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# quiz8ans - again We have y 1 = e − x and y 2 = xe − x...

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Math 222, Quiz 8 Name: Instructions: Answer the following questions fully, showing work where necessary. 1) Solve the following equation, finding the particular solution by “guess- ing”: y ′′ + 2 y + 1 = e x The complementary equation is y ′′ + 2 y + 1 = 0. The auxillary equa- tion to this is r 2 + 2 r + 1 = 0. This factors as ( r + 1) 2 = 0, so that the complementary equation is c 1 e x + c 2 xe x . Since 1 is not a root of the auxillary equation, our “guess” should be Ae x . Plugging this in, you should find A = 1 4 . Thus, the particular solution is 1 4 e x . 2) Now solve the equation by finding the particular solution by variation of parameters (obviously, you don’t need to find the complementary solution
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Unformatted text preview: again). We have y 1 = e − x and y 2 = xe − x . Thus, we have to solve the equations: v ′ 1 e − x + v ′ 2 xe − x = 0 v ′ 1 (-e − x ) + v ′ 2 ( e − x-xe − x ) = e x Solving this gives v ′ 1 =-xe 2 x and v ′ 2 = e 2 x . Then v 1 =-xe 2 x 2 + e 2 x 4 and v 2 = e 2 x 2 . Then y p = v 1 y 1 + v 2 y 2 = e x 4 , as above. Bonus (1 pt.): What book is currently number one on the New York Times NonFction best seller list? I am America (and so can you!) by Stephen Colbert 1...
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