Lecture 5 Notes

# I two distinct real roots b2 4ac 0 r1 r2 y t

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Unformatted text preview: C2 ). = (C1 t + C2 )e−2t y (t) = C1 te−2t + C2 e−2t y2 (t) y1 (t) Is this the general solution? Calculate the Wronskian: W (e −2t −2t , te )(t) Reduction of order y1 (t) = e−2t . For the equation y + 4y + 4y = 0 , say you know ￿￿ Guess ￿ y2 (t) = v (t)e −2t (where v (t) = C1 t + C2 ). = (C1 t + C2 )e−2t y (t) = C1 te−2t + C2 e−2t y2 (t) y1 (t) Is this the general solution? Calculate the Wronskian: W (e −2t −2t , te )(t) = ￿ y1 (t)y2 (t) − ￿ y1 (t)y2 (t) Reduction of order y1 (t) = e−2t . For the equation y + 4y + 4y = 0 , say you know ￿￿ Guess ￿ y2 (t) = v (t)e −2t (where v (t) = C1 t + C2 ). = (C1 t + C2 )e−2t y (t) = C1 te−2t + C2 e−2t y2 (t) y1 (t) Is this the general solution? Calculate the Wronskian: W (e −2t −2t , te )(t) = ￿ y1 (t)y2 (t) − ￿ y1 (t)y2 (t) = e−4t Reduction of order y1 (t) = e−2t . For the equation y + 4y + 4y = 0 , say you know ￿￿ Guess ￿ y2 (t) = v (t)e −2t (where v (t) = C1 t + C2 ). = (C1 t + C2 )e−2t y (t) = C1 te−2t + C2 e−2t y2 (t) y1 (t) Is this the general solution? Calculate the Wronskian: W (e −2t −2t , te )(t) = ￿ y1 (t)y2 (t) − ￿ y1 (t)y2 (t) = e−4t ￿= 0 Reduction of order y1 (t) = e−2t . For the equation y + 4y + 4y = 0 , say you know ￿￿ Guess ￿ y2 (t) = v (t)e −2t (where v (t) = C1 t + C2 ). = (C1 t + C2 )e−2t y (t) = C1 te−2t + C2 e−2t y2 (t) y1 (t) Is this the general solution? Calculate the Wronskian: W (e −2t −2t , te )(t) = ￿ y1 (t)y2 (t) − So yes! ￿ y1 (t)y2 (t) = e−4t ￿= 0 Summary (3.1-3.4) • For the general case, ay + by + cy = 0 , by assuming y (t) = e ￿￿ ￿ we get the characteristic equation: ar + br + c = 0 2 rt Summary (3.1-3.4) • 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. rt Summary (3.1-3.4) • 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 ) rt 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 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 ii.A repeated real root: b2 - 4ac = 0. ( r ) r2 t 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 Sum...
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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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