G251ex1(fa05)

G251ex1(fa05) - Math 251 Sections 1/2 October 12, 2005...

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Unformatted text preview: Math 251 Sections 1/2 October 12, 2005 First Exam There are 9 questions on this exam. Question 1 is worth 20 points. Questions 2 through 9 are worth 10 points each. The total number of points is 100. If a question has multiple parts, then the points assigned to the question are divided equally among the parts, unless otherwise indicated. Show all your work . Partial credit may be given. The use of calculators, books, or notes is not permitted on this exam. Please turn off your cell phone before starting this exam. Time limit 1 hour and 15 minutes. SOLUTIONS 1. a. Suppose y ( t ) is a solves the ODE: y = e y . What ODE does y ( t + 1) solve? ANS. This is an autonomous equation. So a solution shifted satisfies the SAME ODE. For the initial value problems in parts b. and c. state whether or not a unique solution is guaranteed to exist. If the answer is yes and if it is possible to determine the largest interval on which the solution exists without actually solving the equation , then do so. b. ( t- 1) y- ( y + 1) 2 / 3 = 0 , y (- 2) =- 1 ANS. NO GUARANTEE because the partial derivative with respect to y fails to be continuous at (- 2 ,- 1). c. ( t + t 2 ) y + y = 0 , y (- 2) = 0 ANS. This is a linear equation with coeffs being discontinuous at 0 and- 1. So a unique solution exists on I = (- ,- 1). In parts d. and e. assume that L [ y ] = y 00 + py + qy and that p ( t ) and q ( t ) are continuous functions of t on the entire real axis (- , ). d. One of the following functions CANNOT be a solution of L [ y ] = 0 on (- , ). Circle it: 1 t t- sin 2 t 2 t- sin 2 t 2- sin 2 t ANS. Zero is a solution. The function 2 t- sin 2 t has the same initial conditions at 0. It CANNOT be a solution. e. If L [ y 1 ] = 0, and L [ y 2 ] = t , then one of the following solves: L [ y ] = 2 t . Circle it. y 1 + y 2 2 y 1 3 y 1- y 2 4 y 1 + 2 y 2 ANS. By the superposition principle L [4 y 1 + 2 y 2 ] = 2 t . In parts f. and g. consider each differential equation on the left and match it with the statement on the right that best describes the long time behavior of a NONZERO solution iii. f. y 00- 4 y + 5 y = 0 iv. g. y 00 + 4 y + 4 y = 0 i. approaches either...
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G251ex1(fa05) - Math 251 Sections 1/2 October 12, 2005...

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