MATH201 Lecture 7: 7. Higher Order Equations.pdf

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Jan. 23, 2012: many examples here are taken from the textbook. General solution is given by a r2 + b r + c = 0(cid:20) r1, r2: r1(cid:2) r2, both real. General solution is given by r2 4 r + 5 = 0(cid:20) r1,2 = ( 4) ( 4)2 4 5. 2 (cid:17) = 0(cid:20) e c2 = 0(cid:20) c2 = 0. 2 (cid:17) = 1(cid:20) 2 e c2 c1 e = 1. an y(n) +(cid:10) + a1 y + a0 y = 0 y(t) = e2t (c1 cos t + c2 sin t). C1 = e , c2 = 0. y = e e2t cos t. y 4 y + 5 y = 0, y(cid:16) y (cid:16) : basic information. Now y (t) = 2 e2t (c1 cos t + c2 sin t) + e2t ( c1 sin t + c2 cos t). Math 201 lecture 07: higher order equations: write down the characteristic equation, solve it.

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