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Unformatted text preview: Math 251 February 24, 2005 First Exam NAME: Section #: There are 9 questions on this exam. Question 9 is worth 12 points. Each other question is worth 11 points. The points assigned to each part of the question are indicated at the start of the part. 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. 1. a. 9pts Find the general solution to the following ODE ty + 2 y = 2 t > This is a linear equation. 2pts It must be put into the following form: y + 2 t y = 2 t 3pts p = 2 t R 2 t dt = 2 ln t = ln t 2 The integrating factor is = e R p dt = t 2 3pts Multiplying through by the integrating factor gives ( yt 2 ) = 2 t 1pts Integrating gives yt 2 = t 2 + C Finally y = 1 + Ct 2 (Do not remove points of leaving in implicit form) b. 2pts Find the solution of the above equation which satisfies y (1) = 0. 2pts Plug in t = 1 and y = 0 to find C = 1 2. a. 3pts Verify that the following ODE is exact: 2 t + e y + ( te y cos y ) dy dt = 0 (Show your work.) 1pts Identify: F t = 2 t + e y and F y = te y cos y 2pts Check F ty = e y and F yt = e y b. 6pts Find the general solution to the ODE in Part a. (You may leave your answer in implicit form.) 2pts We see that F = t 2 + e y + h ( y ). 2pts Differentiating this with respect to y : F y = 0 + e y + dh dy and comparing gives dh dy = cos y 2pts We see that F = t 2 + e y sin y . 1pts The form of the general solution is F ( t, y ) = C ....
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This note was uploaded on 07/23/2008 for the course MATH 251 taught by Professor Chezhongyuan during the Spring '08 term at Pennsylvania State University, University Park.
 Spring '08
 CHEZHONGYUAN
 Math, Differential Equations, Equations, Partial Differential Equations

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