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Unformatted text preview: 2 MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 23 Upon multiplying the ﬁrst equation by g (t) and the second by −g (t), we ﬁnd the following system of
f (t) g (t) c1 +
g (t) g (t) c2 = 0
−f (t) g (t) c1 + −g (t) g (t) c2 = 0
f (t) g (t) − f (t) g (t) c1
We conclude that
W f, g (t) · c1 = W f, g (t) · c2 = 0
for all t.
Since not both c1 and c2 are zero, this implies W f, g (t) = 0 for all t. Clearly this is a contradiction, so f
and g must be linearly independent. This shows p2 =⇒ p3 . Now we show that p3 =⇒ p1 , so that
f and g are linearly independent. If there did not exist t0 such that W f, g (t0 ) = 0, then W f, g (t) = 0
for all t. In particular, the Quotient Rule for Derivatives gives the expression
W f, g (t)
d g (t)
f (t) g (t) − f (t) g (t)
dt f (t)
f (t) must aconstant. Hence g = λ f so that f and g are linearly dependent. Clearly this is a contradiction,
so W f, g (t0 ) = 0 for some t0 . This shows p3 =⇒ p1 .
We have shown that p1 =⇒ p2 and p2 =⇒ p3 and p3 =⇒ p1 . The three statements p1 =⇒ p3 and
p3 =⇒ p2 and p2 =⇒ p1 follow from the compositions
p1 =⇒ p2 =⇒ p3
p3 =⇒ p1 =⇒ p2
p2 =⇒ p3 =⇒ p1 This shows that pi =⇒ pj holds for all i and j . This completes the proof.
Example. Consider the functions f (t) = et and g (t) = e2t . We will show that f and g are linearly independent using two diﬀerent methods.
First we show that they are independent using the original deﬁnition. Consider the equation
c1 f (t) + c2 g (t) = 0 for all t. The function y (t) = c1 f (t) + c2 g (t) must be the zero function, so in particular its derivative y (t) =
c1 f (t) + c2 g (t) is also the zero function. Hence we have the system of equations
c1 et +
+ 2 c2 e2t =0
=0 =⇒ c1
+ 2 c2 =0
=0 Upon subtracting the ﬁrst equation from the second, we see that c1 = c2 = 0. Since this is the only solution
to the equation c1 f + c2 g = 0, we see that f and g must be linearly independent.
Second we show that they are independent using the Wronskian. The Wronskian of f and g is the function
W f, g (t) = f g − f g = et 2 e2t − et e2t = 2 e3t − e3t = e3t = 0.
This this function is nonzero for all t, we see that f and g must be linearly independent.
Fundamental Solutions Revisited. We return to the homogeneous equation
a(t) y + b(t) y + c(t) y = 0.
Say that we have two solutions y1 = y1 (t) and y2 = y2 (t). When is the linear combination
y (t) = c1 y1 (t) + c2 y2 (t)
the general solution to the diﬀerential equation? If y1 and y2 are linearly independent, then y = c1 y1 + c2 y2
is the general solution.
We prove this as follows. Say that we have the initial conditions
y (t0 ) = y0 and
y (t0 ) = y0 . We showed in the previous lecture that if the quantity
W0 = y1 (t0 ) y2 (t0 ) − y1 (t0 ) y2 (t0 ) ...
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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