LECTURE 4: SEPARABLE EQUATIONS
MINGFENG ZHAO
September 10, 2014
Recall that
Z
I.
y 0 = f (x) =⇒ y =
II.



 1) y(x) ≡ a for some constant a such that f (a) = 0
0
Z
y = f (y) =⇒
1


dy.
 2) x =
f (y)
f (x) dx
Remark 1. “≡” means “y is a constant function”.
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).
Example 1. Draw the slope field of y 0 = xy.
y\x
0
1
2
3
−1 −2
−3
0
0
0
0
0
0
0
1
0
1
2
3
−1 −2
−3
2
0
2
4
6
−2 −4
−6
3
0
3
6
9
−3 −6
−9
−1
0
−1
−2
−3
1
2
3
−2
0
−2
−4
−6
2
4
6
−3
0
−3
−6
−9
3
6
9
1
0
2
MINGFENG ZHAO
Figure 1. Slope field of y 0 = xy
Theorem 1 (Picard’s Theorem on Existence and Uniqueness). If f (x, y) is continuous with respect to x and y, and
∂f
(x, y) exists and is continuous near some (x0 , y0 ), then a solution to
∂y
y 0 = f (x, y),
y(x0 ) = y0 ,
exists for some small interval containing x0 , and is unique.
Example 2. Solve y 0 = (x2 + y)(y − 2), y(1) = 2.
It’s easy to see that y(x) ≡ 2 is a solution to y 0 = (x2 + y)(y − 2), y(1) = 2. Let f (x, y) = (x2 + y)(y − 2), then
• f (x, y) is continuous.
∂f
(x, y) = y − 2 + x2 + y = 2y + x2 − 2 is continuous.
•
∂y
By Picard’s Theorem on Existence and Uniqueness, Theorem 1, then the only solution to y 0 = (x2 + y)(y − 2),
y(1) = 2 is:
y(x) ≡ 2 .
2
2
Problem 1. Solve y 0 = x2 ex − y 2 ex + 1, y(1) = 1.
SEPARABLE EQUATIONS
Question 1. How to solve y 0 = xy?
LECTURE 4: SEPARABLE EQUATIONS
In general, let’s consider
y0 =
dy
= f (x)g(y).
dx
1) If g(a) = 0 for some constant a, then y(x) ≡ a is a solution.
2) If g(y) 6= 0, then
dy
1
·
= f (x).
g(y) dx
Integrate both sides with respect to x, then
Z
Z
dy
1
·
dx = f (x) dx.
g(y) dx
By the change of variables, then
Z
(You can just think the cancellation in
1
dy =
g(y)
Z
f (x) dx.
dy
dy
= dy).
dx, that is,
dx
dx
dx
In summary,



 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)
Z
Remark 2. In the formula
1
dy =
g(y)
Z
f (x) dx, we call y is an implicit solution to y 0 = f (x)g(y).
Example 3. Solve y 0 = xy.
Let f (x) = x and g(y) = y, then
1) y(x) ≡ 0 is a solution.
2) If y 6= 0, then
Z
1
dy =
y
Z
x dx.
That is,
ln |y| =
1 2
x + C.
2
So
y = ±e
x2
2
+C
.
Since C is arbitrary constant, then
y = Ce
x2
2
for some nonzero constant.
3
4
MINGFENG ZHAO
In summary,
The general solution of y 0 = xy is y = Ce
x2
2
.
xy
.
y2 + 1
y
Let f (x) = x and g(y) = 2
, then
y +1
Example 4. Solve y 0 =
1) y(x) ≡ 0 is a solution.
2) If y 6= 0, then
Z
y2 + 1
dy =
y
Z
x dx.
That is,
Z 1
y+
y
Z
dy =
x dx.
So
1
1 2
y + ln |y| = x2 + C.
2
2
xy
0
In summary, the general solution of y = 2
is
y +1
either y ≡ 0, or
1
1 2
y + ln |y| = x2 + C .
2
2
Example 5. 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 1 − x2 + y 2 − x2 y 2 = (1 − x2 ) + y 2 (1 − x2 ) =
(1 − x2 )(1 + y 2 ), then
y0 =
Let f (x) =
(1 − x2 )(1 + y 2 )
.
x2
1 − x2
and g(y) = 1 + y 2 , then
x2
Z
Z Z
1
1
1 − x2
dy =
dx =
− 1 dx.
1 + y2
x2
x2
Hence
arctan y = −
1
− x + C.
x
So the general solution to x2 y 0 = 1 − x2 + y 2 − x2 y 2 is:
1
y = tan − − x + C .
x
Since y(1) = 0, that is, 0 = tan(−1 − 1 + C) = tan(C − 2), then
C = 2 + kπ,
for some integer k.
LECTURE 4: SEPARABLE EQUATIONS
5
Therefore, the solution to x2 y 0 = 1 − x2 + y 2 − x2 y 2 , y(1) = 0 is:
1
y = tan − − x + 2 .
x
Example 6. Find the general solution to y 0 =
x
Let f (x) = − and g(y) = y 2 , then
3
−xy 2
.
3
1) y(x) ≡ 0 is a solution.
2) If y 6= 0, then
Z
−
1
dy =
y2
Z
x
dx.
3
So
x2
1
=
+ C.
y
6
That is,
y=
In summary, the general solution of y 0 =
x2
6
1
6
= 2
.
x + 6C
+C
−xy 2
is
3
either y ≡ 0, or y =
6
.
x2 + C
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