lecture_15 (dragged) 1

# Lecture_15(dragged 1

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Unformatted text preview: y2 ￿ ￿ ￿ ￿ ￿ ￿ ￿￿ ￿￿ ￿ ￿ = a(t) y1 y2 − y1 y2 + b(t) y1 y2 − y1 y2 + c(t) y1 y2 − y1 y2 ￿ ￿￿ ￿ ￿￿ ￿ ￿￿ ￿ = y1 a(t) y2 + b(t) y2 + c(t) y2 − a(t) y1 + b(t) y1 + c(t) y1 y2 = 0. Hence W = W (t) satisﬁes a ﬁrst order diﬀerential equation. Since this equation is separable, we may solve by dividing both sides by W (t): dW = −b(t) W dt 1 dW b(t) =− W dt a(t) a(t) d b(t) ln |W | = − dt a(t) =⇒ ln |W | = − ￿ t b(τ ) dτ + (constant). a(τ ) Upon exponentiating, we can express W = W (t) in the form ￿ ￿t ￿ b(τ ) W (t) = W0 exp − dτ t0 a(τ ) where W0 = W (t0 ). Since the exponential is never zero, we see that W (t) ￿= 0 for any t if and only if W (t0 ) ￿= 0 for some t0 . Example. Consider the constant coeﬃcient equation a y ￿￿ + b y ￿ + c y = 0. If the characteristic equation a r2 + b r + c = 0 has two real roots r1 and r2 , we know that two solutions to the diﬀerential equation are y1 (t) = er1 t and y2 (t) = er2 t . We have seen that the Wronskian of these two functions is ￿ ￿ W (t) = y1 (t) y2 (t) − y1 (t) y2 (t) = (r2 − r1 ) e(r1 +r2 )t ....
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## This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue University-West Lafayette.

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