SSM-Ch03 - Chap 3 Sec 3.1 Higher Order Linear ODEs...

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 third-order 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

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.