MA 2326
Assignment 5
Due 5 March 2015
Id: 2326-s2015-5.m4,v 1.1 2015/02/25 11:47:26 john Exp john
1. Show that
W ( x, x
0
) =
1
3 x 2 x −
0
2
2 xx − 2
0
+ 2 x − 1 x
0
− x 2 x −
0
1
2 x − 2 x
0
2 xx − 1
0
− x −
+ x − 2
1 x 2
0 x 2
0 is a fundamental matrix for
A ( x ) =
0 1
2 x − 2 0
.
Use this to find the solution to the inhomogenous initial value problem x 2 y ′′ ( x )
−
2 y ( x ) = 1 , y ( x
0
) = y
0
, y ′ ( x
0
) = v
0 for x > 0.
Solution: Direct computation shows that
W ′ ( x, x
0
) =
1
3
2 xx − 2
0
2 x −
0
2
−
2 x −
+ 4 x −
3
2 y 2 xx − 1
0 y 2 x −
0
1
−
+ x − 2
2 x −
3 x x 2
0
2
0 and
A ( x ) W ( x, x
0
) =
1
3
2 xx − 2
0
2 x −
0
2
−
2 x −
+ 4 x −
3
2 x x
0
0
2
2 xx x −
0
− 1
0
1
−
+ x − 2
2 x −
3 x 2
0 x 2
0 so
W ′ ( x, x
0
) = A ( x ) W ( x, x
0
) .
1

Id: 2326-s2015-5.m4,v 1.1 2015/02/25 11:47:26 john Exp john 2
Also,
W ( x, x ) =
1 0
0 1
.
Thus W is a fundamental matrix for A . We therefore have y ( x ) = ( 1 0 ) W ( x, x
0
) y
0 v
0
+
Z x x
0
( 1 0 ) W ( x, t ) or y ( x ) =
=
=
( x 2 x − 2
0
+ 2 x − 1 x
0
) y
0
+ ( x 2 x − 1
0
+
( x 2
" x
− x 2 t − 2
− 2
0
+ 2 x − 1 t
3
# t = x
6
+ 2 x − 1 t = x
0 x
0
) y
0
+ ( x 2 x − 1
0
− x − 1 x 2
0
) v
0
+
( x 2
Z x x
0 x −
0
2 x 2 t −
1
− x −
1 t 2
3 t − 2 dt
3
+ 2 x −
1 x
0
) y
0
+ ( x 2 x −
0
1
− x −
1 x 2
0
) v
0
− x − 1 x 2
0
) v
0
+ x 2 x − 2
0
+ 2 x − 1 x
0
3
−
3
.
6
2. Find a non-zero quadratic polynomial which satisfies
0
1 /t 2
(1
− x 2 ) y ′′ ( x )
−
2 xy ′ x + 6 y ( x ) = 0 and then find a second, linearly independent, solution.
Solution: We want a solution of the form y ( x ) = αx 2 + βx + γ.
dt,
Subsituting into the differential equation,
2 α (1
− x 2 )
−
2 x (2 αx + β ) + 6( αx 2 + βx + γ ) = 0
The coefficient of x 2 on the left hand side is zero for any values of α ,
β and γ . To make the coefficients of x and 1 equal to zero as well we need
4 β = 0 , 6 γ
−
2 α = 0 .
The solutions we are looking for are therefore y
1
( x ) = 3 x 2
−
1 and its multiples. To find a second solution we solve the first order linear equations
(1
− x 2 ) w ′ ( x )
−
2 xw ( x ) = 0

Id: 2326-s2015-5.m4,v 1.1 2015/02/25 11:47:26 john Exp john 3 and y
1
( x ) y ′
2
( x )
− y ′
1
( x ) y
2
( x ) = w ( x ) for w and y
2
. The first has as its solution w ( x ) =
1 w (0)
− x 2
.
The second equation then becomes
(3 x 2
−
1) y ′
2
( x )
−
6 xy
2
( x ) =
1 w (0)
− x 2
.
This has the solution y
2
( x ) = y i
( x ) + y h
( x ) where y i
( x ) = y
2
(0) exp
Z x
0
6 t
3 t 2
−
1 dt =
− y
2
(0)(3 x 2
−
1) and y i
( x ) =
=
=
=
Z x
0 exp
= w (0)
= w w w
Z x
(0)(3
(0)(3
(0)
8
0
8
8
Z x w (0)(3 x
4 x x
3
2
2
2 s
(3 s x
−
−
−
2
3
1)
1) t 2
6 t
3 x
−
2
Z x
0
1
− dt
2
1)
−
1) 2 (1
Z x
3
0
(1
−
1
− s
3 s
2 ds
2
)
+ 1 s
(3 s 2
−
1
1)
2
2 w (0)
)(3 s 2
−
1)
!
ds
( s
−
1 /
√
3) 2
1 + x
1
+ s 2 ds
1
−
1
+
6 x
( s + 1 /
√
3) 2 log
1
− x
1 + x
−
3 x 2
−
1
−
1 log
1
− x
−
6 x
−
1 s
−
1
1
+ s + 1
!
ds
The simplest choice is w (0) = 8, y (0) = 0: y
2
( x ) = (3 x 2
−
1) log
1 + x
1
− x −
6 x.