solutions1ac

solutions1ac - 2 x dx = e 1 2 log x = x 1 2 : x 1 2 dy dx +...

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Math 33b, Quiz 1ac, January 15, 2008 Name: UCLA ID: 1. Using any technique possible, find the solution to the initial-value problem 2 dy dx + y x = 1 x 2 ,y (1) = 4 . Solution. We can rewrite this equation as a linear differential equation: dy dx + 1 2 x y = 1 2 x 2 . We multiply both sides by the integrating factor e R 1
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Unformatted text preview: 2 x dx = e 1 2 log x = x 1 2 : x 1 2 dy dx + 1 2 x-1 2 y = 1 2 x-3 2 d dx ( yx 1 2 ) = 1 2 x-3 2 yx 1 2 =-x-1 2 + C y =-1 x + Cx-1 2 Using the initial condition y (1) = 4, we get that 4 =-1 + C (1). Hence C = 5 and y ( x ) =-1 x + 5 x-1 2 . 1...
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This note was uploaded on 04/02/2008 for the course MATH 33B taught by Professor Staff during the Spring '07 term at UCLA.

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