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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 (3.13.4)
• 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 (3.13.4)
• 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 (3.13.4)
• 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 (3.13.4)
• 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 (3.13.4)
• 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 (3.13.4)
• 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.
 Spring '13
 EricCytrynbaum
 Differential Equations, Algebra, Equations

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