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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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.

Ask a homework question - tutors are online