LECTURE 7: EXACT EQUATIONS AND INTEGRATE FACTORS, AND AUTONOMOUS
EQUATIONS
MINGFENG ZHAO
September 23, 2015
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 II: Now we only need to compute f (y):
Z
Take the partial derivative with respect to y on both sides of φ(x, y) =
Z
φy (x, y) =
M (x, y) dx + f (y), we get
My (x, y) dx + f 0 (y).
Z
0
Since φy (x, y) = N (x, y), so we get f (y) = N (x, y) −
My (x, y) dx ( this is a function of y), which implies
that
Z Z
f (y) =
N (x, y) − My (x, y) dx dy.
In summary, we get
0
For exact equation: M (x, y) + N (x, y) y = 0 =⇒
Z
Z Z
M (x, y) dx +
N (x, y) − My (x, y) dx dy = C .
Example 1. Solve 2x + y 2 + 2xyy 0 = 0.
Let M (x, y) = 2x + y 2 and N (x, y) = 2xy, then My (x, y) = 2y and Nx (x, y) = 2y. So My (x, y) = Nx (x, y) = 2y, that
is, 2x + y 2 + 2xyy 0 = 0 is exact. Then there exists some function φ(x, y) such that
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MINGFENG ZHAO
• φx (x, y) = M (x, y) = 2x + y 2 .
• φy (x, y) = N (x, y) = 2xy.
• The solution to 2x + y 2 + 2xyy 0 = 0 is given by φ(x, y) = C.
Since φx (x, y) = 2x + y 2 then
Z
φ(x, y) =
(2x + y 2 ) dx + f (y) = x2 + xy 2 + f (y).
Take the partial derivative with respect to y on the both sides of φ(x, y) = x2 + xy 2 + f (y), then
φy (x, y) = 2xy + f 0 (y).
Since φy (x, y) = 2xy, then 2xy = 2xy + f 0 (y). So f 0 (y) = 0, which implies that f (y) = 0. So we know that
φ(x, y) = x2 + xy 2 .
Hence the general solution to 2x + y 2 + 2xyy 0 = 0 is:
x2 + xy 2 = C .
Use the integrating factor r(x) to solve general equations
In general, M (x, y) + N (x, y)y 0 = 0 is not exact, that is, My (x, y) 6= Nx (x, y). Multiply r(x) on the both sides of
M (x, y) + N (x, y)y 0 = 0, then
r(x)M (x, y) + r(x)N (x, y)y 0 = 0.
We are looking for an integrating factor r(x), that is, r(x)M (x, y) + r(x)N (x, y)y 0 = 0 is exact, then
∂
∂
[r(x)M (x, y)] =
[r(x)N (x, y)].
∂y
∂x
So we get r(x)My (x, y) = r0 (x)N (x, y) + r(x)Nx (x, y), that is,
r0 =
In this case, we need
then we can solve it.
My (x, y) − Nx (x, y)
· r.
N (x, y)
My (x, y) − Nx (x, y)
is a function of only x. So since r(x)M (x, y) + r(x)N (x, y)y 0 = 0 is exact,
N (x, y)
LECTURE 7: EXACT EQUATIONS AND INTEGRATE FACTORS, AND AUTONOMOUS EQUATIONS
Example 2. Solve y + (x2 y − x)y 0 = 0, x > 0.
Let M (x, y) = y and N (x, y) = x2 y − x, then
and Nx = 2xy − 1.
My = 1,
So y + (x2 y − x)y 0 = 0 is not exact. Multiply r(x) on the both sides of y + (x2 y − x)y 0 = 0, then
r(x)y + r(x)(x2 y − x)y 0 = 0.
We are looking for an integrating factor r(x), that is, r(x)y + r(x)(x2 y − x)y 0 = 0 is exact, then
∂
∂
[r(x)y] =
[r(x)(x2 y − x)].
∂y
∂x
So we get
r(x) = r0 (x)(x2 y − x) + r(x)(2xy − 1).
So we have
r(x)[2 − 2xy] = r0 (x)(x2 y − x) = r0 (x)x(xy − 1).
Hence 2r = −xr0 , that is, r0 = − x2 r, which implies that
r(x) = e−
That is, the equation
R
2
x
dx
= e−2 ln x =
1
x2
y
x2 y − x 0
+
y = 0 is exact. So there exist some φ(x, y) such that
x2
x2
y
.
x2
2
x y−x
1
• φy (x, y) =
=y− .
2
x
x
• the solution to the differential equation is φ(x, y) = C.
• φx (x, y) =
Since φx (x, y) =
y
, then
x2
Z
φ(x, y) =
y
y
dx + f (y) = − + f (y).
x2
x
Take the partial derivative with respect to y on the both sides of the above identity, then
φy (x, y) = −
Since φy (x, y) = y −
1
+ f 0 (y).
x
1
1
1
y2
, then y − = − + f 0 (y). So f 0 (y) = y. We can take f (y) = . So we get
x
x
x
2
φ(x, y) = −
y
y2
+ .
x
2
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MINGFENG ZHAO
So the solution is:
−
y
y2
+
=C .
x
2
Phase diagram of the autonomous equation
An autonomous equation is of the form:
dy
= f (y).
dx
Then
Case I: y(x) ≡ a for some constant a such that f (a) = 0. In this case, y(x) ≡ a is called an equilibrium solution, and a
is called a critical point of y 0 = f (y).
Case II: If f (y) 6= 0, then
Z
1
dy = x.
f (y)
For the slope field of an autonomous equation, the slopes are the same as long as they have same y-value. That is to
say, if we know the slope field along a single vertical line(such single vertical line is called phase line/diagram/portrait),
then we know the slope field in the entire xy-plane. On this vertical line, we mark all critical points, and then
we draw arrows between critical points: If f (y) > 0, we draw an up arrow “↑”; If f (y) < 0, we draw a
down arrow “↓”.
Theorem 1. Let f be differentiable, f 0 is continuous, and y be a solution to y 0 = f (y), then either y 0 (x) = 0 for all x,
or, y 0 (x) > 0 for all x, or y 0 (x) < 0 for all x.
Example 3. Find equilibrium points of y 0 = y(1 − y), draw the phase diagram, and describe the solutions y1 , y2 , y3 ,
y4 satisfy the following initial conditions:
1) y1 (0) = 0,
2) y2 (0) = −1,
3) y3 (0) = 2,
4) y4 (0) =
1
.
2
Let f (y) = y(1 − y). Solve y(1 − y) = 0, then y = 0 or y = 1, that is, all equilibrium points of y 0 = y(1 − y) are:
0
and
1.
For the initial value problems:
1) If y1 (0) = 0, since f (0) = 0(1 − 0) = 0, then y1 is a constant function, that is,
y1 (x) ≡ 0.
LECTURE 7: EXACT EQUATIONS AND INTEGRATE FACTORS, AND AUTONOMOUS EQUATIONS
2) If y2 (0) = −1, since f (−1) = (−1)[1 − (−1)] = −2 < 0, then
y2 (x) is strictly decreasing .
3) If y3 (0) = 2, since f (2) = 2(1 − 2) = −2 < 0, then
y3 (x) is strictly decreasing.
4) If y4 (0) = 12 , since f ( 12 ) = 12 (1 − 12 ) =
1
4
> 0, then
y4 (x) is strictly increasing.
Figure 1. Phase diagram of y 0 = y(1 − y)
Figure 2. Sketches of yi (x), i = 1, 2, 3, 4
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MINGFENG ZHAO
Remark 1. In general, for any initial day y(x0 ) = y0 if x0 is between two critical points, then the solution must be the
whole real line, like the case y4 (0) = 1/2 in Example 3.
If x0 is not between two critical points, for example, consider the problem: y 0 = y 2 , y(0) = 1. It’s easy to see that
1
the solution is: y(x) =
. So the solution interval is (−∞, 1). However, we see that y(x) → ∞ as x → 1− .
1−x
Problems you can do:
Lebl’s Book [2]: All exercises except the part of classification of critical points on Page 40 and Page 41.
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.
[2] Jiri Lebl. Notes on Diffy Qs: Differential Equations for Engineers. Createspace, 2014.
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