LECTURE 6: EXACT EQUATIONS AND INTEGRATE FACTORS
MINGFENG ZHAO
September 15, 2014
Recall that



 1) y(x) ≡ a for some constant a such that g(a) = 0
0
Z
Z
I. y = f (x)g(y) =⇒
1


dy = f (x) dx.
 2)
g(y)
II. y 0 + p(x)y = f (x) =⇒ r(x) = e
R
p(x) dx
, y = e−

 y 0 + p(x)y = f (x)
Rx
p(t)
III.
=⇒ r(x) = e x0
 y(x ) = y
0
dt
R
p(x) dx
, y=e
−
Rx
x0
Z
R
e
p(x) dx
p(t) dt
Z
f (x) dx
x Rt
e
x0
p(s) ds
f (t) dt + y0
x0
0
To solve y 0 + p(x)y 0 = f (x):
1) Understand the meaning of the integrating fact r(x), that is,
r(x)f (x) = r(x)[y 0 + p(x)y 0 ] =
d
[r(x)y].
dx
2) Find the integrating factor r(x), that is,
r(x) = e
R
p(x) dx
.
3) Solve the differential equation:
Z
r(x)y =
r(x)f (x) dx.
That is,
y=
1
·
r(x)
Z
r(x)f (x) dx = e−
R
p(x) dx
Z
·
f (x)e
R
1
y 0 + xy = 3, y(0) = 0.
x2 + 1
Rewrite the equations:
Example 1. Solve
y 0 + x(x2 + 1)y = 3(x2 + 1),
1
y(0) = 0.
p(x) dx
dx.
2
MINGFENG ZHAO
Let p(x) = x(x2 + 1) and f (x) = 3(x2 + 1), then
r(x) = e
Since
Rx
0
p(t) dt
=e
Rx
0
t(t2 +1) dt
=e
x4
4
2
+ x2
.
1
d
[r(x)y] = r(x)f (x), so the solution to 2
y 0 + xy = 3, y(0) = 0 is:
dx
x +1
4
y=e
x
Z
2
− x4 − x2
t4
t2
e 4 + 2 3(t2 + 1) dt .
·
0
EXACT EQUATIONS
By the chain rule, we have
d
∂φ
∂φ
dy
φ(x, y) =
(x, y) +
(x, y) .
dx
∂x
∂y
dx
So
φx (x, y) + φy (x, y)
dy
= 0 =⇒ φ(x, y) = C .
dx
That is,
dφ(x, y) :=
∂φ
∂φ
(x, y) dx +
(x, y) dy = 0 =⇒ φ(x, y) = C .
∂x
∂y
If we have the following identity:
dφ(x, y) = M (x, y) dx + N (x, y) dy = 0,
where
M (x, y) =
∂φ
,
∂x
and N (x, y) =
∂φ
.
∂y
In this case, we have
∂2φ
∂N
∂M
=
=
.
∂y
∂x∂y
∂x
Definition 1. Consider the differential equation:
M (x, y) + N (x, y)y 0 = 0.
We say that the differential equation M (x, y) + N (x, y)y 0 = 0 is exact if
∂M
∂N
=
.
∂y
∂x
LECTURE 6: EXACT EQUATIONS AND INTEGRATE FACTORS
3
Example 2. Consider the differential equation:
y 0 = f (x)g(y).
Rewrite the equation:
f (x) −
Let M (x, y) = f (x) and N (x, y) = −
1
y 0 = 0.
g(y)
1
, then
g(y)
My = Nx = 0.
So
f (x) −
1
y 0 = 0 is exact .
g(y)
Question 1. How can we solve an exact equation?
YES, we can solve the exact equation:
1) Since φx (x, y) = M (x, y), we integrate with respect to x, we get
Z
φ(x, y) = M (x, y) dx + f (y).
2) Since φy (x, y) = N (x, y), we integrate with respect to y, we get
Z
φ(x, y) = N (x, y) dy + g(x).
3) Now we only need to determine f (y) or g(x):
Z
– Find f (y): Take the partial derivative with respect to y on both sides of φ(x, y) =
M (x, y) dx + f (y),
we get
Z
My (x, y) dx + f 0 (y).
N (x, y) = φy (x, y) =
So we get
f (y) =
Z Z
N (x, y) − My (x, y) dx dy.
Z
– Find g(y): Take the partial derivative with respect to x on both sides of φ(x, y) =
we get
Z
M (x, y) = φx (x, y) =
Nx (x, y) dx + g 0 (x).
So we get
Z g(x) =
Z
M x, y) −
Nx (x, y) dy
dx.
N (x, y) dy + g(x),
4
MINGFENG ZHAO
In summary, we get
For exact equation: M (x, y) + N (x, y) y 0 = 0 =⇒
Z
M (x, y) dy +
Z Z
N (x, y) − My (x, y) dx dy = C .
To solve the exact differential equation: M (x, y) + N (x, y) y 0 = 0:
1) Since the equation is exact, then there exists some φ(x, y) such that
φx (x, y) = M (x, y),
and φy (x, y) = N (x, y).
2) Since φx (x, y) = M (x, y), then
Z
φ(x, y) =
M (x, y) dx + f (y).
3) Take the partial derivative with respect to y, since φy (x, y) = N (x, y), then
Z
N (x, y) = φy (x, y) =
My (x, y) dx + f 0 (y),
which implies that
Z Z
f (y) =
N (x, y) − My (x, y) dy.
4) So the general solution to M (x, y) + N (x, y) y 0 = 0 is:
Z
Z Z
M (x, y) dx +
N (x, y) − My (x, y) dy = 0.
Example 3. Solve 3y + ex + [3x + cos(y)]y 0 = 0.
Let M (x, y) = 3y + ex and N (x, y) = 3x + cos(y), then
My (x, y) = 3,
and Nx (x, y) = 3.
So [3y + ex ] + [3x + cos(y)] y 0 = 0 is exact. Then we can find some φ(t, y) such that
φx (t, y) = M (x, y),
and φy (x, y) = N (x, y)
Since φx (x, y) = M (t, y) = 3y + ex then
Z
φ(x, y) =
Z
M (t, y) dt + f (y) =
(3y + ex ) dt + f (y) = 3xy + ex + f (y)
Take the partial derivative with respect to y, since φy (x, y) = N (x, y) = 3x + cos(y), then
3x + cos(y) = N (x, y) = φy (x, y) = 3x + f 0 (y).
LECTURE 6: EXACT EQUATIONS AND INTEGRATE FACTORS
5
So f 0 (y) = cos(y), which implies that
f (y) = sin(y).
So the general solution to 3y + ex + [3t + cos(y)]y 0 = 0 is:
3ty + ex y + sin(y) = C .
Example 4. Solve 3x2 y + 8xy 2 + (x3 + 8x2 y + 12y 2 )y 0 = 0, y(2) = 1.
Let M (x, y) = 3x2 y + 8ty 2 and N (x, y) = x3 + 8x2 y + 12y 2 , then
My (x, y) = 3x2 + 16xy,
and Nx (x, y) = 3x2 + 16ty.
So (3x2 y + 8xy 2 ) + (x3 + 8x2 y + 12y 2 ) y 0 = 0 is exact. Hence there exists some φ(x, y) such that
φx (x, y) = M (x, y),
and φy (x, y) = N (x, y)
Since φx (x, y) = M (x, y), then
Z
Z
φ(x, y) = M (x, y) dx + f (y) = (3x2 y + 8xy 2 ) dx + f (y) = x3 y + 4x2 y 2 + f (y).
Take the partial derivative with respect to y, we get
x3 + 8x2 y + 12y 2 = N (x, y) = φy (x, y) = x3 + 8x2 y + f 0 (y)
So f 0 (y) = 12y 2 , which implies that
f (y) = 4y 3 .
Hence
φ(x, y) = x3 + 4x2 y 2 + 4y 3
Therefore, the general solution to 3x2 y + 8xy 2 + (x3 + 8x2 y + 12y 2 )y 0 = 0 is:
x3 + 4x2 y 2 + 4y 3 = C.
Since y(2) = 1, then C = 28. Hence the solution to 3x2 y + 8xy 2 + (x3 + 8x2 y + 12y 2 )y 0 = 0, y(2) = 1 is:
x3 + 4x2 y 2 + 4y 3 = 28 .
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca