Unformatted text preview: µ ( 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 µ ( x ) = exp ±Z tan x dx ² = exp ( − ln  cos x  ) =  cos x  − 1 . 45...
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Derivative, Trigraph, DT DT DT

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