LECTURE 6: EXACT EQUATIONS AND INTEGRATE FACTORS
MINGFENG ZHAO
September 21, 2015
Exact equations
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 My (x, y) = Nx (x, y).
Remark 1. If the differential equation M (x, y) + N (x, y)y 0 = 0 is exact, that is, My (x, y) = Nx (x, y), then there exists
a function φ(x, y) such that
• φx (x, y) = M (x, y).
• φy (x, y) = N (x, y).
• The solution to M (x, y) + N (x, y)y 0 = 0 is given by φ(x, y) = C .
To solve an exact equation:
Let M (x, y) + N (x, y)y 0 = 0 be an exact equation, that is, My (x, y) = Nx (x, y), then we can find φ(x, y) such that
• φx (x, y) = M (x, y).
• φy (x, y) = N (x, y).
• The solution to M (x, y) + N (x, y)y 0 = 0 is given by φ(x, y) = C.
Now let’s find φ(x, y):
Step I: Since φx (x, y) = M (x, y), for any fixed y, we integrate with respect to x, we get
Z
φ(x, y) =
M (x, y) dx + f (y),
for some f (y).
Step I’: Since φy (x, y) = N (x, y), for any fixed x, we integrate with respect to y, we get
Z
φ(x, y) =
N (x, y) dy + g(x),
Step II: Now we only need to compute Either f (y) OR g(x):
1
for some g(x).
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MINGFENG ZHAO
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
φy (x, y) =
My (x, y) dx + f 0 (y).
Since φy (x, y) = N (x, y), so we get f 0 (y) = N (x, y) −
Z
My (x, y) dx ( this is a function of y), which implies
that
f (y) =
Z Z
N (x, y) − My (x, y) dx dy.
Z
– Find g(x): Take the partial derivative with respect to x on both sides of φ(x, y) =
N (x, y) dy + g(x),
we get
Z
φx (x, y) =
Nx (x, y) dy + g 0 (x).
Since φx (x, y) = M (x, y), then we get g 0 (x) = M (x, y) −
Z
Nx (x, y) dy (this is a function of x), which
implies that
Z Z
g(x) =
M x, y) − Nx (x, y) dy dx.
In summary, we get
Z
0
For exact equation: M (x, y) + N (x, y) y = 0 =⇒
Z Z
M (x, y) dx +
N (x, y) − My (x, y) dx dy = C .
Example 1. 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 My (x, y) = Nx (x, y) = 3,
that is, [3y + ex ] + [3x + cos(y)] y 0 = 0 is exact. Then there exists some function φ(x, y) such that
• φx (x, y) = M (x, y) = 3y + ex .
• φy (x, y) = N (x, y) = 3x + cos(y).
• The solution to M (x, y) + N (x, y)y 0 = 0 is given by φ(x, y) = C.
Since φx (x, y) = 3y + ex then
Z
φ(x, y) =
(3y + ex ) dx + f (y) = 3xy + ex + f (y)
Take the partial derivative with respect to y on the both sides of φ(x, y) = 3xy + ex + f (y), then
φy (x, y) = 3y + f 0 (y).
LECTURE 6: EXACT EQUATIONS AND INTEGRATE FACTORS
3
Since φy (x, y) = 3x + cos(y), then 3x + cos(y) = N (x, y) = φy (x, y) = 3x + f 0 (y). So f 0 (y) = cos(y), which implies
that f (y) = sin(y). So we know that
φ(x, y) = 3xy + ex + sin(y).
Hence the general solution to 3y + ex + [3x + cos(y)]y 0 = 0 is:
3xy + ex y + sin(y) = C .
Example 2. Solve 3x2 y + 8xy 2 + (x3 + 8x2 y + 12y 2 )y 0 = 0, y(2) = 1.
Let M (x, y) = 3x2 y + 8xy 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 function φ(x, y) such that
• φx (x, y) = M (x, y) = 3x2 y + 8xy 2 .
• φy (x, y) = N (x, y) = x3 + 8x2 y + 12y 2 .
• The solution to M (x, y) + N (x, y)y 0 = 0 is given by φ(x, y) = C.
Since φx (x, y) = 3x2 y + 8xy 2 , then
Z
φ(x, 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
φy (x, y) = x3 + 8x2 y + f 0 (y).
Since φy (x, y) = x3 + 8x2 y + 12y 2 , then
x3 + 8x2 y + 12y 2 = 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.
4
MINGFENG ZHAO
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 .
Example 3. Consider the separable differential equation y 0 = f (x)g(y) with g(y) 6= 0.
1
1
Rewrite the differential equation, we get f (x) −
y 0 = 0. Let M (x, y) = f (x) and N (x, y) = −
, then
g(y)
g(y)
1
My (x, y) = Nx (x, y) = 0. So f (x) −
y 0 = 0 is exact. Then there exists some φ(x, y) such that
g(y)
• φx (x, y) = M (x, y) = f (x).
1
• φy (x, y) = N (x, y) = −
.
g(y)
• The solution to M (x, y) + N (x, y)y 0 = 0 is given by φ(x, y) = C.
Since φx (x, y) = f (x), then φ(x, y) =
R
f (x) dx + F (y). Take the partial derivative with respect to y, we get
φy (x, y) = F 0 (y).
Since φy (x, y) = −
1
1
, then F 0 (y) = −
, which implies that F (y) = −
g(y)
g(y)
Z
φ(x, y) =
Z
f (x) dx −
Z
1
dy. Hence we get
g(y)
1
dy.
g(y)
Therefore, the solution to y 0 = f (x)g(y) is:
Z
Z
f (x) dx −
1
dy = C .
g(y)
Question 1. How can we solve for the general differential equation M (x, y) + N (x, y) y 0 = 0 which may not be exact?
In general, consider differential equation M (x, y) + N (x, y) y 0 = 0. Multiply some function r(x, y) on both sides of
M (x, y) + N (x, y)y 0 = 0, we get
[rM ] + [rN ] y 0 = 0.
We are looking for a special function r(x, y) such that the above equation is exact (in this case, we call r(x, y) an
integrating factor), that is,
(1)
∂(rN )
∂(rM )
=
.
∂y
∂x
Now [rM ] + [rN ]y 0 = 0 is exact, we can solve it now.
LECTURE 6: EXACT EQUATIONS AND INTEGRATE FACTORS
5
Remark 2. For (1), by the product rule, we get
rMy + M ry = rNx + N rx .
We have the following two special cases:
I. If the integrating factor r is a function of x, that is, ry ≡ 0, then rMy = rNx + N rx , that is, r(x) satisfies
R My −Nx
My − Nx
My − Nx
dx
N
rx = r ·
. In this case, we only need
is a function of only x. Hence r(x) = e
.
N
N
II. If the integrating factor r is a function of y, that is, rx ≡ 0, then rMy + M ry = rNx . That is, r(y) satisfies
R Nx −My
Nx − M y
Nx − My
dy
M
. In this case, we only need
is a function of only y. So we get r(y) = e
ry = µ ·
.
M
M
Remark 3. In the exams for this course, we
Example 4. Solve
Let M (x, y) =
that is,
only consider the case that r is a function of only x.
y2
+ 2yex + (y + ex )y 0 = 0.
2
y2
+2yex and N (x, y) = y +ex , then My (x, y) = y +2ex and Nx (x, y) = ex . Then My (x, y) 6= Nx (x, y),
2
y2
+ 2yex + (y + ex )y 0 = 0 is not exact. Multiply r(x) on the both sides of the equation, we get
2
2
y
r(x)
+ 2yex + r(x)(y + ex )y 0 = 0
2
We are looking for an integrating factor r(x), that is,
2
y
∂
∂
r(x)
+ 2yex
=
[r(x)(y + ex )] .
∂y
2
∂x
So we get
r(x)(y + 2ex ) = r0 (x)(y + ex ) + r(x)ex .
So we have r0 (x) = r(x), which implies that r(x) = ex . Hence, we get an exact equation:
2
y
ex
+ 2yex + ex (y + ex ) y 0 = 0.
2
Then there exists some φ(x, y) such that
2
y
x
x
• φx (x, y) = e
+ 2ye .
2
• φy (x, y) = ex (y + ex ).
y2
• The solution to
+ 2yex + (y + ex )y 0 = 0 is φ(x, y) = C.
2
2
y
Since φx (x, y) = ex
+ 2yex , then
2
2
Z
y
y2
φ(x, y) = ex
+ 2yex dx + f (y) = ex + ye2x + f (y).
2
2
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MINGFENG ZHAO
Take the partial derivative with respect to y, since φy (x, y) = ex (y + ex ), then we get
ex (y + ex ) = φy (x, y) = yex + e2x + f 0 (y).
Then f 0 (y) = 0. So we can take f (y) = 0, which implies that
φ(x, y) =
Hence the solution to
y2 x
e + ye2x .
2
y2
+ 2yex + (y + ex )y 0 = 0 is:
2
y2 x
e + e2x y = C .
2
Example 5. Solve y 0 + p(x)y = f (x).
Rewrite the equation, we have [p(x)y − f (x)] + y 0 = 0. Let M (x, y) = p(x)y − f (x) and N (x, y) = 1, then My (x, y) =
p(x) and Nx (x, y) = 0. So My (x, y) 6= Nx (x, y), that is, [p(x)y − f (x)] + y 0 = 0 is not exact. Multiply r(x) on the both
sides of [p(x)y − f (x)] + y 0 = 0, we have
r(x)[p(x)y − f (x)] + r(x)y 0 = 0.
We are looking for an integrating factor r(x), that is,
∂
∂
[r(x)[p(x)y − f (x)]] =
[r(x)].
∂y
∂x
R
So we get r(x)p(x) = r0 (x), then r(x) = e
e
R
p(x) dx
p(x) dx
, which implies that we get an exact equation:
[p(x)y − f (x)] + e
R
p(x) dx
y 0 = 0.
Then there exists some φ(x, y) such that
R
• φx (x, y) = r(x)[p(x)y − f (x)] = [p(x)y − f (x)]e
R
• φy (x, y) = r(x) = e
p(x) dx
p(x) dx
.
.
0
• The solution to y + p(x)y = f (x) is φ(x, y) = C.
Since φy (x, y) = e
R
p(x) dx
, then
R
φ(x, y) = ye
p(x) dx
+ g(x)
for some g(x).
Take the partial derivative with respect to y, we get
R
φx (x, y) = ye
Since φx (x, y) = r(x)[p(x)y − f (x)] = [p(x)y − f (x)]e
[p(x)y − f (x)]e
R
p(x) dx
R
p(x) dx
· p(x) + g 0 (x).
p(x) dx
R
= ye
, then
p(x) dx
· p(x) + g 0 (x),
LECTURE 6: EXACT EQUATIONS AND INTEGRATE FACTORS
that is, g 0 (x) = −f (x)e
R
p(x) dx
Z
. So we get g(x) = −
R
φ(x, y) = ye
f (x)e
p(x) dx
R
p(x) dx
Z
−
7
, which implies that
R
f (x)e
p(x) dx
.
Hence, the general solution to y 0 + p(x)y = f (x) is:
y = e−
R
p(x) dx
Z
·
f (x)e
R
p(x) dx
dx .
Problems you can do:
Braun’s Book [1]: All Exercises on Page 66 and Page 67. Read all materials in Section 1.9.
References
[1] Martin Braun. Differential Equations and Their Applications: An Introduction to Applied Mathematics. Springer, 1992.
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