Chap
.
3
Higher Order Linear ODEs
Sec
.
3
.
1
Homogeneous Linear ODEs
Example 5
.
Basis
,
Wronskian
. In pulling out the exponential functions, you see that their product
equals
e
0
u003d
1; indeed,
e
u2212
2
x
e
u2212
x
e
x
e
2
x
u003d
e
u2212
2
x
u2212
x
u002b
x
u002b
2
x
u003d
e
0
u003d
1.
In subtracting columns as indicated in the text, you get
1
0
0
0
u2212
2
1
3
4
4
u2212
3
u2212
3
0
u2212
8
7
9
16
.
You can now see the thirdorder determinant shown in the example. This determinant is
simplified as indicated and then developed by the second row:
1
2
4
u2212
3
0
0
7
2
16
u003d u002b
3
2
4
2
16
u003d
3
ue0a2
32
u2212
8
ue0a3
u003d
72.
Problem Set 3
.
1
.
Page 111
1
.
Basis
. A general solution is obtained by four successive integrations,
y
u2032u2032u2032
u003d
c
1
,
y
u2032u2032
u003d
c
1
x
u002b
c
2
,
y
u2032
u003d
1
2
c
1
x
2
u002b
c
2
x
u002b
c
3
,
y
u003d
1
6
c
1
x
3
u002b
1
2
c
2
x
2
u002b
c
3
x
u002b
c
4
,
and
y
is a linear combination of the four functions of the given basis, as wanted.
5
.
Basis
.
Setting
y
u2032u2032
u003d
z
, you have
z
u2032u2032
u002b
9
z
u003d
0. A general solution is
z
u003d
c
1
cos 3
x
u002b
c
2
sin 3
x
.
By integration you get
y
u2032
u003d
u222b
z dx
u003d
1
3
c
1
sin 3
x
u2212
1
3
c
2
cos 3
x
u002b
c
3
so that you obtain a general solution of the given ODE by another integration ,
y
u003d
u222b
y
u2032
dx
u003d
u2212
1
9
c
1
cos 3
x
u2212
1
9
c
2
sin 3
x
u002b
c
3
x
u002b
c
4
.
You see that it involves the given basis cos 3
x
, sin 3
x
,
x
, 1; it is a linear combination of these four
functions with arbitrary coefficients.
7
.
Linear independence
.
e
x
and
e
u2212
x
are solutions of
y
u2032u2032
u2212
y
u003d
0. Conclude that 1,
e
x
,
e
u2212
x
are solutions of
the linear ODE
y
u2032u2032u2032
u2212
y
u2032
u003d
0 with continuous (even constant) coefficients, so that you can apply Theorem
3. Calculate the Wronskian
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26
Ordinary Differential Equations (ODEs)
Part A
W
u003d
1
e
x
e
u2212
x
0
e
x
u2212
e
u2212
x
0
e
x
e
u2212
x
u003d
e
x
u2212
e
u2212
x
e
x
e
u2212
x
u003d
1
u2212
ue0a2
u2212
1
ue0a3
u003d
2.
Conclude that the functions are linearly independent. Hence they form a basis of solutions of
y
u2032u2032u2032
u2212
y
u2032
u003d
0.
That
e
x
e
u2212
x
u003d
1 is irrelevant since you have to consider linear combinations, not products of given
functions.
9
.
Linear dependence
can often be shown by using a functional relation. In the present problem you have
ln
x
2
u003d
2 ln
x
and conclude that the functions are linearly dependent. Indeed, for instance, a linear
combination with coefficients not all zero is, denoting the given functions by
y
1
u003d
ln
x
,
y
2
u003d
ln
x
2
,
y
3
u003d
ue0a2
ln
x
ue0a3
2
,
y
1
u2212
1
2
y
2
u002b
0
y
3
u003d
0.
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 Spring '12
 M.lee
 Linear Algebra, Vector Space, basis, general solution

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