Lecture 5 Notes

5 our next goal is to gure out how to nd solutions to

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Unformatted text preview: mary (3.1-3.4) • For the general case, ay + by + cy = 0 , by assuming y (t) = e ￿￿ ￿ rt we get the characteristic equation: ar + br + c = 0 2 • There are three cases. i. Two distinct real roots: b2 - 4ac > 0. ( r1, r2 ) y (t) = C1 e r1 t + C2 e r2 t ii.A repeated real root: b2 - 4ac = 0. ( r ) y (t) = C1 e + C2 te rt rt iii.Two complex roots: b2 - 4ac < 0. ( r1,2 = α±iβ ) Summary (3.1-3.4) • For the general case, ay + by + cy = 0 , by assuming y (t) = e ￿￿ ￿ rt we get the characteristic equation: ar + br + c = 0 2 • There are three cases. i. Two distinct real roots: b2 - 4ac > 0. ( r1, r2 ) y (t) = C1 e r1 t + C2 e r2 t ii.A repeated real root: b2 - 4ac = 0. ( r ) y (t) = C1 e + C2 te rt rt iii.Two complex roots: b2 - 4ac < 0. ( r1,2 = α±iβ ) y = eαt (C1 cos(β t) + C2 sin(β t)) Second order, linear, constant coeff, homogeneous • Find the general solution to the equation y − 6y + 8y = 0 ￿￿ ￿ (A) y (t) = C1 e−2t + C2 e−4t (B) y (t) = C1 e2t + C2 e4t (C) y (t) = e2t (C1 cos(4t) + C2 sin(4t)) (D) y (t) = e−2t (C1 cos(4t) + C2 sin(4t)) (E) y (t) = C1 e + C2 te 2t 4t Second order, linear, constant coeff, homogeneous • Find the general solution to the equation y − 6y + 8y = 0 ￿￿ ￿ (A) y (t) = C1 e−2t + C2 e−4t (B) y (t) = C1 e2t + C2 e4t (C) y (t) = e2t (C1 cos(4t) + C2 sin(4t)) (D) y (t) = e−2t (C1 cos(4t) + C2 sin(4t)) (E) y (t) = C1 e + C2 te 2t 4t Second order, linear, constant coeff, homogeneous • Find the general solution to the equation y − 6y + 9y = 0 ￿￿ ￿ (A) y (t) = C1 e3t (B) y (t) = C1 e3t + C2 e3t (C) y (t) = C1 e3t + C2 e−3t (D) y (t) = C1 e + C2 te (E) y (t) = C1 e + C2 v (t)e 3t 3t 3t 3t Second order, linear, constant coeff, homogeneous • Find the general solution to the equation y − 6y + 9y = 0 ￿￿ ￿ (A) y (t) = C1 e3t (B) y (t) = C1 e3t + C2 e3t (C) y (t) = C1 e3t + C2 e−3t (D) y (t) = C1 e + C2 te (E) y (t) = C1 e + C2 v (t)e 3t 3t 3t 3t Second order, linear, constant coeff, homogeneous • Find the general solution to the equation y − 6y + 10y = 0 ￿￿ ￿ (A) y (t) = C1 e3t + C2 et (B) y (t) = C1 e3t + C2 e−t (C) y (t) = C1 cos(3t) + C2 sin(3t) (D) y (t) = e (C1 cos(3t) + C2 sin(3t)) (E) y (t) = e (C1 cos(t) + C2 sin(t)) t 3t Second order, linear, constant coeff, homogeneous • Find the general solution to the equation y − 6y + 10y = 0 ￿￿ ￿ (A) y (t) = C1 e3t + C2 et (B) y (t) = C1 e3t + C2 e−t (C) y (t) = C1 cos(3t) + C2 sin(3t) (D) y (t) = e (C1 cos(3t) + C2 sin(3t)) (E) y (t) = e (C1 cos(t) + C2 sin(t)) t 3t Second order, linear, constant coeff, nonhomogeneous (3.5) • Our next goal is to ﬁgure out how to ﬁnd solutions to nonhomogeneous equations like this one: y ￿￿ − 6y ￿ + 8y = sin(2t) • But ﬁrst, a bit more on the connections between matrix algebra and differential equations . . . Som...
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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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