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Unformatted text preview: (E) r1 = 2+4i, r2 = 24i
(C) r1 = 12i, 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 = 12i
(D) r1 = 2+4i, r2 = 24i
(B) r1 = 1+2i, r2 = 12i
(E) r1 = 2+4i, r2 = 24i
(C) r1 = 12i, 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.
 Spring '13
 EricCytrynbaum
 Differential Equations, Algebra, Equations

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