MA 2327
Assignment 3
Due 29 October 2015
Id:
2327-f2015-3.m4,v 1.2 2015/10/27 16:50:38 john Exp john
1. Solve the following initial value problems
(a)
x′ (t) = − tan(t) x(t) + sec t,
x(0) = 1.
Solution: This is x′ (t) = a(t)x(t) + b(t) with a(t) = − tan t and
b(t) = sec t, so we just substitute those, together with the initial
values t0 = 0, x0 = 1, into the general solution formula
x(t) = exp
Z
Z
t
t0
a(s) ds x0 +
t
a(s) ds = −
r
Z
t
t0
exp
Z
r
t
Z
Z
t
t0
exp
Z
t
tan t = log
r
t
exp
Z
t
exp
Z
a(s) ds =
r
t0
t
a(s) ds b(r) dr.
r
cos t
,
cos r
cos t
,
cos r
a(s) ds = cos t,
cos t
sec r dr = cos t
a(s) ds b(r) dr =
0 cos r
= cos t tan t = sin t,
Z
t
x(t) = cos t + sin t.
1
Z
0
t
sec2 r dr
Id:
2327-f2015-3.m4,v 1.2 2015/10/27 16:50:38 john Exp john 2
(b)
(1 − t2 )x′ (t) + 2tx(t) = 1,
x(0) = 1.
Solution: This is x′ (t) = a(t)x(t) + b(t) with a(t) = −2t/(1 − t2 )
and b(t) = 1/(1 − t2 ). Also t0 = 0 and x0 = 1.
Z
r
t
a(s) ds = −
Z
t
r
t
exp
Z
t
exp
Z
a(s) ds =
t0
exp
=
Z
Z
r
t
0
t
1 − t2
,
1 − r2
a(s) ds = 1 − t2 .
t0
t
!
r
Z
2t
1 − t2
,
=
log
1 − t2
1 − r2
a(s) ds b(r) dr
1 − t2
dr
(1 − r 2 )2
1/4
1/4
1/4
1/4
= (1 − t )
+
−
+
2
2
(r − 1)
(r + 1)
r−1 r+1
0
2t
1+t
1 − t2
+ log
=
4
1 − t2 1 −
t
t 1
1+t
2
= + (1 − t ) log
.
2 4
1−t
2
Z
t
!
dr
2. Assume, as we will see later in lecture, that for all x0 , x1 , . . . , xn−1 the
homogeneous initial value problem
cn (x)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = 0
x(t0 ) = x0 ,
x′ (t0 ) = x1 ,
...
x(n−1) (t0 ) = xn−1
has a unique solution in the interval I when t0 ∈ I, ck is continuous
in I and cn has no zeros in I. Here x(k) denotes the k’th derivative of x.
Show the following:
(a) If y and z are solutions of
cn (t)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = 0
then so is αy + βz for any constants α and β. In other words, the
set of solutions is a vector space.
Id:
2327-f2015-3.m4,v 1.2 2015/10/27 16:50:38 john Exp john 3
Solution: Set x = αy + βz. Then
cn (t)x(n) (t) + · · · + c0 (t)x(t)
= cn (t)(αy (n) (t) + βz (n) (t)) + · · · + c0 (t)(αy(t) + βz(t))
= α(cn (t)y (n) (t) + · · · + c0 (t)y(t)) + β(cn (t)z (n) (t) + · · · + c0 (t)z(t))
= 0.
(b) If y and z are solutions of
cn (t)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = f (t)
then y − z is a solution of
cn (t)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = 0
Solution: Set x = y − z. Then
cn (t)x(n) (t) + · · · + c0 (t)x(t)
= cn (t)(y (n) (t) − z (n) (t)) + · · · + c0 (t)(y(t) − z(t))
= (cn (t)y (n) (t) + · · · + c0 (t)y(t)) − (cn (t)z (n) (t) + · · · + c0 (t)z(t))
= f (t) − f (t) = 0.
(c) If y is a solution of
cn (t)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = 0
and z is a solution of
cn (t)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = f (t)
then y + z is a solution of
cn (t)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = f (t)
Solution: Set x = y + z. Then
cn (t)x(n) (t) + · · · + c0 (t)x(t)
= cn (t)(y (n) (t) + z (n) (t)) + · · · + c0 (t)(y(t) + z(t))
= (cn (t)y (n) (t) + · · · + c0 (t)y(t)) + (cn (t)z (n) (t) + · · · + c0 (t)z(t))
= 0 + f (t) = f (t).
Id:
2327-f2015-3.m4,v 1.2 2015/10/27 16:50:38 john Exp john 4
(d) The initial value problem
cn (t)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = f (t)
x(t0 ) = x0 ,
x′ (t0 ) = x1 ,
...
x(n−1) (t0 ) = xn−1
has at most one solution.
Solution: Suppose y and z are solutions and let w = x − y. By
(2b), w satisfies
cn (t)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = 0.
But
w(t0 ) = x(t0 ) − y(t0 ) = x0 − x0 = 0
and similarly w ′ (t0 ) = 0, . . . , w (n−1) (t0 ) = 0. By hypothesis the
initial value problem
cn (x)x(n) (t) + · · · + c1 (t)x′ (t) + c0 x(t) = 0
x(t0 ) = 0,
x′ (t0 ) = 0,
...
x(n−1) (t0 ) = 0
has a unique solution. Since w and 0 both solve this initial value
problem w = 0 or, equivalently, y = z. Thus there is at most one
solution to
cn (t)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = f (t)
x(t0 ) = x0 ,
x′ (t0 ) = x1 ,
...
x(n−1) (t0 ) = xn−1 .
(e) Suppose that the differential equation
cn (t)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = f (t)
has at least one solution. Then the initial value problem
cn (t)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = f (t)
x(t0 ) = x0 ,
x′ (t0 ) = x1 ,
...
x(n−1) (t0 ) = xn−1
has a solution for any values of x0 , x1 , . . . , xn−1 .
Solution: Let y be some solution to
cn (t)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = f (t).
Set
y0 = y(t0 ),
y1 = y ′(t0 ),
...
yn−1 = y (n−1) (t0 )
Id:
2327-f2015-3.m4,v 1.2 2015/10/27 16:50:38 john Exp john 5
and
z0 = x0 − y0 ,
z1 = x1 − y1 ,
...
zn−1 = xn−1 − yn−1 .
By hypothesis there is a z which solves
cn (x)x(n) (t) + · · · + c1 (t)x′ (t) + c0 x(t) = 0
with
z(t0 ) = z0 ,
z ′ (t0 ) = z1 ,
...
z (n−1) (t0 ) = zn−1 .
But then, by (2c), x = y + z solves
cn (t)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = f (t).
Also,
x(t0 ) = y(t0 ) + z(t0 ) = y0 + x0 − y0 = x0
and similarly x′ (t0 ) = x1 , . . . , x(n−1) (t0 ) = xn−1 . Thus we have
our solution to the initial value problem
cn (t)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = f (t)
x(t0 ) = x0 ,
x′ (t0 ) = x1 ,
...
x(n−1) (t0 ) = xn−1 .