LECTURE 3: SLOPE FIELDS
MINGFENG ZHAO
September 14, 2015
Separable differential equations
For a separable differential equation:
y0 =
dy
= f (x)g(y).
dx
There are two cases:
Case I: If g(a) = 0 for some constant a, then y(x) ≡ a is a solution.
1
dy
= f (x)g(y), then
dy = f (x) dx, which implies that
Case II: If g(y) 6= 0, since
dx
g(y)
Z
1
dy =
g(y)
Z
f (x) dx.
In summary, we have



 1) y(x) ≡ a for some constant a such that g(a) = 0
0
Z
Z
y = f (x)g(y) =⇒
.
1


dy = f (x) dx.
 2)
g(y)
Example 1. Solve x2 y 0 = 1 − x2 − y 2 + x2 y 2 , y(1) = 0.
Fist, let’s find the general solution to x2 y 0 = 1 − x2 − y 2 + x2 y 2 . Since x2 y 0 = 1 − x2 − y 2 + x2 y 2 , then
y0 =
1 − x2 − y 2 + x2 y 2
.
x2
Since 1 − x2 − y 2 + x2 y 2 = (1 − x2 ) − y 2 (1 − x2 ) = (1 − x2 )(1 − y 2 ), then
(1 − x2 )(1 − y 2 )
dy
1 − x2
= y0 =
=
· (1 − y 2 ).
dx
x2
x2
Case I: If 1 − y 2 = 0, that is, y = ±1. So y(x) = ±1 are two solutions.
1
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MINGFENG ZHAO
dy
1 − x2
1
1 − x2
(1 − x2 )(1 − y 2 )
2
Case II: If 1 − y 2 6= 0, since
=
·
(1
−
y
),
then
dy
=
dx, which implies that
=
x2
x2
1 − y2 x2
Z dx 2
Z
1−x
1
1
1
1
1
1
1
1 − x2
dy =
dx. Since
=− 2
= 2 − 1, then we get
=
−
and
2
2
2
2
1−y
x
1−y
y −1
2 y+1 y−1
x
x
1
1 y + 1 ln = − − x + C.
2
y−1
x
In summary, the general solution to x2 y 0 = 1 − x2 − y 2 + x2 y 2 is:
1 y + 1 1
y(x) = 1, or y(x) = −1, or
ln = − − x + C.
2
y − 1
x
1
1
1 y + 1 = − − x + C, we get 0 = ln 1 = −1 − 1 + C.
Since y(1) = 0, then we only need look at the solution ln 2
y − 1
x
2
So C = 2. Therefore, the solution to x2 y 0 = 1 − x2 − y 2 + x2 y 2 , y(1) = 0 is:
1
1 y + 1 =− −x+2 .
ln 2
y − 1
x
Remark 1. For Example 1, later we are going to see that
1
y+1
1
ln −
=− −x+2 .
2
y−1
x
Direction/slope fields
Definition 1. The direction/slope field of y 0 = f (x, y) is a picture on the xy-plane such that for each point (x, y) on
the plane, one draws a short line segment with the slope f (x, y) at the point (x, y).
To draw the slope field of y 0 = f (x, y):
1) select points in the xy-plane,
2) compute the numbers f (x, y) at the selected points (x, y),
3) at each selected point (x, y), draw a short tangent line whose slope is f (x, y).
LECTURE 3: SLOPE FIELDS
3
Remark 2. For the differential equation y 0 = f (x, y), by the definition of the slope fields, we know that for any y = g(x)
which is a solution to y 0 = f (x, y), then at any point (x0 , y0 ) on the curve y = g(x), the line segment is the tangent line
of the curve y = g(x) at the point (x0 , y0 ).
Example 2. y 0 = 3x2 . (Recall that the general solution is: y = x3 + C.)
Figure 1. Slope field of y 0 = 3x2 with graphs of solutions satisfying y(0) = 0, and y(0) = ±1
Remark 3. The slope field of y 0 = f (x) have the same direction/shape vertically.
Example 3. y 0 = −y. (Recall that the general solution is: y = Ce−x .)
Figure 2. Slope field of y 0 = −y with graphs of solutions satisfying y(0) = ±2, and y(0) = ±3
Remark 4. The slope field of y 0 = f (y) have the same direction/shape horizontally.
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MINGFENG ZHAO
Figure 3. Slope field of y 0 = xy with graphs of solutions satisfying y(0) = ±0.2, and y(0) = ±1
Example 4. y 0 = xy. (In Lecture 2, we knew that the general solution is: y = Ce
x2
2
.)
Remark 5. For the directions of the short tangent line segment:
• “/” means positive slope
• “\” means a negative slope
• “−” means a zero slope
• “|” means infinity slope.
Theorem of the existence and uniqueness for the first order differential equations with
the initial value condition
Question 1. For the initial value problem:

 y 0 = f (x, y)
 f (x ) = y .
0
0
1) Does a solution exist?
2) Is the solution unique if it exists?
The answer to Question 1 really depends on the smoothness of f with respect to x and y.
Example 5 (Non-existence). Can you find a solution to y 0 =
1
for y(0) = 0?
x
LECTURE 3: SLOPE FIELDS
The general solution to y 0 =
1
is:
x
Z
5
1
dx = ln |x| + C.
x
y=
1
for y(0) = 0 exists, say g(x) is a solution. Then g(x) = ln |x| + C for some constant C, but
x
which is not defined at x = 0. Hence
If the solution to y 0 =
The solution to y 0 =
1
for y(0) = 0 does NOT exist.
x
2
Example 6 (Non-uniqueness). y 0 = 3y 3 , y(0) = 0.
1) y1 (x) ≡ 0 is a solution.
2) Let y2 (x) = x3 , then y2 (0) = 0, and
y20 (x) = 3x2 ,
2
2
3y 3 = 3(x3 ) 3 = 3x2 = y20 (x).
In summary,
2
Both y1 and y2 are solutions to y 0 = 3y 3 , y(0) = 0 .
p
Example 7 (Non-uniqueness). y 0 = 2 |y|, y(0) = 0.
1) y1 (x) ≡ 0 is a solution.

 x2 ,
if x ≥ 0
2) Let y2 (x) =
, then y2 (0) = 0 and
 −x2 , if x < 0.

 2x,
y20 =
 −2x,
if x > 0
.
if x < 0.
So
p
y20 = 2 |y2 |,
for all x 6= 0.
On the other hand,
lim
h&0
y2 (h) − y2 (0)
h2
= lim
= lim h = 0
h&0 h
h&0
h
y2 (h) − y2 (0)
−h2
= lim
= lim −h = 0.
h%0
h%0
h%0
h
h
p
p
So y20 (0) exists and y20 (0) = 0 = 2 |y2 (0)|. Hence, y2 is also a solution to y 0 = 2 |y|, y(0) = 0.
lim
In summary,
p
Both y1 and y2 are solutions to y 0 = 2 |y|, y(0) = 0 .
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MINGFENG ZHAO
Theorem 1 (Picard’s Theorem on Existence and Uniqueness). Let (x0 , y0 ) be a fixed point in RN , if f (x, y) is continuous
∂f
with respect to x and y near some (x0 , y0 ), and
(x, y) exists and is continuous with respect to x and y near the point
∂y
(x0 , y0 ), then a solution to

 y 0 = f (x, y),
 y(x ) = y ,
0
0
exists for some small interval containing x0 , and is unique.
In Theorem 1, that solution is called a local solution on that small interval containing x0 . In this course, we always
say that the domain of a function should be an interval.
Problems you can do:
Lebl’s Book [1]: All exercises on Page 21, and read Example 1.2.2 on Page 25.
References
[1] 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