differential equations - math.uh.eduetgen/ch1-slides-sum20-notes.pdfdifferential equations 1. basic...
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DIFFERENTIAL
EQUATIONS
1. Basic Terminology
A differential equation is an equation
that contains an unknown function to-
gether with one or more of its deriva-
tives.
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Examples:
1. y′ = 2x + cos x
2.dy
dt= ky (exponential growth/decay)
3. x2y′′ − 2xy′ + 2y = 4x3
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4.∂2u
∂x2+
∂2u
∂y2= 0 (Laplace’s eqn.)
5.d3y
dx3− 4
d2y
dx2+ 4
dy
dx= 0
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TYPE:
If the unknown function depends on a
single independent variable, then the
equation is an
ordinary differential equation (ODE).
If the unknown function depends on
more than one independent variable,
then the equation is a
partial differential equation (PDE).
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ORDER:
The order of a differential equation is
the order of the highest derivative of
the unknown function appearing in
the equation.
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Examples:
1. y′ = 2x + cos x
2.dy
dt= ky (exponential growth/decay)
3. x2y′′ − 2xy′ + 2y = 4x3
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4.∂2u
∂x2+
∂2u
∂y2= 0 (Laplace’s eqn.)
5.d3y
dx3− 4
d2y
dx2+ 4
dy
dx= 0
6.d2y
dx2+2x sin
(
dy
dx
)
+3exy =d3
dx3(e2x)
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SOLUTION:
A solution of a differential equation
is a function defined on some domain
D such that the equation reduces to
an identity when the function is substi-
tuted into the equation.
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Examples:
1. y′ = 2x + cos x
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2. y′ = ky
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3. y′′ − 2y′ − 8y = 4e2x
Is y = 2e4x−
12 e2x a solution?
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y′′ − 2y′ − 8y = 4e2x
Is y = e−2x + 2e3x a solution?
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4. x2y′′ − 4xy′ + 6y = 3x4
Is y = 32 x4 + 2x3 a solution?
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x2y′′ − 4xy′ + 6y = 3x4
Is y = 2x2 + x3 a solution?
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5.∂2u
∂x2+
∂2u
∂y2= 0
u = ln√
x2 + y2 Solution?
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∂2u
∂x2+
∂2u
∂y2= 0
u = cosx sinh y, u = 3x − 4y
Solutions??
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# Finding solutions
6. Find a value of r, if possible,
such that y = erx is a solution of
y′′ − 3y′ − 10y = 0.
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7. Find a value of r, if possible,
such that y = xr is a solution of
x2y′′ + 2x y′ − 6y = 0.
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8. Find a value of r, if possible,
such that y = xr is a solution of
y′′ −1
xy′ −
3
x2y = 0.
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From now on, all differential equa-
tions are ordinary differential equa-
tions.
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2. n-PARAMETER FAMILY OF
SOLUTIONS / GENERAL SOLU-
TION
Example: Solve the differential equa-
tion:
y′′′ − 12x + 6e2x = 0
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NOTE: To solve a differential equa-
tion having the special form
y(n)(x) = f(x),
simply integrate f n times,
and EACH integration step produces
an arbitrary constant;
there will be n independent arbi-
trary constants.
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Intuitively, to find a set of solutions of
an n-th order differential equation
F[
x, y, y′, y′′, . . . , y(n)]
= 0
we “integrate” n times, with each in-
tegration step producing an arbitrary
constant of integration (i.e., a param-
eter). Thus, ”in theory,” an n-th order
differential equation has an n-parameter
family of solutions.
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SOLVING A DIFFERENTIAL EQUA-
TION:
To solve an n-th order differential equa-
tion
F(
x, y, y′, y′′, . . . , y(n))
= 0
means to find an n-parameter family of
solutions. (Note: Same n.)
NOTE: An “n-parameter family of
solutions” is more commonly called the
GENERAL SOLUTION.24
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Examples: Find the general solution:
1. y′ − 3x2− 2x + 4 = 0
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y = x3 + x2− 4x + C
-3 -2 -1 1 2 3x
-10
10
20
30
y
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2. y′′ + 2sin 2x = 0
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y =1
2sin2x + C1x + C2
-3 -2 -1 1 2 3x
-�
- �-
�-2
2
��y
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3. y′′′ − 3y′′ + 3y′ − y = 0
Answer: y = C1ex + C2xex + C3x2ex
4. x2y′′ − 4xy′ + 6y = 3x4
Answer: y = C1x2 + C2x3 + 32 x4
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PARTICULAR SOLUTION:
If specific values are assigned to the
arbitrary constants in the general solu-
tion of a differential equation, then the
resulting solution is called a particular
solution of the equation.
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Examples:
1. y′′ = 6x + 8e2x
General solution:
y = x3 + 2e2x + C1x + C2
Particular solutions:
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2. x2y′′ − 2xy′ + 2y = 4x3
General solution:
y = C1x + C2x2 + 2x3
Particular solutions:
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# 3. THE DIFFERENTIAL EQUA-
TION OF AN n-PARAMETER FAM-
ILY:
Given an n-parameter family of curves.
The differential equation of the fam-
ily is an n-th order differential equation
that has the given family as its general
solution.
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Examples:
1. y2 = Cx3 + 4 is the general solu-
tion of a DE.
a. What is the order of the DE?
b. Find the DE?
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2. y = C1x + C2x3 is the general
solution of a DE.
a. What is the order of the DE?
b. Find the DE?
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General strategy for finding the dif-
ferential equation
Step 1. Differentiate the family n
times. This produces a system of n+1
equations.
Step 2. Choose any n of the equa-
tions and solve for the parameters.
Step 3. Substitute the “values” for
the parameters in the remaining equa-
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Examples:
The given family of functions is the
general solution of a differential equa-
tion.
(a) What is the order of the equation?
(b) Find the equation.
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1. y = Cx3− 2x
(a)
(b)
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2. y = C1e2x + C2e3x
(a)
(b)
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3. y = C1 cos 3x + C2 sin 3x
(a)
(b)
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4. y = C1x4 + C2x + C3
(a)
(b)
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5. y = C1 + C2x + C3x2
(a)
(b)
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4. INITIAL-VALUE PROBLEMS:
1. Find a solution of
y′ = 3x2 + 2x + 1
which passes through the point (−2,4);
that is, satisfies y(−2) = 4.
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y = x3 + x2 + x + C (the general solu-
tion)
-�
-3 -2 -1 1 2 3x
- � �-
� �-20
20
� �� �
y
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y = x3 + x2 + x + 10 (the particular
solution that satisfies the
- � -3 -2 -1 1 2 3x
- � �
-20
20
� �
y
� �
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2. y = C1 cos 3x + C2 sin 3x is the
general solution of
y′′ + 9y = 0.
a. Find a solution which satisfies
y(0) = 3
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b. Find a solution which satisfies
y(0) = 3, y′(0) = 4
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y = 3cos 3x + 43 sin 3x is the solution
of
y′′ + 9y = 0, y(0) = 3, y′(0) = 4.
-2 -1 1 2x
-2
-1
1
2
y
H0,3L
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c. Find a solution which satisfies
y(0) = 4, y(π) = 4
d. Find a solution which satisfies
y(0) = 4, y(π) = −4
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An n-th order initial-value problem
consists of an n-th order differential
equation
F[
x, y, y′, y′′, . . . , y(n)]
= 0
together with n (initial) conditions of
the form
y(c) = k0, y′(c) = k1, y′′(c) = k2, . . . ,
y(n−1)(c) = kn−1
where c and k0, k1, . . . , kn−1 are
given numbers.
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NOTES:
1. An n-th order differential equation
can always be written in the form
F[
x, y, y′, y′′, · · · , y(n)]
= 0
by bringing all the terms to the left-
hand side of the equation.
2. The initial conditions determine
a particular solution of the differential
equation.
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Strategy for Solving an Initial-Value
Problem:
Step 1. Find the general solution of
the differential equation.
Step 2. Use the initial conditions to
solve for the arbitrary constants in the
general solution.
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Examples:
1. Find a solution of the initial-value
problem
y′ = 4x + 6e2x, y(0) = 5
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2. y = C1e−2x + C2e4x is the general
solution of
y′′ − 2y′ − 8y = 0
Find a solution that satisfies the initial
conditions
y(0) = 3, y′(0) = 2
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3. y = C1x + C2x3 is the general
solution of
y′′ −3
xy′ +
3
x2y = 0
a. Find a solution which satisfies
y(1) = 2, y′(1) = −4.
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b. Find a solution which satisfies
y(0) = 0, y′(0) = 2.
c. Find a solution which satisfies
y(0) = 4, y′(0) = 3.
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EXISTENCE AND UNIQUENESS:
The fundamental questions in a course
on differential equations are:
1. Does a given initial-value problem
have a solution? That is, do solutions
to the problem exist ?
2. If a solution does exist, is it
unique ? That is, is there exactly one
solution to the problem or is there more
than one solution?57