Now we nd the particular solution corresponding to

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Unformatted text preview: the roots: 2 (4 r − 1) + 122 = 0 =⇒ Using Euler’s Formula, we find that Hence the general solution is 4 r − 1 = ±12 i =⇒ r= 1 4 ± 3 i. ert = et/4 cos 3t ± i et/4 sin 3t. y (t) = c1 et/4 cos 3t + c2 et/4 sin 3t for constants c1 and c2 . Now we find the particular solution corresponding to the initial conditions. We have the derivatives ￿ ￿ y (0) = c1 y (t) = (c1 cos 3t + c2 sin 3t) et/4 =⇒ ￿ t/4 t...
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This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue University-West Lafayette.

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