# test2 - APPLIED MATH 33: Practice Exam 2 (Solution) 1. Find...

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APPLIED MATH 33: Practice Exam 2 (Solution) 1. Find the general solution to the diferential equation y 0 + y 0 6 y = e 2 x 2 . Solution: To ±nd a particular solution, we will split the equation into two. y 0 + y 0 6 y = e 2 x (1) y 0 + y 0 6 y = 2 . (2) The method we are going to use is the method o² undetermined coeﬃcients. Guess a particular solution to equation (1) o² ²orm y = Ae 2 x y 0 =2 Ae 2 x ,y 0 =4 Ae 2 x Substituting into (1) we have y 0 + y 0 6 y = e 2 x (4 A +2 A 6 A ) 0 . Try y = Axe 2 x y 0 =( A Ax ) e 2 x 0 =(4 A +4 Ax ) e 2 x Substituting into (1) we have e 2 x (4 A Ax + A Ax 6 Ax )=5 Ae 2 x = e 2 x A = 1 5 There²ore, a particular solution to equation (1) is y = 1 5 xe 2 x . For equation (2), guess a particular solution is a constant y B . Hence y 0 =0 0 ,and y 0 + y 0 6 y = 6 B = 2 B = 1 3 There²ore, a particular solution to equation (2) is y 1 3 . A particluar to the original non-homogenous equation is there²ore y p = 1 5 xe 2 x + 1 3 . The corresponding non-homogenous equation is y 0 + y 0 6 y characteristic equation r 2 + r 6=0 r 1 = 3 ,r 2 . A set o² ²undamental solutions is y 1 = e 3 x , y 2 = e 2 x . The general solution is there²ore y = c 1 y 1 + c 2 y 2 + y p = c 1 e 3 x + c 2 e 2 x + 1 5 xe 2 x

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## This note was uploaded on 03/03/2010 for the course MATH 198 taught by Professor Xia during the Spring '08 term at Vanderbilt.

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test2 - APPLIED MATH 33: Practice Exam 2 (Solution) 1. Find...

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