# 13 where a is a symmetric matrix and f is a constant

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Unformatted text preview: b2 + · · · + cn bn . To ﬁnish solving the initial value problem (2.12), we need to determine the constants c1 , c2 , . . . , cn so that c1 b1 + c2 b2 + · · · + cn bn = f . It is here that our formula (2.11) comes in handy; using it we obtain x = (f · b1 )b1 eλ1 t + · · · + (f · bn )bn eλn t . Example. Suppose 5 dx =4 dt 0 we want to solve the initial value problem 40 3 1 . 5 0 x, x(0) = f , where f= 01 4 We saw in Section 2.1 that A has the eigenvalues λ1 = 1 with multiplicity two and λ2 = 9 with multiplicity one. Moreover, the orthogonal matrix 1 1 0 √2 √2 − 1 B = 0 √1 √2 2 10 0 has the property that 1 B −1 AB = 0 0 0 1 0 0 0 . 9 Thus the general solution to the matrix diﬀerential equation dx/dt = Ax is c1 et x = B c2 et = c1 b1 et + c2 b2 et + c3 b3 e9t , c3 e9t 46 where 0 b1 = 0 , 1 √ 1 b2 = 2 − √1 , 2 0 √ 1 2 1 b3 = √2 . 0 Setting t = 0 in our expression for x yields x(0) = c1 b1 + c2 b2 + c3 b3 . To solve the initial value problem, we employ (2.11) to see that √ 1 0 3 3 2 √ − c1 = 0 · 1 = 4, c2 = √1 · 1 = 2, 2 1 4 4 0 √ 1 3 2 √ 1 c3 = √2 · 1 = 2 2. 4 0 Hence the solution is √ √ 1 1 0...
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## This document was uploaded on 01/12/2014.

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