Lecture 5 Notes

# Guess that y2 t v ty1 t for some as yet unknown v

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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 &gt; 0. ( r1 ≠ r2 ) ii.A repeated real root: b2 - 4ac = 0. iii.Two complex roots: b2 - 4ac &lt; 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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