . We want to find a and b such that ∂P ∂y = ∂Q ∂x . Doing the algebra, we get ∂P ∂y = 2( b + 1) x a y b cos x ( b + 1) x a +1 y b sin x ∂Q ∂x = 3( a + 1) x a y b cos x 3 x a +1 y b sin x For these expressions to be equal, we must have 3 = ( b + 1) and 2( b + 1) = 3( a + 1), so b = 2 and a = 1. Hence the desired integrating factor is xy 2 . Multiplying the entire differential equation by xy 2 , we get (2 xy 3 cos x x 2 y 3 sin x ) dx + (3 x 2 y 2 cos x ) dy = 0 . We seek a function f such that ∂f ∂y = (3 x 2 y 2 cos x ) and ∂f ∂x = (2 xy 3 cos x x 2 y 3 sin x ). Integrating the first equation by y , we get f ( x, y ) = x 2 y 3 cos x + φ ( x ) . Differentiating by x , we now have ∂f ∂x = 2 xy 3 cos x x 2 y 3 sin x + φ ( x ) . We conclude that φ ( x ) = 0, so we can choose φ ( x ) = 0 and f ( x, y ) = x 2 y 3 cos x . The solution to the differential equation is thus x 2 y 3 cos x = C for some constant C . Using the initial condition y ( π ) = 1, we have π 2 1 3 cos π = C , so C = π 2 and the solution is x 2 y 3 cos x = π 2 . 2 2. (12 pts) Consider the differential equation x = sin x. a. (5 pts) Draw a phaseline diagram for this differential equation. Identify all of the equilibrium points, and determine their stability....
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This note was uploaded on 05/01/2008 for the course MATH 33B taught by Professor Staff during the Spring '07 term at UCLA.
 Spring '07
 staff
 Math

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