LECTURE 4: FIRST ORDER LINEAR DIFFERENTIAL EQUATIONS AND THE
INTEGRATING FACTORS
MINGFENG ZHAO
September 16, 2015
Theorem of the existence and uniqueness for the first order differential equations with
the initial value condition
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
(x, y) exists and is continuous with respect to x and y near the point
with respect to x and y near some (x0 , y0 ), and
∂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.
Remark 1. For the uniqueness part in Theorem 1, if we have two solution curves y1 (x) and y2 (x) to the differential
equation y 0 = f (x, y), if these two curves y1 (x) and y2 (x) meet at one point, then y1 (x) = y2 (x) for all x belonging to
the domain.
Figure 1. This can NOT happen
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2
MINGFENG ZHAO
Example 1. Let f be differentiable and f 0 be continuous. Assume f (a) = 0 and y(x) satisfies y 0 = f (y) and y(x0 ) = a.
Then y(x) ≡ a .
p
Example 2. Is it possible to solve the equation y 0 = y |x| for y(0) = 0? Is the solution unique?
p
p
It’s easy to see that y(x) ≡ 0 is a solution to y 0 = y |x| for y(0) = 0. Let f (x, y) = y |x|, then f (x, y) is continuous
with respect to x and y, and
p
∂f
(x, y) = |x| is continuous near (0, 0).
∂y
By Theorem 1, we know that
y(x) ≡ 0 is the only solution to y 0 = y
p
|x| for y(0) = 0 .
Example 3. Let A be a constant, solve y 0 = y 2 , y(0) = A.
Let f (x, y) = y 2 , then fy (x, y) = 2y and f (x, y) = y 2 are continuous, which implies that we can apply Theorem 1.
Since y(x) = 0 is a solution to y 0 = y 2 , then
• If A = 0, by Theorem 1, then y(x) = 0 for all x.
• If A > 0, by Theorem 1, then y(x) > 0 for all x.
• If A < 0, by Theorem 1, then y(x) < 0 for all x.
Since the general solution to y 0 = y 2 is either y(x) ≡ 0, or y = −
1
.
x+C
Case I: If A = 0, then y(x) ≡ 0.
1
1
, that is, C = − . Hence y =
C
A
For the domain of the solution:
1
– If A > 0, then x < .
A
1
– If A < 0, then x > .
A
Case II: If A 6= 0, then A = y(0) = −
1
A
1
.
−x
In summary,



y(x) ≡ 0,
if A = 0







1
1
y(x) = 1
, for all x < , if A > 0
A
−
x

A





1
1


, for all x > , if A < 0.
 y(x) = 1
A
A −x
LECTURE 4: FIRST ORDER LINEAR DIFFERENTIAL EQUATIONS AND THE INTEGRATING FACTORS
3
Example 4. For the problem x2 y 0 = 1 − x2 − y 2 + x2 y 2 , y(1) = 0.
Since x2 y 0 = 1 − x2 − y 2 + x2 y 2 , then
y0 =
1 − x2
1 − x2 − y 2 + x2 y 2
=
· (1 − y 2 ).
2
x
x2
1 − x2
· (1 − y 2 ), it’s easy to know that f (x, y) and fy (x, y) are continuous near (1, 0), so we can apply
x2
1 − x2
1 − x2 − y 2 + x2 y 2
Theorem 1. It’s easy to see that y(x) = −1 and y(x) = 1 are two solutions to y 0 =
=
·(1−y 2 ).
2
x
x2
Since y(1) = 0, by Theorem 1, we know that −1 < y(x) < 1 for all x. Otherwise, we say y(x0 ) ≥ 1 for some x0 , since
Let f (x, y) =
y(1) = 0, by the continuity, we know that there exists some x0 ∈ (1, x0 ) such that y(x0 ) = 1, which implies that y(x) is a
1 − x2
· (1 − y 2 ) and y(x0 ) = 1, by Theorem 1, we know that y(x) = 1, which contradicts
solution to the problem y 0 =
x2
with y(1) = 0. So we must have y(x) < 1. Similar argument, we know that y(x) > −1.
Example 5 (Newton’s Law of Cooling). Bob made a cup of coffee, and Bob likes to drink coffee only once it will not
burn him at 60 degrees. Initially at time t = 0 minutes, Bob measured the temperature and the coffee was 89 degrees
Celsius. One minute later, Bob measured the coffee again and it had 85 degrees. The temperature of the room (the
ambient temperature) is 22 degrees. When should Bob start drinking?
Newton’s Law of Cooling:
The rate of changing of the temperature of the object is proportional to the difference
between the ambient temperature and the temperature of the object.
That is,
The rate of changing of the temperature of the object
= constant.
The ambient temperature − The temperature of the object
Let T (t) be the temperature, A(t) be the ambient temperature, where t is time in minute, then
T 0 (t) = k[A(t) − T (t)],
for some constant k.
By the assumption, A = 22, then we have
T 0 = k(22 − T ),
T (0) = 89
and T (1) = 85.
Since the temperatures are changing, then k 6= 0. First, let’s find the general solution to T 0 = k(22 − T ):
1) T (t) ≡ 22 is a solution. (In our initial value problem, we can ignore this case, because the temperature T (t) is
changing.)
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MINGFENG ZHAO
dT
1
= k(22 − T ), then
dT = k dt, which implies that − ln |T − 22| = kt + C. So
dt
22 − T
|T − 22| = e−kt−C , that is, T = 22 ± e−C · e−kt . Hence we have T (t) = 22 + Cekt for some non-zero constant C.
2) If T 6= 22, since
In summary, the general solution of T 0 = k(22 − T ) is
T (t) = 22 + Ce−kt .
Since T (0) = 89 and T (1) = 85, that is,
T (0) = 22 + C = 89
So C = 89 − 22 = 67 and e−k =
and T (1) = 22 + Ce−k = 85.
85 − 22
63
63
63
=
=
, then k = − ln
> 0. Hence
C
67
67
67
63
T (t) = 22 + 67et ln 67 .
Figure 2. Temperature decreases in 90 minutes
63
63
Let’s solve 60 = T (t) = 22 + 67et ln 67 , we get et ln 67 =
ln 38
38
60 − 22
67
=
. Hence t =
≈ 9.212 .
67
67
ln 63
67
Remark 2. In general, the ambient temperature A(t) is a function of t, in the following, we will study how to solve
T 0 (t) = k[A(t) − T (t)].
First order linear differential equations and the integrating factors
A first order linear differential equation is of the form:
y 0 + p(x)y = f (x),
LECTURE 4: FIRST ORDER LINEAR DIFFERENTIAL EQUATIONS AND THE INTEGRATING FACTORS
5
where f (x) is called the non-homogeneous term. If f (x) = 0, then y 0 + p(x)y = 0 is said to be homogeneous.
Let’s solve y 0 + p(x)y = f (x), both sides multiply some function r(x) , we get
r(x)y 0 (x) + r(x)p(x)y(x) = r(x)f (x).
For the left hand side, if we can find some r(x) (which is called an integrating factor) such that
r(x)y 0 (x) + r(x)p(x)y(x) =
d
[r(x)y(x)] .
dx
If we can find such r(x), let v(x) = r(x)y(x), then v 0 = r(x)f (x). So
Z
r(x)y(x) = v = r(x)f (x) dx.
Hence
1
·
y(x) =
r(x)
Z
r(x)f (x) dx.
Question 1. Can we find a r(x) such that
r(x)y 0 (x) + r(x)p(x)y(x) =
d
[r(x)y(x)]?
dx
By the product rule, we get
r(x)y 0 (x) + r(x)p(x)y(x) = r0 (x)y(x) + r(x)y 0 (x).
So r(x)p(x)y(x) = r0 y(x), that is, r0 (x) = p(x)r(x). Then
Z
Z
1
dr = p(x) dx.
r
R
R
R
So ln |r| = p(x) dx, that is, r(x) = e p(x) dx . Hence, r(x) = e p(x)
r(x)y 0 (x) + r(x)p(x)y(x) =
dx
satisfies our need, that is,
d
[r(x)y(x)].
dx
Therefore,
0
y + p(x)y = f (x) =⇒ y = e
−
R
p(x) dx
Z
e
R
p(x) dx
f (x) dx.
In summary,
R
y 0 + p(x)y = f (x) =⇒ r(x) = e
Remark 3. Since r(x) = e
R
p(x) dx
p(x) dx
, y = e−
R
p(x) dx
Z
e
R
p(x) dx
f (x) dx .
, so there are infinitely many integrating factors for the differential equation y 0 +
p(x)y = f (x). But when we find solutions to y 0 + p(x)y = f (x), we only need to find one r(x).
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MINGFENG ZHAO
Example 6. Recall that the Newton’s law of cooling gives the following differential equation:
T 0 = k[A(t) − T ].
That is, T 0 + kT = kA(t). Multiply r(t) on the both sides, we get r(t)T 0 + kr(t)T = kr(t)A(t). We are looking for
an integrating factor r(t), that is, r(t)T 0 + kr(t)T = [r(t)T ]0 = r(t)T 0 + r0 (t)T , then r0 (t) = kr(t), which implies that
R
r(t) = ekt . So we get [ekt T ]0 = ekt kA(t). Hence we have ekt T = ekt kA(t) dt, that is,
Z
T (t) = ke−kt A(t)ekt dt .
2
Example 7. Solve y 0 + 2xy = ex−x , y(0) = −1.
2
2
Multiply r(x) on the both sides of y 0 + 2xy = ex−x , we get r(x)y 0 + 2xr(x)y = r(x)ex−x . We are looking for an
integrating factor r(x), that is, r(x)y 0 + 2xr(x)y = [r(x)y]0 = r(x)y 0 + r0 (x)y, then r0 (x) = 2xr(x), which implies that
Z
2
2
2
2
2
r(x) = ex . So we get [ex y]0 = ex ex−x = ex . Hence we have ex y = ex dx = ex + C, that is,
2
2
y = ex−x + Ce−x .
Since y(0) = −1, then −1 = 1 + C, that is, C = −2. So
2
2
2
y = ex−x − 2e−x is the solution to y 0 + 2xy = ex−x , y(0) = −1 .
Problems you can do:
Lebl’s Book [2]: All exercises on Page 30 and Page 31.
Braun’s Book [1]: Exercises 4-15 on Page 80, and 17-19 on Page 81, and all Exercises from Page 8 to Page 10.
Read all materials in Section 1.2.
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