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Unformatted text preview: ( t ) = t 3 , multiply by t 3 back, and come up with the original equation.) We now use the initial condition, x (2) = 0, to nd C . 0 = x (2) = 1 2(2) + C 2 3 1 4 + C 8 = 0 C = 2 . Hence, the solution is x = 1 / (2 t ) 2 / ( t 3 ). 21. Putting the equation in standard form yields dy dx + sin x cos x y = 2 x cos x dy dx + (tan x ) y = 2 x cos x. Therefore, P ( x ) = tan x and so ( x ) = exp Z tan x dx = exp ( ln  cos x  ) =  cos x  1 . 45...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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