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Unformatted text preview: e 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!
• Reduction of order  a method for guessing another solution. rt Reduction of order
• You have one solution y1 (t) and you want to ﬁnd another independent
one, y2 (t) . Reduction of order
• You have one solution y1 (t) and you want to ﬁnd another independent
one, y2 (t) . • Guess that y2 (t) = v (t)y1 (t) for some as yet unknown v (t) . If you
can ﬁnd v (t) this way, great. If not, gotta try something else. Reduction of order
• You have one solution y1 (t) and you want to ﬁnd another independent
one, y2 (t) . • Guess that y2 (t) = v (t)y1 (t) for some as yet unknown v (t) . If you
can ﬁnd v (t) this way, great. If not, gotta try something else.
• Example  y + 4y + 4y = 0.
Only one root to the characteristic equation, r=2, so we only get one solution that way: y1 (t) = e −2t . Reduction of order
• You have one solution y1 (t) and you want to ﬁnd another independent
one, y2 (t) . • Guess that y2 (t) = v (t)y1 (t) for some as yet unknown v (t) . If you
can ﬁnd v (t) this way, great. If not, gotta try something else.
• Example  y + 4y + 4y = 0.
Only one root to the characteristic equation, r=2, so we only get one solution that way:
• Use Reduction of order to ﬁnd a second solution. y2 (t) = v (t)e−2t y1 (t) = e −2t . Reduction of order
y1 (t) = e−2t .
For the equation y + 4y + 4y = 0 , say you know
Reduction of order
y1 (t) = e−2t .
For the equation y + 4y + 4y = 0 , say you know
Guess y2 (t) = v (t)e −2t . Reduction of order
y1 (t) = e−2t .
For the equation y + 4y + 4y = 0 , say you know
Guess y2 (t) = v (t)e −2t .
y2 (t) = v (t)e−2t − 2v (t)e−2t Reduction of order
y1 (t) = e−2t .
For the equation y + 4y + 4y = 0 , say you know
Guess y2 (t) = v (t)e −2t .
y2 (t) = v (t)e−2t − 2v (t)e−2t
y2 (t) = v (t)e−2t − 2v (t)e−2t − 2v (t)e−2t + 4v (t)e−2t Reduction of order
y1 (t) = e−2t .
For the equation y + 4y + 4y = 0 , say you know
Guess y2 (t) = v (t)e −2t .
y2 (t) = v (t)e−2t − 2v (t)e−2t
y2 (t) = v (t)e−2t − 2v (t)e−2t − 2v (t)e−2t + 4v (t)e−2t
y 2 + 4y2 + 4y2 = Reduction of order
y1 (t) = e−2t .
For the equation y + 4y + 4y = 0 , say you know
Guess y2 (t) = v (t)e −2t . ⇒
y2 (t) = v (t)e−2t − 2v (t)e−2t 4y2 (t) = 4v (t)e−2t
y2 (t) = v (t)e−2t − 2v (t)e−2t − 2v (t)e−2t + 4v (t)e−2t
y 2 + 4y2 + 4y2 = Reduction of order
y1 (t) = e−2t .
For the equation y + 4y + 4y = 0 , say you know
Guess y2 (t) = v (t)e −2t . ⇒ ⇒
y2 (t) = v (t)e−2t − 2v (t)e−2t
4y2 (t) = 4v (t)e−2t 4y2 (t)...
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 Spring '13
 EricCytrynbaum
 Differential Equations, Algebra, Equations

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