Chem Differential Eq HW Solutions Fall 2011 179

# Chem Differential Eq HW Solutions Fall 2011 179 - Section...

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Section A.2 Linear Ordinary Diﬀerential Equations with Constant Coeﬃcients A179 61. To solve y ±± - 4 y =0 ,y (0) = 0 ± (0) = 3, start with the general solution y ( x )= c 1 cosh 2 x + c 2 sinh 2 x. Then y (0) = 0 c 1 cosh 0 + c 2 sinh 0 = 0 c 1 = 0; so y ( x c 2 sinh 2 x. y ± (0) = 3 2 c 2 cosh 0 = 3 c 2 = 3 2 y ( x 3 2 sinh 2 x. 65. To solve y ±± - 5 y ± +6 y = e x (0) = 0 ± (0) = 0, use the general solution from Exercise 27 (modify it slightly): y = c 1 e 2 x + c 2 e 3 x + 1 2 e x . Then y (0) = 0 c 1 + c 2 = - 1 2 ; y ± (0) = 0 2 c 1 +3 c 2 = - 1 2 c 2 = 1 2 ; c 1 = - 1 y ( x - e 2 x + 1 2 e 3 x + 1 2 e x . 69. Because of the initial conditions, it is more convenient to take y = c 1 cos[2( x - π 2 )] + c 2 sin[2( x - π 2 )] as a general solution of
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## This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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