LECTURE 2: INTEGRALS AS SOLUTIONS
MINGFENG ZHAO
September 05, 2014
For the Exercise 0.2.11, you need to know the following formulas:
y 0 = ky
=⇒ y = Cekx
y 00 = k 2 y
=⇒ y = C1 ekx + C2 e−kx
y 00 = −k 2 y
=⇒ y = C1 cos(kx) + C2 sin(kx).
The first formula, we will learn it in this lecture. The last two formulas will be given in the Chapter 2.
Recall that in the last lecture, we know
y 0 = f (x) =⇒ y =
Z
f (x) dx
and


Z x
 dy = f (x, y)
dx
=⇒ y =
f (t) dt + y0 .

x0
 y(x0 ) = y0 .
Example 1. Find the general solution(i.e., all solution) of y 0 = 3x2 .
Z
y=
Example 2. Find the general solution of y 0 =
3x2 dx = x3 + C .
1
.
x2 − 1
Z
y=
x2
1
dx.
−1
Take the partial fraction decomposition:
1
1
=
2
x −1
2
1
1
−
x−1 x+1
.
Then
Z
y
1
dx
x2 − 1
Z
1
1
1
=
−
dx
2 x−1 x+1
=
1
2
MINGFENG ZHAO
=
=
So the general solution of y 0 =
x2
1
(ln |x − 1| − ln |x + 1|) + C
2
1 x − 1 + C.
ln 2
x + 1
1
is
−1
1 x − 1 y = ln +C .
2
x + 1
2
Example 3. Solve y 0 = e−x , y(0) = 1.
x
Z
2
e−t dt + 1 .
y=
0
Example 4. Solve
dx
= sin(t) + t, x(0) = 20.
dt
Z
x=
0
t
1
1
t
[sin(y) + y] dy + 20 = − cos(y)|0 + t2 + 20 = − cos(t) + t2 + 21 .
2
2
Example 5. Find the general solution of y 00 = x.
Let v(x) = y 0 (x), then v 0 = x. So
Z
v=
x dx =
1 2
x + C1 .
2
That is,
y0 =
1 2
x + C1 .
2
So
y=
1 3
x + C1 x + C2 .
6
In general, for an nth order differential equation, the general solution to this differential equation will have n free
constants.
Example 6. Solve y 00 = sin(x) for y(0) = 0, y 0 (0) = 2.
Let v(x) = y 0 (x), then v 0 = sin(x) and v(0) = y 0 (0) = 2. So
Z x
x
v=
sin(t) dt + 2 = − cos(t)|0 + 2 = − cos(x) + 1 + 2 = 3 − cos(x).
0
That is,
y 0 = 3 − cos(x).
LECTURE 2: INTEGRALS AS SOLUTIONS
3
Since y(0) = 0, then
x
Z
[3 − cos(t)] dt = 3x − sin(x) .
y=
0
t
Example 7. Suppose a car drives at a speed e 2 m/s, where t is time in a second. How far did the car get in 2 seconds
(starting at t = 0)? How far in 10 seconds?
Let x(t) be the distance traveled in t seconds, then
speed = derivative of distance = x0 .
By the assumption,
t
x0 = e 2
and x(0) = 0.
Then
Z
t
y
t
e 2 dy = 2e 2 − 2.
x=
0
So
x(2) = 2e − 2 ≈ 3.436
and
x(10) = 2e5 − 2 ≈ 294.826 .
Example 8. Suppose that the car accelerates at a rate of t2 m/s2 . At time t = 0, the car is at the 1 meter mark, and
is traveling at 10 m/s. Where is the car at time t = 10?
Let x(t) be the mark at time t, then
acceleration = second derivative of distance = x00
By the assumption,
x00 = t2 ,
x(0) = 1
x0 (0) = 10.
and
Let v(t) = x0 (t), then
v 0 = t2
and v(0) = x0 (0) = 10.
So
Z
v=
t
y 2 dy + 10 =
0
1 3
t + 10.
3
That is,
x0 =
1 3
t + 10
3
and
x(0) = 1.
So
Z t
x=
0
1 3
y + 10
3
dy + 1 =
1 4
t + 10t + 1.
12
4
MINGFENG ZHAO
Hence
x(10) =
1
2803
· 104 + 10 · 10 + 1 =
≈ 934.333 .
12
3
III. Let’s consider the differential equation
y0 =
dy
= f (y)
dx
−→ autonomous equation.
1) If f (a) = 0 for some constant a, then y(x) ≡ a is a solution. In this case, a is called an equilibrium point.
2) If f (y) 6= 0, by the Inverse Function Theorem from Calculus, switch the roles of x and y, we get
dx
1
=
.
dy
f (y)
So
Z
x=
1
dy.
f (y)
In summary,



 1) y(x) ≡ a for some constant a such that f (a) = 0
0
Z
y = f (y) =⇒
.
1


dy.
 2) x =
f (y)
In Lecture 3, we will know that if f is a “nice” function, y is a solution of y 0 = f (y) with y(x0 ) = a and f (a) = 0,
then y(x) ≡ a.
Example 9. Solve y 0 = ky, where k is a nonzero constant
1) y(x) ≡ 0 is a solution.
2) If y 6= 0, then
Z
x=
1
1
dy = ln |y| + C.
ky
k
So
|y| = ek(x−C) = e−kC · ekx =⇒ y = ±e−kC · ekx .
Since C is arbitrary constant, then ±e−kC can take any nonzero constant. So
y = Cekx ,
for some nonzero constant C.
LECTURE 2: INTEGRALS AS SOLUTIONS
5
In summary, the general solution of y 0 = ky is
y = Cekx .
Example 10. Find the general solution of y 0 = y 2 .
1) y ≡ 0 is a solution.
2) If y 6= 0, then
Z
x=
1
1
1
dy = − + C =⇒ y =
.
y2
y
C −x
So the general solution of y 0 = y 2 is
either y = 0, or y =
1
.
C −x
Example 11 (Unlimited Population/Exponential Growth Model). Suppose there are 100 bacteria at time 0 and 200
bacteria 10 seconds later. How many bacteria will there will be 1 minute from time 0?
ASSUMPTION for the unlimited population/exponential growth model:
The rate of growth of the population is proportional to the size of the population.
That is,
The rate of growth of the population
= constant.
The size of the population
Let P (t) be the population of bacteria in t seconds, then
dP
= kP,
dt
for some constant k.
By the assumption, we have
P 0 = kP,
P (0) = 100
and P (10) = 200.
We have known that the general solution of P 0 = kP is
P (t) = Cekt .
Since P (0) = 100 and P (10) = 200, that is,
P (0) = C = 100
and P (10) = Ce10k = 100e10k = 200.
So e10k = 2, then 10k = ln 2, that is,
k=
ln 2
.
10
6
MINGFENG ZHAO
Hence
ln 2
P (t) = 100e 10 t .
So
P (60) = 100e6 ln 2 = 100 · 26 = 6400 .
Figure 1. Bacteria growth in the first 60 seconds
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