Unformatted text preview: refore the supposition that
W (y1 , y2 ) = 0 at some point on I is false and W (y1 , y2 ) = 0 for points
on I .
We now know what we mean by essentially diﬀerent. Let y1 (t), y2 (t) be two l.i.
solutions of (6.15). Then the linear combination y (t) = Ay1 (t) + By2 (t) satisﬁes
(6.15) and can satisfy the initial conditions y (t0 ) = y0 , dy (t0 ) = y0 for all t0 ∈ I
dt
and all choices of real numbers y0 , y0 .
Let us summarize:
For this course, our main interest is in identifying linearly independent solutions
y1 (t), y2 (t) of (6.15) with coeﬃcients p(t), q (t) continuous on the interval I of interest. For this purpose, the linear independence of y1 (t), y2 (t) on I is equivalent to
the nonvanishing of the Wronskian on I .
The equivalence is broken if our interval includes points at which p(t) and/or q (t)
are discontinuous.
Example: Take f (t) = t, g (t) = t2 on the interval I = (−1, 1).
Then t and t2 are certainly l.i. on (−1, 1). [Since k1 t + k2 t2 = 0 for all t ∈ (−1, 1)
implies that k1 = k2 = 0. Put for example t = 1/2 or t = 1/4.] They are also
solutions to
2
2
y y + 2 y = 0.
t
t
But the Wronskian, W (f, g ) = t2 , vanishes at t = 0. Note that p, q are discontinuous
t 2
at t = 0 and exp t0 t dt is not deﬁned if t and t0 have opposite signs.
Example: Take f (t) = t3 , g (t) = t2 t on the interval I = (−1, 1).
Then W ≡ 0 for all t ∈ I yet t3 and t2 t are l.i. on I . [Take k1 t3 + k2 t2 t = 0 for
20 all t ∈ (−1, 1). Put t = 1/2, −1/2 ⇒ k1 = k2 = 0.] Also t3 and t2 t are solutions
to
3
3
y − y + 2 y = 0
t
t
t 3
But again p, q are discontinuous at t = 0 and exp t0 t dt is not deﬁned if t and t0
have opposite signs.
In many other contexts (e.g. writing any vector in R3 as a linear combination of
three l.i. vectors; writing any function which is piecewise continuous on (−L, L) as
π
π
a linear combination of the l.i. basis ‘vectors’, {cos nLx , sin nLx }∞ , we will need
n=0
the notion of linear independence. The ‘Wronskian’ will not always be the most
useful way to test. 21 Homework Lectures 1115
Question 1:
For each of the following equations, give the
(a) order and say whether each is
(b) linear or nonlinear
(c) autonomous or nonautonomous
(i) dy
dt + 3ty = 5 dy
dx (ii) + y 2 = 3y + 10 (iii) d3 y
dt3 + y dy + cos t = 0
dt (iv) d2 y
dt2 + 0.2 dy + y = cos ω t
dt (v) d2 y
dx2 (vi) dy
dx (vii) + xy = 0
= −x
y d4 y
dt4 =0 Question 2:
Solve the following initial value problems:
(i) dy
dt (ii) + y = e−t , dy
dt = y (1 − y ), (iii) d2 y
dt2 (iv) dy
dx (v) dy
dx y (0) = 1
(a) y (0) = 2; (b) y (0) = 1 .
2 + 2 dy + 5y = cos 2t,
dt
= −x,
y √
= + y, y (0) = dy
dt (0) = 0. y (1) = 1.
y (0) = 0. Is there more than one solution? 23 Question 3:
A tank initially contains 100 gallons of brine with a concentration of 1/2 lb. per
gallon. At time t = 0, an inlet valve is turned on which carries a new brine solution
into the tank at a rate of 2 gallons per mi...
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This note was uploaded on 01/22/2014 for the course MATH 254 taught by Professor Indik during the Fall '08 term at Arizona.
 Fall '08
 INDIK

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