LECTURE 1: INTRODUCTION
MINGFENG ZHAO
September 03, 2014
In Calculus I, given a function y(t), we can compute y 0 , y 00 , y 000 , · · · .
1) y(t) = et , then
y 0 = et =⇒ y 0 = y .
In fact,
y 00 = y 000 = · · · = et = y.
2) y(t) = cos(t), then
y 0 = − sin(t)
y 00 = − cos(t) =⇒ y 0 = −y .
Moreover,
y 000 = sin(t)
y (4) = cos(t) =⇒ y (4) = y.
3) y(t) = sin(t), then
y 0 = cos(t)
y 00 = − sin(t) =⇒ y 0 = −y .
Moreover,
y 000 = − cos(t)
y (4) = sin(t) =⇒ y (4) = y.
4) Let k, C be constants, y(t) = Cekt , then
y 0 = Ckekt = k · Cekt =⇒ y 0 = ky .
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MINGFENG ZHAO
5) Let k, C1 , C2 be constants, y(t) = C1 ekt + C2 e−kt , then
y 0 = C1 kekt − C2 ke−kt
y 00 = C1 k 2 ekt + C2 k 2 e−kt =⇒ y 00 = k 2 y .
6) Let k, C1 , C2 be constants, y(t) = C1 cos(kt) + C2 sin(kt), then
y 0 = −C1 k sin(kt) + C2 k cos(kt)
y 00 = −C1 k 2 cos(kt) − C2 k 2 sin(kt) =⇒ y 00 = −k 2 y .
Definition 1. A differential equation is an equation involving the derivatives of the dependent variable.
Example 1. Some differential equations:
y0 = t
y 0 = 2y + et
y 00 + 2y 0 + 3y = et
dx
+ x = 2 cos(t)
dt
dP
= kP, k is a constant.
dt
Definition 2. If the highest order of derivative appearing in a differential equation is n, then we say that this differential
equations is of nth order.
Example 2.
y0 = t
−→
first order
y 0 = 2y + et
−→
first order
y 00 + 2y 0 + 3y = et
−→
second order
dx
+ x = 2 cos(t)
−→
dt
dP
= kP, k is a constant −→
dt
y (4) = y
−→
first order
first order
forth order.
Definition 3. A function is called a solution to a differential equation, if we get an identity when plugging the function
and its derivatives into the differential equation.
Example 3.
dx
+ x = 2 cos(t).
dt
LECTURE 1: INTRODUCTION
3
Let x(t) = cos(t) + sin(t) + Ce−t , where C is a constant, then
dx
= − sin(t) + cos(t) − Ce−t .
dt
So
LHS
RHS
=
dx
+x
dt
=
− sin(t) + cos(t) − Ce−t + cos(t) + sin(t) + Ce−t
=
2 cos(t)
=
2 cos(t).
Hence
LHS = RHS =⇒ x(t) = cos(t) + sin(t) + Ce−t is a solution to
dx
+ x = 2 cos(t) .
dt
2
dy
Example 4.
= 1 − y2 .
dt
Let y(t) = sin(t), then
dy
= cos(t)
dt
So
LHS
RHS
2
=
dy
dt
=
cos2 (t)
=
1 − y2
=
1 − sin2 (t)
=
cos2 (t)
Hence
LHS = RHS =⇒ y(t) = sin(t) is a solution to
NonExample 1. x000 + 12x00 + 48x0 − 64x = 0.
Let x(t) = et , then
x = x0 = x00 = x000 = et
dy
dt
2
= 1 − y2 .
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MINGFENG ZHAO
So
LHS
= x000 + 12x0 + 48x0 − 64x
= et + 12et + 48et − 64et
= et [1 − 12 + 48 − 64]
= −28et
RHS
=
0.
Hence
LHS 6= RHS =⇒ x(t) = et is NOT a solution to x000 + 12x0 + 48x0 − 64x = 0 .
Usually, there are many solutions to a differential equation, not just one. But if we put some extra “good” conditions,
the solution may be unique. Roughly speaking, for an nth order differential equation, if we give n “good” conditions,
the solution to this differential equation may be unique.
Example 5. Verify that x(t) = Ce−2t is a solution to x0 = −2x. Find C to solve for the initial condition x(0) = 100.
In fact,
x0 = −2Ce−2t =⇒ x0 = −2x =⇒ x(t) = Ce−2t is a solution to x0 = −2x .
If we want x(0) = 100, that is, 100 = x(0) = Ce(−2)·0 = Ce0 = C, then
C = 100 .
In this case, we say that x(t) = 100e−2t is a solution to the following initial value problem:

 x0 = −2x
 x(0) = 100.
In fact, that’s the only solution.
In this ODE course, we are going to learn how to solve a differential equation. The simplest one is the first order
differential equation:
Differential Equation:
dy
= f (x, y)
dx
and


 dy = f (x, y)
dx
The Initial Value Problem:
.

 y(x0 ) = y0 .
LECTURE 1: INTRODUCTION
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INTEGRALS AS SOLUTIONS
I. Let’s consider the differential equation
y 0 = f (x).
Integrate both sides with respect to x, then
Z
Z
Z
0
y (x) dx = f (x) dx =⇒ y = f (x) dx.
II. Let’s consider the initial value problem:


 dy = f (x, y)
dx

 y(x0 ) = y0 .
– First, according to the differential equation, we must have
Z
y = f (x) dx.
– Secondly, by the Fundamental Theorem of Calculus, then
Z
Z x
y = f (x) dx =
f (t) dt + C, for some constant C.
x0
– Finally, by the initial condition y(x0 ) = y0 , we get C = y0 , that is,
Z x
y=
f (t) dt + y0 .
x0
In summary, we have
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 .
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