MATH 255 Lecture 4: Week4.pdf

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24 Apr 2015
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Week 4 (i) section 2. 2 second-order homogeneous linear equations with constant coe cients: + cy = 0; where a; b; c 2 r: Function erx is a solution of (1) if and only if ar2 + br + c = 0: That is, (cid:0)b (cid:6) p b2 (cid:0) 4ac. If b2 (cid:0) 4ac > 0, (2) has two real roots (cid:0)b + p b2 (cid:0) 4ac. ; r1 = r1 = (1) (2) (3) with r1 > r2, and er1x and er2x are two linearly independent solutions of (1). If b2 (cid:0) 4ac = 0, then (2) has only one root (cid:0)b. While y1(x) = er1x is solution of (1), we claim that y2(x) = xer1x is also a solution of (1). 1 + br1 + c)xer1x + 2ar1er1x + ber1x. 1 + br1 + c = 0 and r1 = b.

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