MA 2327
Assignment 2
Due 19 October 2015
Id:
2327-f2015-2.m4,v 1.2 2015/10/17 15:16:35 john Exp john
1. Find invariants for the following equations
(a)
dx
3tx2 + t3
=− 3
dt
x + 3t2 x
(b)
3t2 − 1
dx
=
dt
2x
(c)
cos x cos t
dx
=
dt
sin x sin t
Solution:
(a) U(t, x) = 14 (x4 + 6t2 x2 + t4 )
(b) U(t, x) = t3 − x2 − t
(c) U(t, x) = cos x sin t
2. Show that p and q defined on the set (t, x) 6= (0, 0) by
p(t, x) =
t2
x
,
+ x2
q(t, x) =
satisfy
∂p ∂q
+
=0
∂x ∂t
1
t2
t
+ x2
Id:
2327-f2015-2.m4,v 1.2 2015/10/17 15:16:35 john Exp john 2
but there is no function U defined on the set (t, x) 6= (0, 0) with
∂U
= p,
∂t
∂U
= −q.
∂x
Hint: What are the values of U on the curve t = cos θ, x = sin θ?
Solution: The partial derivatives are
t2 − x2
∂p
= 2
,
∂x
(t + x2 )2
∂q
x2 − t2
= 2
∂t
(t + x2 )2
so the equation
∂p ∂q
+
=0
∂x ∂t
is clearly satisfied. Suppose there were a function U defined on the set
(t, x) 6= (0, 0) with
∂U
∂U
= p,
= −q.
∂t
∂x
Differentiating,
∂U
d
∂U
d
d
U(cos θ, sin θ) =
(cos θ, sin θ) cos θ +
(cos θ, sin θ) sin θ
dθ
∂t
dθ
∂x
dθ
= − p(cos θ, sin θ) sin θ − q(cos θ, sin θ) cos θ = −1.
So U(cos θ, sin θ) is strictly decreasing, but it must also be periodic with
period 2π since (cos θ, sin θ) and (cos(θ + 2π), sin(θ + 2π)) are the same
point.
3. The equation
dx
= x + et
dt
can be written as
p(t, x)
dx
=
dt
q(t, x)
in a variety of ways. We could, for example, choose
p(t, x) = x + et ,
q(t, x) = 1
or we could choose
p(t, x) = xe−t + 1,
q(t, x) = e−t .
The two are of course equivalent, but show that the first is not integrable, while the second is.
Id:
2327-f2015-2.m4,v 1.2 2015/10/17 15:16:35 john Exp john 3
Note: A factor which when multiplied by the numerator and denominator of an equation makes it integrable is called an integrating factor.
Except in special cases there is unfortunately no algorithm for finding
them.
Solution: If
p(t, x) = x + et , q(t, x) = 1
then
so
∂p
(t, x) = 1,
∂x
∂q
(t, x) = 0
∂t
∂p ∂q
+
= 1 6= 0
∂x ∂t
If
p(t, x) = xe−t + 1,
then
so
∂p
(t, x) = e−t ,
∂x
q(t, x) = e−t .
∂q
(t, x) = −e−t
∂t
∂p ∂q
+
= 0.
∂x ∂t