Lecture 5 Notes

# A y t et c1 cos2t c2 sin2t b y t c1 e c y

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Unformatted text preview: (E) r1 = -2+4i, r2 = -2-4i (C) r1 = 1-2i, r2 = -1+2i Complex roots (Section 3.3) • Example: Find the (real valued) general solution to the equation y + 2y + 5y = 0 ￿￿ ￿ • Step 1: Assume y (t) = e , plug this into the equation and ﬁnd values of r that make it work. rt (A) r1 = 1+2i, r2 = 1-2i (D) r1 = 2+4i, r2 = 2-4i (B) r1 = -1+2i, r2 = -1-2i (E) r1 = -2+4i, r2 = -2-4i (C) r1 = 1-2i, r2 = -1+2i Complex roots (Section 3.3) • Example: Find the (real valued) general solution to the equation y + 2y + 5y = 0 ￿￿ ￿ • Step 2: Real part of r goes in the exponent, imaginary part goes in the trig functions. (A) y (t) = e−t (C1 cos(2t) + C2 sin(2t)) (B) y (t) = C1 e (C) y (t) = C1 cos(2t) + C2 sin(2t) + C3 e (D) y (t) = C1 cos(2t) + C2 sin(2t) (−1+2i)t + C2 e (−1−2i)t −t Complex roots (Section 3.3) • Example: Find the (real valued) general solution to the equation y + 2y + 5y = 0 ￿￿ ￿ • Step 2: Real part of r goes in the exponent, imaginary part goes in the trig functions. (A) y (t) = e−t (C1 cos(2t) + C2 sin(2t)) (B) y (t) = C1 e (C) y (t) = C1 cos(2t) + C2 sin(2t) + C3 e (D) y (t) = C1 cos(2t) + C2 sin(2t) (−1+2i)t + C2 e (−1−2i)t −t Complex roots (Section 3.3) • Example: Find the solution to the IVP y ￿￿ + 2y ￿ + 5y = 0, y (0) = 1, y ￿ (0) = 0 • General solution: (A) (B) (C) (D) y (t) = e (C1 cos(2t) + C2 sin(2t)) −t y (t) = e (2 cos(2t) + sin(2t)) ￿ ￿ 1 −t y (t) = e cos(2t) − sin(2t) 2 1 −t y (t) = e (2 cos(2t) − sin(2t)) 2 1 −t y (t) = e (2 cos(2t) + sin(2t)) 2 −t Complex roots (Section 3.3) • Example: Find the solution to the IVP y ￿￿ + 2y ￿ + 5y = 0, y (0) = 1, y ￿ (0) = 0 • General solution: (A) (B) (C) (D) y (t) = e (C1 cos(2t) + C2 sin(2t)) −t y (t) = e (2 cos(2t) + sin(2t)) ￿ ￿ 1 −t y (t) = e cos(2t) − sin(2t) 2 1 −t y (t) = e (2 cos(2t) − sin(2t)) 2 1 −t y (t) = e (2 cos(2t) + sin(2t)) 2 −t Repeated roots (Section 3.4) • For the general case, ay + by + cy = 0 , by assuming y (t) = e ￿￿ ￿ we get the characteristic equation: ar + br + c = 0 2 rt Repeated roots (Section 3.4) • For the general case, ay + by + cy = 0 , by assuming y (t) = e ￿￿ ￿ we get the characteristic equation: ar + br + c = 0 2 • There are three cases. i. Two distinct real roots: b2 - 4ac > 0. ( r1 ≠ r2 ) ii.A repeated real root: b2 - 4ac = 0. iii.Two complex roots: b2 - 4ac < 0. rt Repeated roots (Section 3.4) • For the general case, ay + by + cy = 0 , by assuming y (t) = e ￿￿ ￿ we get the characteristic equation: ar + br + c = 0 2 • There are three cases. i. Two distinct real roots: b2 - 4ac > 0. ( r1 ≠ r2 ) ii.A repeated real root: b2 - 4ac = 0. iii.Two complex roots: b2 - 4ac < 0. • For case ii ( r1 = r2 =r), we need another independent solution! rt Repeated roots (Section 3.4) • For the general case, ay + by + cy = 0 , by assuming y (t) = e ￿￿ ￿ we get th...
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## This note was uploaded on 02/12/2014 for the course MATH 256 taught by Professor Ericcytrynbaum during the Spring '13 term at The University of British Columbia.

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