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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):
= −b(t) W
a(t) a(t) d
ln |W | = −
a(t) =⇒ ln |W | = − t b(τ )
dτ + (constant).
a(τ ) Upon exponentiating, we can express W = W (t) in the form
W (t) = W0 exp −
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
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.
- Spring '09
- Linear Independence