MA 2327
Assignment 8
Due 14 December 2015
Id:
2327-f2015-8.m4,v 1.1 2015/12/08 16:46:15 john Exp john
1. Find a non-zero quadratic polynomial which satisfies
(1 − t2 )x′′ (t) − 2tx′ t + 6x(t) = 0
and then find a second, linearly independent, solution.
Solution: We want a solution of the form
x(t) = αt2 + βt + γ.
Subsituting into the differential equation,
2α(1 − t2 ) − 2t(2αt + β) + 6(αt2 + βt + γ) = 0
The coefficient of t2 on the left hand side is zero for any values of α,
β and γ. To make the coefficients of t and 1 equal to zero as well we
need
4β = 0, 6γ − 2α = 0.
The solutions we are looking for are therefore
x1 (t) = 3t2 − 1
and its multiples. To find a second solution we solve the first order
linear equations
(1 − t2 )u′(t) − 2tu(t) = 0
and
x1 (t)x′2 (t) − x′1 (t)x2 (t) = u(t)
1
Id:
2327-f2015-8.m4,v 1.1 2015/12/08 16:46:15 john Exp john 2
for u and x2 . The first has as its solution
u(t) =
u(0)
.
1 − t2
The second equation then becomes
(3t2 − 1)x′2 (t) − 6tx2 (t) =
u(0)
.
1 − t2
This has the solution
x2 (t) = xi (t) + xh (t)
where
xi (t) = x2 (0) exp
t
Z
0
6t
dt = −x2 (0)(3t2 − 1)
3t2 − 1
and
u(0)
6t
dt
ds
(1 − s2 )(3s2 − 1)
0
s 3t2 − 1
Z t
3t2 − 1
ds
= u(0)
0 (3s2 − 1)2 (1 − s2 )
!
Z
3s2 + 1
1
u(0)(3t2 − 1) t
3 2
ds
+
=
4
(3s − 1)2 1 − s2
0
!
Z
u(0)(3t2 − 1) t
1
1
1
1
√
√
=
+
−
ds
+
8
0
(s − 1/ 3)2 (s + 1/ 3)2 s − 1 s + 1
6t
1+t
u(0)(3t2 − 1)
log
−
=
8
1 − t 3t2− 1
1+t
u(0)
3t2 − 1 log
− 6t
=
8
1−t
xi (t) =
Z
t
exp
Z
t
The simplest choice is u(0) = 8, x2 (0) = 0:
1+t
x2 (t) = (3t − 1) log
− 6t.
1−t
2