5. Preface
This set of lecture notes for ordinary and partial differential equations grew
out of the course Engineering Mathematics I have taught at Cankaya Univer¸
sity since 1999. It is a one-semester course for second year students. The main
audience for this text, of course, is students. Presentation is user-friendly.
There are more examples and fewer theorems than usual.
The material is based on a solid background in calculus. The student
is assumed to be familiar with algebra, trigonometry, functions and graphs,
series, differentiation, and most importantly, integration techniques of various kinds. It is my (and my students’) sad experience that if you cannot
differentiate and integrate, you cannot solve differential equations. Knowledge of Linear Algebra, except for the determinants of a simple nature, is
not assumed.
There are 14 chapters. Each chapter can be covered in one week. After
a summary of methods and solved exercises, there are a number of end of
chapter exercises with answers. The exercises that take exceptionally longer
times are marked with a star. ( ) Nobody can learn how to solve problems
by watching someone else solve problems. So I advise the students to try
each problem on their own.
I would like to thank all my students who helped me write this book
by the valuable feedback they provided. In particular, special thanks are for
˙
Nuh Co¸kun, Nevrez Imamo˜lu, Nilg¨n Din¸arslan and I¸ıl Lelo˜lu who have
s
g
u
c
s
g
made a very extensive and meticulous check of the whole manuscript.
You may send all kinds of comments, suggestions and error reports to
sermutlu@cankaya.edu.tr.
Assist. Prof. Dr. Emre Sermutlu
ix
6. 2
CHAPTER 1. FIRST ORDER ODE
are partial differential equations. (Partial Differential Equations are usually
much more difficult)
Order: The order of a differential equation is the order of the highest derivative that occurs in the equation.
A first order differential equation contains y , y and x so it is of the form
F (x, y, y ) = 0 or y = f (x, y).
For example, the following differential equations are first order:
Chapter 1
First Order Differential
Equations
y + x2 y = ex
xy = (1 + y 2 )
y 2 = 4xy
While these are second order:
The subject of differential equations is an important part of applied mathematics. Many real life problems can be formulated as differential equations.
In this chapter we will first learn the basic concepts and classification of
differential equations, then we will see where they come from and how the
simplest ones are solved. The concepts and techniques of calculus, especially
integration, will be necessary to understand differential equations.
1.1
Definitions
Ordinary Differential Equation: An ordinary differential equation is an
equation that contains derivatives of an unknown function y(x).
Partial Differential Equation: A partial differential equation is a differential equation involving an unknown function of two or more variables, like
u(x, y).
For example,
y − 4y + y = 0
y 2 + 1 = x2 y + sin x
are ordinary differential equations.
uxx + uyy = 0
u2 + u2 = ln u
x
y
1
y − x2 y + y = 1 + sin x
y + 6yy = x3
General and Particular Solutions: A general solution of a differential
equation involves arbitrary constants. In a particular solution, these constants are determined using initial values.
As an example, consider the differential equation y = 2x.
y = x2 + c is a general solution ,
y = x2 + 4 is a particular solution .
Example 1.1 Find the general solution of the differential equation y = 0.
Then find the particular solution that satisfies y(0) = 5, y (0) = 3.
y = 0 ⇒ y = c ⇒ y = cx + d. This is the general solution.
y (0) = 3 ⇒ c = 3, y(0) = 5 ⇒ d = 5
Therefore y = 3x + 5. This is the particular solution.
Explicit and Implicit Solutions: y = f (x) is an explicit solution,
F (x, y) = 0 is an implicit solution. We have to solve equations to obtain y
for a given x in implicit solutions, whereas it is straightforward for explicit
solutions.
For example, y = e4x is an explicit solution of the equation y = 4y.
x3 + y 3 = 1 is an implicit solution of the equation y 2 y + x2 = 0
7. 1.2. MATHEMATICAL MODELING
1.2
3
Mathematical Modeling
4
CHAPTER 1. FIRST ORDER ODE
Example 1.4 Solve the initial value problem y + y 2 xex = 0, y(0) = 2
Differential equations are the natural tools to formulate, solve and understand many engineering and scientific systems. The mathematical models of
most of the simple systems are differential equations.
y = −y 2 xex
dP
= αP
dt
ex dx
Example 1.3 The downward acceleration of an object in free fall is g. Find
the height as a function of time if the initial height is y0 and initial speed is
v0 .
2=
d2 y
= −g
dt2
1
= xex − ex + c
y
⇒
c=
1
xex − ex +
3
2
3
2
Example 1.5 Find the general solution of the differential equation
y + y 2 = 1.
dy
+ y2 = 1
dx
1
y = − gt2 + v0 t + y0
2
⇒
dy
= 1 − y2
dx
dy
=
1 − y2
Separable Equations
If we can separate x and y in a first order differential equation and put them
to different sides as g(y)dy = f (x)dx, it is called a separable equation. We
can find the solution by integrating both sides. (Don’t forget the integration
constant!)
f (x)dx + c
⇒
xex
1
−1 + c
y=
dy
= −gt + v0
dt
g(y)dy =
xex dx
1
− ex + c
This is the general solution. Now we will use the condition y(0) = 2 to
determine the constant c.
y=
where P0 = P (0)
1.3
dy
= xex dx
y2
Using integration by parts, we have u = x, dv = ex dx, du = dx, v = ex
therefore
1
= xex −
y
P = P0 eαt
−
dy
=
y2
−
Example 1.2 The rate of growth of a population is proportional to itself.
Find the population as a function of time.
⇒
(1.1)
1
2
1
1
+
1−y 1+y
⇒
dy
= dx
1 − y2
dx
dy =
1
1+y
ln
=x+c
2
1−y
1+y
= e2x+2c
1−y
dx
8. 1.4. TRANSFORMATIONS
5
6
CHAPTER 1. FIRST ORDER ODE
After some algebra, we obtain
y=
u3/2
= 3 ln x + c
3/2
ke2x − 1
ke2x + 1
where k = e2c
u=
Example 1.6 Solve the initial value problem y = x3 e−y , y(1) = 0.
ey dy =
ey =
9
ln x + c1
2
e0 =
Let’s use the substitution u = x + y. Then,
1
+c
4
y = u − x, ⇒
1.4
x4 3
+
4
4
(u + 6)dx = (−u − 3)(du − dx)
3dx = (−u − 3)du
Sometimes a change of variables simplifies a differential equation just as
y
substitutions simplify integrals. For example if y = f
, the substitution
x
y
u = will make the new equation separable.
x
y
+3
x
x
.
y
√
u du =
3dx =
3x = −
3x = −
If y = ux, then y = u x + u and u x + u = u + 3
ux=3
dy = du − dx
and the equation can be expressed in terms of u and x.
Transformations
Example 1.7 Solve y =
2/3
Example 1.8 Solve the differential equation (x + y + 6)dx = (−x − y − 3)dy.
3
c=
4
y = ln
2/3
x3 dx
x4
+c
4
⇒
y(1) = 0
y=x
9
ln x + c1
2
1
u
3dx
x
1
u
This is an implicit solution.
(−u − 3) du
u2
− 3u + c
2
(x + y)2
− 3(x + y) + c
2
9. EXERCISES
7
Exercises
CHAPTER 1. FIRST ORDER ODE
Answers
Solve the following differential equations.
1) y 3 y + x3 = 0
2) y + 4x3 y 2 = 0
y
3) xy = x + y Hint: y = f
x
4) (x2 + y 2 ) dx + xydy = 0 Hint: y = f
8
y
x
1) x4 + y 4 = c
1
2) y = 4
x +c
3) y = x(ln |x| + c)
c
x2
4) y 2 = 2 −
x
2
5) y = − ln c +
2
5) y = xey−x
1 + ln x
6) y =
4y 3
7) y = 3x2 sec2 y
8) y = y(y + 1)
9) y + 2y = y 2 + 1
10) (1 + y 2 )dx + x2 dy = 0
y
x
12) y = eax+by
13) y = x2 y 2 − 2y 2 + x2 − 2
14) y = −
2x + y
x
Solve the following initial value problems:
15) (y 2 + 5xy + 9x2 )dx + x2 dy = 0, y(1) = −4
16) y 3 y + x3 = 0, y(0) = 1
17) y = −2xy, y(0) = 3
18) y = 1 + 4y 2 , y(0) = 0
19) (x2 + 1)1/2 y = xy 3 , y(0) = 2
20)
dx
x x2
= − , x(0) = 1
dt
5 25
2
6) y 4 = x ln x + c
7) 2y + sin 2y = 4x3 + c
ex
8) y =
c − ex
9) y = 1 −
1
x+c
10) y = tan c +
11) y = a
e−x
2
1
x
11) y = cxa
12)
eax e−by
+
=c
a
b
13) y = tan
x3
− 2x + c
3
14) y = −x +
c
x
15) y =
16)
17)
18)
19)
x
− 3x
ln x − 1
x4 + y 4 = 1
2
y = 3e−x
1
y = 2 tan 2x
√
y = ( 9 − 2 x2 + 1)−1/2
4
20) x =
5et/5
4 + et/5
10. 10
CHAPTER 2. EXACT EQUATIONS
So, the solution of this equation is very simple, if du is zero, u must be a
constant, therefore
x4 + x2 y 2 + y 4 = c
∂N
∂M
=
is necessary and sufficient for the
∂y
∂x
equation M (x, y)dx + N (x, y)dy = 0 to be exact.
Method of Solution: To solve M dx + N dy = 0,
Theorem 2.1: The condition
Chapter 2
• Check for Exactness
Exact and Linear Differential
Equations
• If the equation is exact, find u by integrating either M or N .
u=
M dx + k(y) or u =
N dy + l(x)
Note that we have arbitrary functions as integration constants.
In this chapter, we will learn how to recognize and solve three different types
of equations: Exact, linear, and Bernoulli. All of them are first order equations, therefore we expect a single integration constant in the solution.
At this stage it seems like there’s a special trick for every different kind
of question. You will gain familiarity with exercise and experience.
• Determine the arbitrary functions using the original equation. The
solution is u(x, y) = c
Example 2.1 Solve the equation 3y 2 dx + (3y 2 + 6xy)dy = 0.
Let’s check for exactness first.
2.1
∂(3y 2 )
= 6y,
∂y
Exact Equations
A first order differential equation of the form
M (x, y)dx + N (x, y)dy = 0
The equation is exact.
(2.1)
u(x, y) =
is called an exact differential equation if there exists a function u(x, y) such
that
∂u
∂u
= M,
=N
(2.2)
∂x
∂y
In other words, du = M dx + N dy, so M dx + N dy is a total differential.
For example, the equation
(4x3 + 2xy 2 )dx + (4y 3 + 2x2 y)dy = 0
is exact, and
u = x4 + x2 y 2 + y 4
9
∂(3y 2 + 6xy)
= 6y
∂x
3y 2 dx + k(y) = 3y 2 x + k(y)
∂u
= 6yx + k (y) = 3y 2 + 6xy
∂y
k (y) = 3y 2
⇒
k(y) = y 3
We do not need an integration constant here because u(x, y) = c already
contains one
u(x, y) = 3y 2 x + y 3 = c
11. 2.2. INTEGRATING FACTORS
2.2
11
Integrating Factors
12
CHAPTER 2. EXACT EQUATIONS
But this equation is more difficult than the one we started with. If we make
a simplifying assumption that F is a function of one variable only, we can
solve for F and obtain the following theorem:
Consider the equation
Theorem 2.2: Consider the equation P dx + Qdy = 0. Define
P dx + Qdy = 0
(2.3)
that is not exact. If it becomes exact after multiplying by F , i.e. if
F P dx + F Qdy = 0
R=
(2.4)
is exact, then F is called an integrating factor. (Note that P, Q and F are
functions of x and y)
1
For example, ydx − xdy = 0 is not exact, but F = 2 is an integrating
x
factor.
Example 2.2 Solve (2xex − y 2 )dx + 2ydy = 0. Use F = e−x .
a) If R depends only on x, then F (x) = exp
factor.
˜
b) If R depends only on y, then F (y) = exp
factor.
∂Q ∂P
−
∂x
∂y
R=
˜
R(y)dy
is an integrating
8x2 y + 2 − 6x2 y − 1
2x2 y + 1
1
= 3
=
3y + x
2x
2x y + x
x
F (x) = e
R(x)dx
= eln x = x
Multiply the equation by x to obtain the exact equation
∂(2x − y 2 e−x )
∂(2ye−x )
= −2ye−x ,
= −2ye−x
∂y
∂x
(4x3 y 2 + 2yx)dx + (2x4 y + x2 )dy = 0
Now the equation is exact. We can solve it as we did the previous example
and obtain the result
u(x, y) =
x2 + y 2 e−x = c
How To Find an Integrating Factor: Let P dx+Qdy = 0 be a differential
equation that is not exact, and let F = F (x, y) be an integrating factor. By
definition,
(2.5)
(2.6)
R(x) dx is an integrating
Example 2.3 Solve (4x2 y 2 + 2y)dx + (2x3 y + x)dy = 0
(2x − y 2 e−x )dx + 2ye−x dy = 0
Fy P + F P y = Fx Q + F Qx
1
˜
and R =
P
The equation is not exact.
The equation is not exact. Let’s multiply both sides by e−x . The new
equation is:
⇒
∂P
∂Q
−
∂y
∂x
∂(2x3 y + x)
∂(4x2 y 2 + 2y)
= 8x2 y + 2,
= 6x2 y + 1
∂y
∂x
∂(2y)
∂(2xex − y 2 )
= −2y,
=0
∂y
∂x
(F P )y = (F Q)x
1
Q
(4x3 y 2 + 2yx) dx + k(y) = x4 y 2 + yx2 + k(y)
∂u
= 2x4 y + x2 + k (y) = 2x4 y + x2
∂y
u(x, y) = x4 y 2 + x2 y = c
⇒
k(y) = 0
12. 2.3. LINEAR FIRST ORDER EQUATIONS
2.3
13
Linear First Order Equations
2.4
If a first order differential equation can be written in the form
y + p(x)y = r(x)
CHAPTER 2. EXACT EQUATIONS
Bernoulli Equation
The equation
y + p(x)y = g(x)y a
(2.7)
it is called a linear differential equation. If r(x) = 0, the equation is homogeneous, otherwise it is nonhomogeneous.
We can express the equation (2.7) as [p(x)y − r(x)]dx + dy = 0. This is
not exact but it has an integrating factor:
R = p(x), F = e
14
p dx
is called Bernoulli equation. It is nonlinear. Nonlinear equations are usually
much more difficult than linear ones, but Bernoulli equation is an exception.
It can be linearized by the substitution
u(x) = [y(x)]1−a
(2.14)
(2.8)
Then, we can solve it as other linear equations.
Method of Solution:
• Given a first order linear equation, express it in the following form:
y + p(x)y = r(x)
p dx
y +e
p dx
py = re
2
p dx
y −
ex
2x
y=
3
3xy 2
p(x) dx to
obtain
e
Example 2.5 Solve the equation
(2.9)
• Multiply both sides by the integrating factor F (x) = exp
(2.10)
Here a = −2 therefore u = y 1−(−2) = y 3 ⇒ u = 3y 2 y
Multiplying both sides of the equation by 3y 2 we obtain
2
• Express the left hand side as a single parenthesis.
e
p dx
y
= re
p dx
3y 2 y − 2xy 3 =
(2.11)
y(x) = e−h
eh r dx + c
(2.12)
2
ex
x
⇒
u − 2xu =
e
−2x dx
= e−x
2
2
Multiplying both sides by e−x , we get
2
p dx.
2
e−x u − 2xe−x u =
Example 2.4 Solve y + 4y = 1
2
The integrating factor is F = e
equation by e4x to obtain
4 dx
(e−x u) =
4x
= e . Multiply both sides of the
2
e4x y + 4e4x y =e4x
e−x u = ln x + c
⇒
⇒
1
y = + ce−4x
4
1
x
1
x
u = (ln x + c)ex
(e4x y) =e4x
e4x
e4x y =
+c
4
ex
x
This equation is linear. Its integrating factor is
• Integrate both sides. Don’t forget the integration constant. The solution is:
where h =
(2.13)
y = (ln x + c)ex
2
1/3
2
13. EXERCISES
15
16
Exercises
Answers
Solve the following differential equations. (Find an integrating factor if
necessary)
1) (yex + xyex + 1)dx + xex dy = 0
2) (2r + 2 cos θ)dr − 2r sin θdθ = 0
3) (sin xy + xy cos xy)dx + (x2 cos xy)dy = 0
4) 2 cos ydx = sin ydy
5) 5dx − ey−x dy = 0
6) (2xy + 3x2 y 6 ) dx + (4x2 + 9x3 y 5 ) dy = 0
7) (3xey + 2y) dx + (x2 ey + x) dy = 0
1
5
8) y + y =
x
x
9) y +
1
1
y=
x ln x
ln x
c
− 1 e−x
x
r2 + 2r cos θ = c
x sin xy = c
F = e2x , e2x cos y = c
F = ex , 5ex − ey = c
F = y 3 , x 2 y 4 + x3 y 9 = c
F = x, x3 ey + x2 y = c
c
1
y= + 5
5 x
1) y =
2)
3)
4)
5)
6)
7)
8)
9) y =
x+c
ln x
10) y = −1 +
x4
12) y = 4 − 5e− 4
Reduce to linear form and solve the following equations:
2 sin x 1/2
13) y − 4y tan x =
y
cos3 x
x
y
5 ln x 4/5
25
y=
y
15) y +
x
x5
13) y =
14) y =
c − ln cos x
cos2 x
1
2
2
− x + ce−2x
15) y =
y
1
=− 9 3
x
xy
x ln x − x + c
x5
16) y =
14) y + y = −
1
c
+ 4
8
x
x
19) x = y −2
2
1/4
1
3
cosh 3y + c
Hint: x ↔ y
20) y =
20) 2xyy + (x − 1)y 2 = x2 ex
5
17) y = arcsin[c(x − 1)]
1
c
+ 3
18) F = y, x =
2y y
tan y
17) y =
x−1
18) y 2 dx + (3xy − 1)dy = 0
19) y (sinh 3y − 2xy) = y
c
cos x
11) y = x4 cos x + c cos x
10) y − y tan x = tan x
11) y + y tan x = 4x3 cos x
12) y + x3 y = 4x3 , y(0) = −1
16) y +
CHAPTER 2. EXACT EQUATIONS
Hint: z = y 2
cxe−x + 1 xex
2
14. 18
Chapter 3
Second Order Homogeneous
Differential Equations
For first order equations, concepts from calculus and some extensions were
sufficient. Now we are starting second order equations and we will learn many
new ideas, like reduction of order, linear independence and superposition of
solutions.
Many differential equations in applied science and engineering are second
order and linear. If in addition they have constant coefficients, we can solve
them easily, as explained in this chapter and the next. For nonconstant
coefficients, we will have limited success.
3.1
Linear Differential Equations
If we can express a second order differential equation in the form
y + p(x)y + q(x)y = r(x)
(3.1)
it is called linear. Otherwise, it is nonlinear.
Consider a linear differential equation. If r(x) = 0 it is called homogeneous, otherwise it is called nonhomogeneous. Some examples are:
y + y 2 = x2 y
Nonlinear
sin xy + cos xy = 4 tan x Linear Nonhomogeneous
x2 y + y = 0
Linear Homogeneous
17
CHAPTER 3. SECOND ORDER EQUATIONS
Linear Combination: A linear combination of y1 , y2 is y = c1 y1 + c2 y2 .
Theorem 3.1: For a homogeneous linear differential equation any linear
combination of solutions is again a solution.
The above result does NOT hold for nonhomogeneous equations.
For example, both y = sin x and y = cos x are solutions to y + y = 0, so
is y = 2 sin x + 5 cos x.
Both y = sin x + x and y = cos x + x are solutions to y + y = x, but
y = sin x + cos x + 2x is not.
This is a very important property of linear homogeneous equations, called
superposition. It means we can multiply a solution by any number, or add
two solutions, and obtain a new solution.
Linear Independence: Two functions y1 , y2 are linearly independent if
c1 y1 + c2 y2 = 0 ⇒ c1 = 0, c2 = 0. Otherwise they are linearly dependent.
(One is a multiple of the other).
For example, ex and e2x are linearly independent. ex and 2ex are linearly
dependent.
General Solution and Basis: Given a second order, linear, homogeneous
differential equation, the general solution is:
y = c1 y1 + c2 y2
(3.2)
where y1 , y2 are linearly independent. The set {y1 , y2 } is called a basis, or a
fundamental set of the differential equation.
As an illustration, consider the equation x2 y − 5xy + 8y = 0. You can
easily check that y = x2 is a solution. (We will see how to find it in the
last section) Therefore 2x2 , 7x2 or −x2 are also solutions. But all these are
linearly dependent.
We expect a second, linearly independent solution, and this is y = x4 . A
combination of solutions is also a solution, so y = x2 + x4 or y = 10x2 − 5x4
are also solutions. Therefore the general solution is
y = c1 x 2 + c2 x 4
and the basis of solutions is {x2 , x4 }.
(3.3)
15. 3.2. REDUCTION OF ORDER
3.2
19
Reduction of Order
3.3
If we know one solution of a second order homogeneous differential equation,
we can find the second solution by the method of reduction of order.
Consider the differential equation
y + py + qy = 0
(3.4)
Suppose one solution y1 is known, then set y2 = uy1 and insert in the equation. The result will be
y1 u + (2y1 + py1 )u + (y1 + py1 + qy1 )u = 0
(3.5)
(3.6)
This is still second order, but if we set w = u , we will obtain a first order
equation:
y1 w + (2y1 + py1 )w = 0
(3.7)
Solving this, we can find w, then u and then y2 .
Example 3.1 Given that y1 = x2 is a solution of
CHAPTER 3. SECOND ORDER EQUATIONS
Homogeneous Equations with Constant
Coefficients
Up to now we have studied the theoretical aspects of the solution of linear homogeneous differential equations. Now we will see how to solve the constant
coefficient equation y + ay + by = 0 in practice.
We have the sum of a function and its derivatives equal to zero, so the
derivatives must have the same form as the function. Therefore we expect
the function to be eλx . If we insert this in the equation, we obtain:
λ2 + aλ + b = 0
But y1 is a solution, so the last term is canceled. So we have
y1 u + (2y1 + py1 )u = 0
20
(3.8)
This is called the characteristic equation of the homogeneous differential
equation y + ay + by = 0.
If we solve the characteristic equation, we will see three different possibilities:
Two real roots, double real root and complex conjugate roots.
Two Real Roots: The general solution is
y = c1 eλ1 x + c2 eλ2 x
(3.9)
2
x y − 3xy + 4y = 0
Example 3.2 Solve y − 3y − 10y = 0
find a second linearly independent solution.
Let y2 = ux2 . Then
y2 = u x2 + 2xu
and
2
Try y = eλx . The characteristic equation is λ2 − 3λ − 10 = 0 with solution
λ1 = 5, λ = −2, so the general solution is
y = c1 e5x + c2 e−2x
y2 = u x + 4xu + 2u
Inserting these in the equation, we obtain
x4 u + x3 u = 0
If w = u then
1
x4 w + x3 w = 0 or w + w = 0
x
1
This linear first order equation gives w = , therefore u = ln x and
x
2
y2 = x ln x
Example 3.3 Solve the initial value problem y −y = 0, y(0) = 2, y (0) = 4
We start with y = eλx as usual. The characteristic equation is λ2 − 1 = 0.
Therefore λ = ±1. The general solution is: y = c1 ex + c2 e−x
Now, we have to use the initial values to determine the constants.
y(0) = 2 ⇒ c1 + c2 = 2 and y (0) = 4 ⇒ c1 − c2 = 4.
By solving this system, we obtain c1 = 3, c2 = −1 so the particular solution
is:
y = 3ex − e−x
16. 3.3. CONSTANT COEFFICIENTS
21
Double Real Root: One solution is eλx but we know that a second order
equation must have two independent solutions. Let’s use the method of
reduction of order to find the second solution.
y − 2ay + a2 y = 0
⇒
y1 = eax
(3.10)
22
CHAPTER 3. SECOND ORDER EQUATIONS
3.4
Cauchy-Euler Equation
The equation x2 y + axy + by = 0 is called the Cauchy-Euler equation. By
inspection, we can easily see that the solution must be a power of x. Let’s
substitute y = xr in the equation and try to determine r. We will obtain
Let’s insert y2 = ueax in the equation.
r(r − 1)xr + arxr + bxr = 0
ax
ax
e u + (2a − 2a)e u = 0
Obviously, u = 0 therefore u = c1 + c2 x. The general solution is
y = c1 eλx + c2 xeλx
(3.12)
Example 3.4 Solve y + 2y + y = 0
λx
r2 + (a − 1)r + b = 0
(3.11)
(3.17)
(3.18)
This is called the auxiliary equation. Once again, we have three different
cases according to the types of roots. The general solution is given as follows:
• Two real roots
2
y = e . The characteristic equation is λ + 2λ + 1 = 0. Its solution is the
double root λ = −1, therefore the general solution is
y = c1 e−x + c2 xe−x
(3.13)
This can be proved using Taylor series expansions.
If the solution of the characteristic equation is
λ1 = α + iβ, λ2 = α − iβ
y = c1 e
(cos βx + i sin βx) + c2 e
αx
(cos βx − i sin βx)
(3.14)
(3.15)
By choosing new constants A, B, we can express this as
y=e
αx
(A cos βx + B sin βx)
y = c1 xr + c2 xr ln x
(3.20)
• Complex conjugate roots where r1 , r2 = r ± si
y = xr [c1 cos(s ln x) + c2 sin(s ln x)]
then the general solution of the differential equation will be
αx
(3.19)
• Double real root
Complex Conjugate Roots: We need the complex exponentials for this
case. Euler’s formula is
eix = cos x + i sin x
y = c1 xr1 + c2 xr2
(3.21)
Example 3.6 Solve x2 y + 2xy − 6y = 0
Insert y = xr . Auxiliary equation is r2 + r − 6 = 0. The roots are
r = 2, r = −3 therefore
y = c1 x2 + c2 x−3
(3.16)
Example 3.5 Solve y − 4y + 29y = 0.
Example 3.7 Solve x2 y − 9xy + 25y = 0
y = eλx . The characteristic equation is λ2 −4λ+29 = 0. Therefore λ = 2±5i.
The general solution is
Insert y = xr . Auxiliary equation is r2 − 10r + 25 = 0. Auxiliary equation
has the double root r = 5 therefore the general solution is
y = e2x (A cos 5x + B sin 5x)
y = c1 x5 + c2 x5 ln x
17. EXERCISES
23
24
Exercises
CHAPTER 3. SECOND ORDER EQUATIONS
Answers
1)
2)
3)
4)
5)
Are the following sets linearly independent?
1) {x4 , x8 }
2) {sin x, sin2 x}
3) {ln(x5 ), ln x}
Use reduction of order to find a second linearly independent solution:
4) x2 (ln x − 1) y − xy + y = 0,
y1 = x
1
5) x2 ln x y + (2x ln x − x)y − y = 0,
y1 =
x
6) y + 3 tan x y + (3 tan2 x + 1)y = 0,
y1 = cos x
Yes
Yes
No
y2 = ln x
y2 = ln x − 1
6) y2 = sin x cos x
7) y = (1 + x)e−x
1
8) y = c1 e−2x + c2 e− 2 x
Solve the following equations:
7) y + 2y + y = 0, y(0) = 1, y (0) = 0
9) y = e8x
5
8) y + y + y = 0
2
10) y = c1 e−12x + c2 xe−12x
9) y − 64y = 0, y(0) = 1,
10) y + 24y + 144y = 0
y (0) = 8
11) y = 4e−x + 3xe−x
7
11) y + 2y + y = 0, y(−1) = e, y(1) =
e
12) 5y − 8y + 5y = 0
π2
13) y + 2y + 1 +
y = 0, y(0) = 1, y (0) = −1
4
14) y − 2y + 2y = 0, y(π) = 0, y(−π) = 0
15) xy + y = 0
16) x2 y − 3xy + 5y = 0
17) x2 y − 10xy + 18y = 0
18) x2 y − 13xy + 49y = 0
19) Show that y1 = u and y2 = u
y −
v
u
+2
v
u
y +
12) y = e0.8x [A cos(0.6x) + B sin(0.6x)]
13) y = e−x cos
14) y = ex sin x
15) y = c1 + c2 ln x
16) y = x2 [c1 cos(ln x) + c2 sin(ln x)]
vdx are solutions of the equation
vu
u2 u
+2 2 −
vu
u
u
π
x
2
y=0
20) Show that y1 = u and y2 = v are solutions of the equation
(uv − vu )y + (vu − uv )y + (u v − v u )y = 0
17) y = c1 x2 + c2 x9
18) y = c1 x7 + c2 x7 ln x
18. 26
CHAPTER 4. NONHOMOGENEOUS EQUATIONS
the nonhomogeneous one. The general solution is of the form
y = yh + yp
(4.3)
Example 4.1 Find the general solution of y − 3y + 2y = 2x − 3 using
yp = x.
Chapter 4
Let’s solve y − 3y + 2y = 0 first. Let yh = eλx . Then
λ2 − 3λ + 2 = 0
Second Order Nonhomogeneous
Equations
which means λ = 2 or λ = 1. The homogenous solution is
yh = c1 ex + c2 e2x
therefore the general solution is:
y = x + c1 ex + c2 e2x
In this chapter we will start to solve the nonhomogeneous equations, and
see that we will need the homogeneous solutions we found in the previous
chapter.
Of the two methods we will learn, undetermined coefficients is simpler,
but it can be applied to a restricted class of problems. Variation of parameters
is more general but involves more calculations.
4.1
Consider the nonhomogeneous equation
(4.1)
Let yp be a solution of this equation. Now consider the corresponding homogeneous equation
y + p(x)y + q(x)y = 0
(4.2)
Let yh be the general solution of this one. If we add yh and yp , the result
will still be a solution for the nonhomogeneous equation, and it must be the
general solution because yh contains two arbitrary constants. This interesting
property means that we need the homogeneous equation when we are solving
25
The solution of y = 0 is simply yh = c1 x + c2 , therefore the general solution
must be
y = − cos x + c1 x + c2
As you can see, once we have a particular solution, the rest is straightforward, but how can we find yp for a given equation?
Example 4.3 Find a particular solution of the following differential equations. Try the suggested functions. (Success not guaranteed!)
General and Particular Solutions
y + p(x)y + q(x)y = r(x)
Example 4.2 Find the general solution of y = cos x using yp = − cos x.
a) y + y = ex ,
Try yp = Aex
b) y − y = ex ,
Try yp = Aex
c) y + 2y + y = x
Try yp = Ax + B
d) y + 2y = x
Try yp = Ax + B
e) y + 2y + y = 2 cos x Try yp = A cos x and yp = A cos x + B sin x
As you can see, some of the suggestions work and some do not.
yp is usually similar to r(x). We can summarize our findings as:
• Start with a set of functions that contains not only r(x), but also all
derivatives of r(x).
• If one of the terms of yp candidate occurs in yh , there is a problem.
19. 4.2. METHOD OF UNDETERMINED COEFFICIENTS
4.2
27
Method of Undetermined Coefficients
28
CHAPTER 4. NONHOMOGENEOUS EQUATIONS
The homogeneous equation is
3y + 10y + 3y = 0
To solve the constant coefficient equation
d2 y
dy
+ a + by = r(x)
2
dx
dx
(4.4)
• Solve the corresponding homogeneous equation, find yh .
• Find a candidate for yp using the following table:
Term in r(x)
Choice for yp
xn
eax
cos bx or sin bx
xn eax
xn cos bx or xn sin bx
An xn + · · · + A1 x + A0
Aeax
A cos bx + B sin bx
(An xn + · · · + A1 x + A0 )eax
(An xn + · · · + A0 ) cos bx
+(Bn xn + · · · + B0 ) sin bx
eax cos bx or eax sin bx
Aeax cos bx + Beax sin bx
xn eax cos bx or xn eax sin bx
(An xn + · · · + A0 )eax cos bx
+(Bn xn + · · · + B0 )eax sin bx
(You don’t have to memorize the table. Just note that the choice
consists of r(x) and all its derivatives)
• If your choice for yp occurs in yh , you have to change it. Multiply it
by x if the solution corresponds to a single root, by x2 if it is a double
root.
• Find the constants in yp by inserting it in the equation.
• The general solution is y = yp + yh
Note that this method works only for constant coefficient equations, and
only when r(x) is relatively simple.
Its solution is
yh = c1 e−3x + c2 e−x/3
To find a particular solution, let’s try yp = Ax + B. Inserting this in the
equation, we obtain:
10A + 3Ax + 3B = 9x
Therefore, A = 3, B = −10. The particular solution is:
yp = 3x − 10
The general solution is:
y = c1 e−3x + c2 e−x/3 + 3x − 10
Example 4.5 Find the general solution of y − 4y + 4y = e2x
The solution of the associated homogeneous equation
y − 4y + 4y = 0
is
yh = c1 e2x + c2 xe2x
Our candidate for yp is yp = Ae2x . But this is already in the yh so we have
to change it. If we multiply by x, we will obtain Axe2x but this is also in yh .
Therefore we have to multiply by x2 . So our choice for yp is yp = Ax2 e2x .
Now we have to determine A by inserting in the equation.
yp = 2Ax2 e2x + 2Axe2x
yp = 4Ax2 e2x + 8Axe2x + 2Ae2x
Example 4.4 Find the general solution of the equation
3y + 10y + 3y = 9x
4Ax2 e2x + 8Axe2x + 2Ae2x − 4(2Ax2 e2x + 2Axe2x ) + 4Ax2 e2x = e2x
20. 4.3. METHOD OF VARIATION OF PARAMETERS
2Ae
2x
=e
2x
⇒
29
CHAPTER 4. NONHOMOGENEOUS EQUATIONS
Therefore the particular solution is
1
1
A = , yp = x2 e2x
2
2
yp (x) = −y1
1
y = yh + yp = c1 e2x + c2 xe2x + x2 e2x
2
4.3
30
y1 r
dx
aW
(4.11)
e−x
Example 4.6 Find the general solution of y + 2y + y = √
x
yh = c1 e−x + c2 xe−x
Method of Variation of Parameters
Consider the linear second order nonhomogeneous differential equation
a(x)y + b(x)y + c(x)y = r(x)
y2 r
dx + y2
aW
W =
(4.5)
e−x
xe−x
−x
−x
−e
e − xe−x
yp = −e−x
If a(x), b(x) and c(x) are not constants, or if r(x) is not among the functions
given in the table, we can not use the method of undetermined coefficients. In
this case, the variation of parameters can be used if we know the homogeneous
solution.
Let yh = c1 y1 + c2 y2 be the solution of the associated homogeneous equation
a(x)y + b(x)y + c(x)y = 0
(4.6)
Let us express the particular solution as:
xe−x e−x
√ dx + xe−x
e−2x x
yp = −e−x
√
x dx + xe−x
= e−2x
e−x e−x
√ dx
e−2x x
1
√ dx
x
4
x3/2
x1/2
+ xe−x
= e−x x3/2
3/2
1/2
3
4
y = yh + yp = c1 e−x + c2 xe−x + e−x x3/2
3
2
Example 4.7 Find the general solution of x y − 5xy + 8y = x5
yp = −e−x
We can find the homogeneous solution of the Cauchy-Euler equation as:
yp = v1 (x)y1 + v2 (x)y2
(4.7)
yh = c1 x4 + c2 x2
There are two unknowns, so we may impose an extra condition. Let’s choose
v1 y1 + v2 y2 = 0 for simplicity. Inserting yp in the equation, we obtain
r
a
= 0
v1 y1 + v2 y2 =
v1 y1 + v2 y2
−y2 r
,
aW
v2 =
y1 r
aW
(4.8)
(4.9)
where W is the Wronskian
W =
y1 y2
y1 y2
= y1 y2 − y2 y1
x4 x 2
4x3 2x
= −2x5
Therefore the particular solution is
The solution to this linear system is
v1 =
W =
(4.10)
yp (x) = −x4
1
yp (x) = x4
2
1 5
yp (x) = x
3
The general solution is
x2 x5
dx + x2
x2 (−2x5 )
1
dx − x2 x2 dx
2
1
y = c1 x 4 + c2 x 2 + x 5
3
x4 x5
dx
x2 (−2x5 )
21. EXERCISES
31
Exercises
1)
2)
3)
4)
5)
6)
32
CHAPTER 4. NONHOMOGENEOUS EQUATIONS
Answers
Find the general solution of the following differential equations
y + 4y = x cos x
y − 18y + 81y = e9x
y = −4x cos 2x − 4 cos 2x − 8x sin 2x − 8 sin 2x
y + 3y − 18y = 9 sinh 3x
y + 16y = x2 + 2x
y − 2y + y = x2 ex
7) 2x2 y − xy + y =
1
x
1) y = c1 sin 2x + c2 cos 2x + 1 x cos x + 2 sin x
3
9
1
2) y = c1 e9x + c2 xe9x + x2 e9x
2
3) y = c1 + c2 x + x cos 2x + 3 cos 2x + 2x sin 2x + sin 2x
1
1
4) y = c1 e3x + c2 e−6x + e−3x + xe3x
4
2
1 2 1
1
5) y = c1 sin 4x + c2 cos 4x + x + x −
16
8
128
1
6) y = c1 ex + c2 xex + x4 ex
12
√
1
7) y = c1 x + c2 x +
6x
8) x y + xy − 4y = x ln x
9) y − 8y + 16y = 16x
10) y = x3
11) y + 7y + 12y = e2x + x
1
1
x2
8) y = c1 x2 + c2 x−2 + x2 ln2 x − x2 ln x +
8
16
64
1
9) y = c1 e4x + c2 xe4x + x +
2
x5
10) y =
+ c1 + c2 x
20
12) y + 12y + 36y = 100 cos 2x
11) y = c1 e−3x + c2 e−4x +
2
2
13) y + 9y = ex + cos 3x + 2 sin 3x
1
7
1 2x
e + x−
30
12
144
12) y = c1 e−6x + c2 xe−6x + 2 cos 2x +
3
sin 2x
2
1 x 1
1
e − x cos 3x + x sin 3x
10
3
6
14) y + 10y + 16y = e−2x
13) y = c1 cos 3x + c2 sin 3x +
15) y − 4y + 53y = (53x)2
1
14) y = c1 e−2x + c2 e−8x + xe−2x
6
16) y + y = (x2 + 1)e3x
15) y = e2x (c1 cos 7x + c2 sin 7x) + 53x2 + 8x −
17) y + y = csc x
16) y = e3x (0.1x2 − 0.12x + 0.152) + c1 sin x + c2 cos x
17) y = c1 sin x + c2 cos x − x cos x + sin x ln | sin x|
18) y = c1 sin x + c2 cos x − cos x ln | sec x + tan x| − sin x ln | csc x + cot x|
18) y + y = csc x sec x
19) y − 4y + 4y =
e2x ln x
x
20) y − 2y + y =
e2x
(ex + 1)2
19) y = c1 e2x + c2 xe2x + xe2x
74
53
(ln x)2
− ln x + 1
2
20) y = c1 ex + c2 xex + ex ln(1 + ex )
22. 34
CHAPTER 5. HIGHER ORDER EQUATIONS
means that all the constants c1 , c2 , . . . , cn are zero, then this set of functions
is linearly independent. Otherwise, they are dependent.
For example, the functions x, x2 , x3 are linearly independent. The functions cos2 x, sin2 x, cos 2x are not.
Given n functions, we can check their linear dependence by calculating
the Wronskian. The Wronskian is defined as
Chapter 5
Higher Order Equations
W (y1 , y2 , . . . , yn ) =
y1
y1
.
.
.
(n−1)
y1
In this chapter, we will generalize our results about second order equations to
higher orders. The basic ideas are the same. We still need the homogeneous
solution to find the general nonhomogeneous solution. We will extend the two
methods, undetermined coefficients and variation of parameters, to higher
dimensions and this will naturally involve many more terms and constants
in the solution. We also need some new notation to express nth derivatives
easily.
...
...
yn
yn
.
.
.
(5.4)
(n−1)
. . . yn
and the functions are linearly dependent if and only if W = 0 at some point.
5.2
Differential Operators
We can denote differentiation with respect to x by the symbol D
Dy =
dy
=y,
dx
D2 y =
d2 y
=y
dx2
(5.5)
etc. A differential operator is
5.1
An n
form
Linear Equations of Order n
th
L = a0 Dn + a1 Dn−1 + · · · + an−1 D + an
order differential equation is called linear if it can be written in the
a0 (x)
dn y
dn−1 y
dy
+ a1 (x) n−1 + · · · + an−1 (x) + an (x)y = r(x)
n
dx
dx
dx
(5.1)
and nonlinear if it is not linear.(Note that a0 = 0)
If the coefficients a0 (x), a1 (x), . . . an (x) are continuous, then the equation has
exactly n linearly independent solutions. The general solution is
y = c1 y1 + c2 y2 + · · · + cn yn
(5.2)
We will only work with operators where coefficients are constant.
We can add, multiply, expand and factor constant coefficient differential operators using common rules of algebra. In this respect, they are like
polynomials. So, the following expressions are all equivalent:
(D − 2)(D − 3)y = (D − 3)(D − 2)y
= (D2 − 5D + 6)y
= y − 5y + 6y
Let’s apply some simple operators to selected functions:
(D − 2)ex = Dex − 2ex
= ex − 2ex = −ex
Linear Independence: If
c1 y1 + c2 y2 + · · · + cn yn = 0
33
(5.3)
(5.6)
23. 5.3. HOMOGENEOUS EQUATIONS
35
36
CHAPTER 5. HIGHER ORDER EQUATIONS
(D − 2)e2x = De2x − 2e2x
= 2e2x − 2e2x = 0
Now we are in a position to solve very complicated-looking homogeneous
equations.
(D − 2)2 xe2x = (D − 2)(D − 2)xe2x
= (D − 2)(e2x + 2xe2x − 2xe2x )
= (D − 2)e2x = 0
Method of Solution:
• Express the given equation using operator notation (D notation).
(D2 − 4) sin(2x) = (D − 2)(D + 2) sin(2x)
= (D − 2)(2 cos(2x) + 2 sin(2x))
= −4 sin(2x) + 4 cos(2x) − 4 cos(2x) − 4 sin(2x)
= −8 sin 2x
• Factor the polynomial.
• Find the solution for each component.
• Add the components to obtain the general solution.
5.3
Homogeneous Equations
Example 5.1 Find the general solution of y (4) − 7y + y − 7y = 0.
Based on the examples in the previous section, we can easily see that:
The general solution of the equation (D − a)n y = 0 is
y = eax (c0 + c1 x + . . . + cn−1 xn−1 )
In operator notation, we have
(D4 − 7D3 + D2 − 7D)y = 0
(5.7)
Factoring this, we obtain
if a is real.
Some special cases are:
D(D − 7)(D2 + 1)y = 0
Dn y = 0 ⇒ y = c0 + c1 x + . . . + cn−1 xn−1
(D − a)y = 0 ⇒ y = eax
(D − a)2 y = 0 ⇒ y = c1 eax + c2 xeax
We know that
(5.8)
We can extend these results to the case of complex roots. If z = a + ib is a
root of the characteristic polynomial, then so is z = a − ib. (Why?)
Consider the equation
(D − a − ib)n (D − a + ib)n y = (D2 − 2aD + a2 + b2 )n y = 0
(5.9)
The solution is
y = eax cos bx(c0 + c1 x + . . . + cn−1 xn−1 )
+eax sin bx(k0 + k1 x + . . . + kn−1 xn−1 )
(5.10)
2
(D + b )y = 0
⇒
Therefore the general solution is
y = c1 + c2 e7x + c3 sin x + c4 cos x
Note that the equation is fourth order and the solution has four arbitrary
constants.
Example 5.2 Solve D3 (D − 2)(D − 3)2 (D2 + 4)y = 0.
Using the same method, we find:
A special case is obtained if a = 0.
2
Dy = 0 ⇒ y = c
(D − 7)y = 0 ⇒ y = ce7x
(D2 + 1)y = 0 ⇒ y = c1 sin x + c2 cos x
y = c1 cos bx + c2 sin bx
(5.11)
y = c1 + c2 x + c3 x2 + c4 e2x + c5 e3x + c6 e3x x + c7 cos 2x + c8 sin 2x
24. 5.4. NONHOMOGENEOUS EQUATIONS
5.4
37
Nonhomogeneous Equations
38
CHAPTER 5. HIGHER ORDER EQUATIONS
Then we will proceed similarly to simplify the steps. Eventually, we will
obtain the system
In this section, we will generalize the methods of undetermined coefficients
and variation of parameters to nth order equations.
Undetermined Coefficients: Method of undetermined coefficients is the
same as given on page 27. We will use the same table, but this time the
modification rule is more general. It should be:
• In case one of the terms of yp occurs in yh , multiply it by xk where k
is the smallest integer which will eliminate any duplication between yp
and yh .
v1 y1
v1 y1
.
.
.
(n−1)
v1 y1
(n)
v1 y1
yp = y1
The homogeneous solution is yh = (c0 + c1 x + c2 x2 + c3 x3 )ex . According to
the table, we should choose yp as Aex + Bxex , but this already occurs in
the homogeneous solution. Multiplying by x, x2 , x3 are not enough, so, we
should multiply by x4 .
yp = Ax4 ex + Bx5 ex
Inserting this in the equation, we obtain:
24Aex + 120Bxex = xex
1 5 x
xe
120
n−1
d y
d y
dy
+ a1 (x) n−1 + · · · + an−1 (x) + an (x)y = r(x)
n
dx
dx
dx
(5.12)
Let the homogeneous solution be yh = c1 y1 + · · · + cn yn
Then the particular solution is yp = v1 y1 + · · · + vn yn
Here, vi are functions of x. Since we have n functions, we can impose n − 1
conditions on them. The first condition will be
v1 y1 + · · · + vn yn = 0
=
=
+
+
(n−1)
vn y n
(n)
vn yn
=
=
0
0
.
.
.
(5.14)
0
r(x)
a0 (x)
v1 dx + · · · + yn
vn dx
(5.15)
x3 y − 6x2 y + 15xy − 15y = 8x6
We can find the homogeneous solution yh = c1 x + c2 x3 + c3 x5 using our
method for Cauchy-Euler equations. Then, the particular solution will be
yp = xv1 + x3 v2 + x5 v3 . Using the above equations, we obtain the system
xv1 + x3 v2 + x5 v3 = 0
v1 + 3x2 v2 + 5x4 v3 = 0
6xv2 + 20x3 v3 = 8x3
Variation of Parameters: The idea is the same as in second order equations, but there are more unknowns to find and more integrals to evaluate.
Consider
a0 (x)
vn yn
vn yn
.
.
.
Example 5.4 Find the general solution of
Therefore A = 0, B = 1/120 and the general solution is
n
+ ···
+ ···
+
+
Then, we will solve this linear system to find vi , and integrate them to
obtain yp .
Example 5.3 Solve the equation (D − 1)4 y = xex .
y = (c0 + c1 x + c2 x2 + c3 x3 )ex +
+ ···
+ ···
(5.13)
The solution of this system is v1 = x4 , v2 = −2x2 , v3 = 1 therefore the
particular solution is
yp = x
x4 dx + x3
(−2x2 ) dx + x5
and the general solution is
y = c1 x + c2 x 3 + c3 x 5 +
8 6
x
15
dx =
8 6
x
15
25. EXERCISES
39
Exercises
CHAPTER 5. HIGHER ORDER EQUATIONS
Answers
1) y = c0 + c1 x + c2 x2 + c3 x3 + c4 x4
1) D5 y = 0
2) y = c1 ex + c2 xex + c3 x2 ex
2) (D − 1)3 y = 0
3) y − 4y + 13y = 0
4) (D − 2)2 (D + 3)3 y = 0
5) (D2 + 2)3 y = 0
d4 y
d2 y
+ 5 2 + 4y = 0
dx4
dx
7) (D2 + 9)2 (D2 − 9)2 y = 0
6)
4
40
3
2
dy
dy
dy
−2 3 +2 2 =0
4
dx
dx
dx
9) y − 3y + 12y − 10y = 0
8)
3) y = c1 e2x cos 3x + c2 e2x sin 3x + c3
4) y = c1 e2x + c2 xe2x + c3 e−3x + c4 xe−3x + c5 x2 e−3x
√
√
√
√
5) y = c1 cos 2x + c2 sin 2x + c3 x cos 2x + c4 x sin 2x
√
√
+ c5 x2 cos 2x + c6 x2 sin 2x
6) y = c1 cos 2x + c2 sin 2x + c3 cos x + c4 sin x
7) y = c1 e3x + c2 xe3x + c3 e−3x + c4 xe−3x + c5 cos 3x + c6 sin 3x
+ c7 x cos 3x + c8 x sin 3x
8) y = c1 + c2 x + c3 ex cos x + c4 ex sin x
10) (D2 + 2D + 17)2 y = 0
9) y = c1 ex + c2 ex cos 3x + c3 ex sin 3x
11) (D4 + 2D2 + 1)y = x2
10) y = c1 e−x sin 4x + c2 e−x cos 4x + c3 xe−x sin 4x + c4 xe−x cos 4x
12) (D3 + 2D2 − D − 2)y = 1 − 4x3
11) y = c1 cos x + c2 sin x + c3 x cos x + c4 x sin x + x2 − 4
√
√
√
13) (2D4 + 4D3 + 8D2 )y = 40e−x [ 3 sin( 3x) + 3 cos( 3x)]
14) (D3 − 4D2 + 5D − 2)y = 4 cos x + sin x
15) (D3 − 9D)y = 8xex
12) y = c1 ex + c2 e−x + c3 e−2x + 2x3 − 3x2 + 15x − 8
√
√
√
13) y = c1 + c2 x + c3 e−x cos 3x + c4 e−x sin 3x + 5xe−x cos 3x
14) y = c1 ex + c2 xex + c3 e2x + 0.2 cos x + 0.9 sin x
3
15) y = c1 + c2 e3x + c3 e−3x + ex − xex
4
26. 42
CHAPTER 6. SERIES SOLUTIONS
term wise, i.e.
∞
(an ± bn )(x − x0 )n
f (x) ± g(x) =
n=0
• In the interval of convergence, the series can be multiplied or divided
to give another power series.
Chapter 6
∞
cn (x − x0 )n
f (x)g(x) =
Series Solutions
n=0
where
cn = a0 bn + a1 bn−1 + · · · + an b0
If none of the methods we have studied up to now works for a differential
equation, we may use power series. This is usually the only choice if the
solution cannot be expressed in terms of the elementary functions. (That
is, exponential, logarithmic, trigonometric and polynomial functions). If the
solution can be expressed as a power series, in other words, if it is analytic,
this method will work. But it takes time and patience to reach the solution.
Remember, we are dealing with infinitely many coefficients!
6.1
Power Series
• In the interval of convergence, derivatives and integrals of f (x) can be
found by term wise differentiation and integration, for example
∞
n an (x − x0 )n−1
f (x) = a1 + 2a2 (x − x0 ) + · · · =
n=1
(n)
(x
• The series ∞ f n! 0 ) (x − x0 )n is called the Taylor Series of the funcn=0
tion f (x). The function f (x) is called analytic if its Taylor series
converges.
Examples of some common power series are:
∞
Let’s remember some facts about the series
ex =
∞
an (x − x0 )n = a0 + a1 (x − x0 ) + a2 (x − x0 )2 + · · ·
n=0
(6.1)
n=0
∞
cos x =
n=0
from calculus.
∞
sin x =
• There is a nonnegative number ρ, called the radius of convergence, such
that the series converges absolutely for |x − x0 | < ρ and diverges for
|x − x0 | > ρ . The series defines a function f (x) = ∞ an (x − x0 )n
n=0
in its interval of convergence.
n=0
1
=
1−x
41
=1+x+
(−1)n x2n
2n!
=1−
x2
+ ···
2!
x2 x4
+
− ···
2!
4!
(−1)n x2n+1
x3 x5
=x−
+
− ···
(2n + 1)!
3!
5!
∞
xn
= 1 + x + x2 + · · ·
(−1)n+1 xn
n
=x−
n=0
∞
ln(1 + x) =
• In the interval of convergence, the series can be added or subtracted
xn
n!
n=1
x2 x3
+
− ···
2
3
27. 6.2. CLASSIFICATION OF POINTS
6.2
43
Classification of Points
44
CHAPTER 6. SERIES SOLUTIONS
Example 6.1 Solve y + 2xy + 2y = 0 around x0 = 0.
First we should classify the point. Obviously, x = 0 is an ordinary point, so
we can use power series method.
Consider the equation
R(x)y + P (x)y + Q(x)y = 0
(6.2)
∞
∞
n
y=
If both of the functions
P (x)
,
R(x)
Q(x)
R(x)
an x , y =
n=0
(6.3)
are analytic at x = x0 , then the point x0 is an ordinary point. Otherwise, x0
is a singular point.
Suppose that x0 is a singular point of the above equation. If both of the
functions
Q(x)
P (x)
, (x − x0 )2
(6.4)
(x − x0 )
R(x)
R(x)
are analytic at x = x0 , then the point x0 is called a regular singular point.
Otherwise, x0 is an irregular singular point.
For example, the functions 1+x+x2 , sin x, ex (1+x4 ) cos x are all analytic
cos x 1 ex 1 + x2
at x = 0. But, the functions
, ,
,
are not.
x
x x
x3
We will use power series method around ordinary points and Frobenius’
method around regular singular points. We will not consider irregular singular points.
∞
nan x
n−1
n(n − 1)an xn−2
, y =
n=1
n=2
Inserting these in the equation, we obtain
∞
∞
n(n − 1)an x
n−2
∞
+ 2x
n=2
nan x
n−1
an x n = 0
+2
n=1
∞
n=0
∞
∞
n(n − 1)an xn−2 +
n=2
2nan xn +
n=1
2an xn = 0
n=0
To equate the powers of x, let us replace n by n + 2 in the first sigma.
(n → n + 2)
∞
∞
∞
n=1
n=0
2an xn = 0
2nan xn +
(n + 2)(n + 1)an+2 xn +
n=0
Now we can express the equation using a single sigma, but we should start
the index from n = 1. Therefore we have to write n = 0 terms separately.
∞
[(n + 2)(n + 1)an+2 + (2n + 2)an ] xn = 0
2a2 + 2a0 +
n=1
6.3
Power Series Method
If x0 is an ordinary point of the equation R(x)y + P (x)y + Q(x)y = 0, then
the general solution is
∞
an (x − x0 )n
y=
−2(n + 1)
−2
an =
an
(n + 2)(n + 1)
(n + 2)
This is called the recursion relation. Using it, we can find all the constants
in terms of a0 and a1 .
a2 = −a0 , an+2 =
(6.5)
2
1
a4 = − a2 = a0
4
2
2
1
a6 = − a4 = − a0
6
6
2
a3 = − a1 ,
3
2
4
a 5 = − a3 = a1
5
15
n=0
The coefficients an can be found by inserting y in the equation and setting
the coefficients of all powers to zero. Two coefficients (Usually a0 and a1 )
must be arbitrary, others must be defined in terms of them. We expect two
linearly independent solutions because the equation is second order linear.
We can find as many coefficients as we want in this way. Collecting them
together, the solution is :
1
1
y = a0 1 − x 2 + x 4 − x 6 + · · ·
2
6
2
4
+ a1 x − x 3 + x 5 + · · ·
3
15
28. 6.3. POWER SERIES METHOD
45
In most applications, we want a solution close to 0, therefore we can neglect
the higher order terms of the series.
Remark: Sometimes we can express the solution in closed form (in terms
of elementary functions rather than an infinite summation) as in the next
example:
Example 6.2 Solve (x − 1)y + 2y = 0 around x0 = 0.
Once again, first we should classify the given point. The function
analytic at x = 0, therefore x = 0 is an ordinary point.
∞
∞
∞
an x n , y =
y=
n=0
2
is
x−1
nan xn−1 , y =
n=1
n(n − 1)an xn−2
n=2
Inserting these in the equation, we obtain
∞
(x − 1)
∞
n(n − 1)an x
n−2
nan xn−1 = 0
+2
n=2
n=1
∞
∞
∞
n(n − 1)an xn−2 +
n(n − 1)an xn−1 −
n=1
n=2
n=2
2nan xn−1 = 0
46
CHAPTER 6. SERIES SOLUTIONS
Exercises
Find the general solution of the following differential equations in the
form of series. Find solutions around the origin (use x0 = 0). Write the
solution in closed form if possible.
1) (1 − x2 )y − 2xy = 0
2) y + x4 y + 4x3 y = 0
3) (2 + x3 )y + 6x2 y + 6xy = 0
4) (1 + x2 )y − xy − 3y = 0
5) (1 + 2x2 )y + xy + 2y = 0
6) y − xy + ky = 0
7) (1 + x2 )y − 4xy + 6y = 0
8) (1 − 2x2 )y + (2x + 4x3 )y − (2 + 4x2 )y = 0
9) (1 + 8x2 )y − 16y = 0
10) y + x2 y = 0
The following equations give certain special functions that are very important in applications. Solve them for n = 1, 2, 3 around origin. Find
polynomial solutions only.
To equate the powers of x, let us replace n by n+1 in the second summation.
∞
∞
n(n − 1)an x
n−1
−
n=2
∞
(n + 1)nan+1 x
n−1
+
n=1
2nan x
n−1
=0
n=1
Now we can express the equation using a single sigma.
11)
12)
13)
14)
(1 − x2 )y − 2xy + n(n + 1)y = 0
y − 2xy + 2ny = 0
xy + (1 − x)y + ny = 0
(1 − x2 )y − xy + n2 y = 0
(Legendre’s Equation)
(Hermite’s Equation)
(Laguerre’s Equation)
(Chebyshev’s Equation)
∞
[(n(n − 1) + 2n)an − n(n + 1)an+1 ] xn−1 = 0
(−2a2 + 2a1 ) +
n=2
a2 = a1 , an+1 =
n2 − n + 2n
an for n
n(n + 1)
2
So the recursion relation is:
an+1 = an
All the coefficients are equal to a1 , except a0 . We have no information about
it, so it must be arbitrary. Therefore, the solution is:
y = a0 + a1 x + x 2 + x 3 + · · ·
x
y = a0 + a1
1−x
Solve the following initial value problems. Find the solution around the
point where initial conditions are given.
15)
16)
17)
18)
xy + (x + 1)y − 2y = 0,
y + 2xy − 4y = 0,
4y + 3xy − 6y = 0,
(x2 − 4x + 7)y + y = 0,
x0
x0
x0
x0
= −1,
= 0,
=0
=2
y(−1) = 1,
y(0) = 1,
y(0) = 4,
y(2) = 4,
y (−1) = 0
y (0) = 0
y (0) = 0
y (2) = 10
19) Find the recursion relation for (p + x2 )y + (1 − q − r)xy + qry = 0
around x = 0. (Here p, q, r are real numbers, p = 0)
20) Solve (1 + ax2 )y + bxy + cy = 0 around x0 = 0
29. EXERCISES
47
Answers
x3 x5
+
+ ···
3
5
n=3
x3 x6 x9
+
−
+ ···
2
4
8
a1 x
+
3
1+ x
2
3) y = a0 1 −
a0
3
1+ x
2
3
1
3
4) y = a0 1 + x2 + x4 − x6 + · · ·
2
8
16
2
2
5) y = a0 1 − x2 + x4 − x6 + · · ·
3
3
6) y = a0 1 −
+a1 x −
2
+ a1 x + x3
3
17
1
+ a1 x − x 3 + x 5 + · · ·
2
40
k − 1 3 (k − 1)(k − 3) 5 (k − 1)(k − 3)(k − 5) 7
x +
x −
x + ···
3!
5!
7!
x3
3
x4 x6
1+x +
+
+ ···
2
6
2
10) y = a0 1 −
⇒
⇒
n=3
⇒
x8
x4
+
+ ···
12 672
+ a1 x
y = a1 x
y = a0 (1 − 3x2 )
5
y = a1 (x − x3 )
3
+ a1 x −
⇒
14) n = 1
n=2
⇒
⇒
⇒
x5
x9
+
+ ···
20 1440
y = a1 x
y = a0 (1 − 2x2 )
2
y = a1 (x − x3 )
3
y = a0 (1 − x)
1
y = a0 (1 − 2x + x2 )
2
3
1
y = a0 (1 − 3x + x2 − x3 )
2
6
y = a1 x
y = a0 (1 − 2x2 )
4
y = a1 (x − x3 )
3
1
1
15) y = 1 − (x + 1)2 − (x + 1)3 − (x + 1)4 − · · ·
3
6
16) y = 1 + 2x2
17) y = 4 + 3x2
1
1
18) y = 4 1 − (x − 2)2 + (x − 2)4 + · · ·
6
72
1
7
+ 10 (x − 2) − (x − 2)3 +
(x − 2)5 + · · ·
18
1080
19) an+2 = −
8
64
9) y = a0 (1 + 8x2 ) + a1 x + x3 − x5 + · · ·
3
15
11) n = 1
n=2
OR
k 2 k(k − 2) 4 k(k − 2)(k − 4) 6
x +
x −
x + ···
2!
4!
6!
7) y = a0 (1 − 3x2 ) + a1 x −
8) y = a0
x4 x7 x10
+
−
+ ···
2
4
8
⇒
n=3
+ a1 x −
⇒
n=3
x5
x10
x15
+
−
+ ···
5
5 · 10 5 · 10 · 15
x11
x16
x6
+
−
+ ···
x−
6
6 · 11 6 · 11 · 16
⇒
n=2
1 1+x
ln
2 1−x
⇒
⇒
13) n = 1
OR y = a0 + a1
2) y = a0 1 −
y=
CHAPTER 6. SERIES SOLUTIONS
12) n = 1
n=2
1) y = a0 + a1 x +
+a1
48
(n − q)(n − r)
an
p(n + 2)(n + 1)
x4
x2
+ c(2a + 2b + c)
2
4!
x6
+ ···
− c(2a + 2b + c)(12a + 4b + c)
6!
x3
x5
+ a1 x − (b + c)
+ (b + c)(6a + 3b + c)
3!
5!
x7
−(b + c)(6a + 3b + c)(20a + 5b + c)
+ ···
7!
20) y = a0 1 − c
30. 50
CHAPTER 7. FROBENIUS’ METHOD
Case 2 - Equal roots: A basis of solutions is
∞
∞
an x n ,
y 1 = xr
y2 = y1 ln x + xr
bn x n
(7.3)
n=1
n=0
Case 3 - Roots differing by an integer: A basis of solutions is
Chapter 7
∞
∞
an xn ,
y 1 = xr 1
bn x n
y2 = ky1 ln x + xr2
n=1
(7.4)
n=0
In this chapter, we will extend the methods of the previous chapter to regular
singular points. The calculations will be considerably longer, but the basic
ideas are the same. The classification of the given point is necessary to make
a choice of methods.
where r1 − r2 = N > 0 (r1 is the greater root) and k may or may not be zero.
In all three cases, there is at least one relatively simple solution of the
form y = xr ∞ an xn . The equation is second order, so there must be a
n=0
second linearly independent solution. In Cases 2 and 3, it may be difficult
to find the second solution. You may use the method of reduction of order.
This is convenient especially if y1 is simple enough. Alternatively, you may
use the above formulas directly, and determine bn one by one using the an
and the equation.
7.1
7.2
Frobenius’ Method
An Extension of Power Series Method
Suppose x0 is a regular singular point. For simplicity, assume x0 = 0. Then
p(x)
q(x)
the differential equation can be written as y +
y + 2 y = 0 where
x
x
p(x) and q(x) are analytic. We can try a solution of the form
∞
y = xr
an x n
(7.1)
n=0
The equation corresponding to the lowest power xr−2 , in other words
r(r − 1) + p0 r + q0 = 0 is called the indicial equation, where p0 = p(0), and
q0 = q(0). Now we can find r, insert it in the series formula, and proceed as
we did in the previous chapter.
We can classify the solutions according to the roots of the indicial equation.
Case 1 - Distinct roots not differing by an integer: A basis of solutions
is
∞
∞
y 1 = xr 1
an x n ,
y2 = xr2
bn x n
n=0
n=0
49
(7.2)
Examples
Example 7.1 Solve 4xy + 2y + y = 0 around x0 = 0.
2
First we should classify the given point. The function 4x is not analytic at
x = 0 therefore x = 0 is a singular point. We should make a further test to
determine whether it is regular or not.
x2
The functions 2x and 4x are analytic therefore x = 0 is a R.S.P., we can
4x
use the method of Frobenius.
∞
∞
an xn+r , y =
y=
n=0
∞
(n + r)an xn+r−1 , y =
n=0
(n + r)(n + r − 1)an xn+r−2
n=0
Note that the summation for the derivatives still starts from 0, because r
does not have to be an integer. This is an important difference between
methods of power series and Frobenius.
Inserting these in the equation, we obtain
∞
∞
(n + r)(n + r − 1)an xn+r−2 + 2
4x
n=0
∞
(n + r)an xn+r−1 +
n=0
an xn+r = 0
n=0
31. 7.2. EXAMPLES
51
∞
∞
4(n + r)(n + r − 1)an x
n+r−1
∞
+
n=0
2(n + r)an x
n=0
n+r−1
an xn+r = 0
+
52
CHAPTER 7. FROBENIUS’ METHOD
For simplicity, we may choose a0 = 1. Then
an =
We want to equate the powers of x, so n → n + 1 in the first two terms.
∞
∞
∞
4(n + r + 1)(n + r)an+1 xn+r +
n=−1
2(n + r + 1)an+1 xn+r +
n=−1
an xn+r = 0
Therefore the second solution is :
∞
n=0
Now we can express the equation using a single sigma, but the index of the
common sigma must start from n = 0. Therefore we have to write n = −1
terms separately.
[4r(r−1)+2r]a0 xr−1 +
{[4(n + r + 1)(n + r) + 2(n + r + 1)]an+1 + an } xn+r = 0
n=0
We know that a0 = 0, therefore 4r2 − 2r = 0. This is the indicial equation.
Its solutions are r = 0, r = 1 . Therefore this is Case 1.
2
If r = 0, the recursion relation is
n=0
The general solution is y = c1 y1 + c2 y2
2
First we should classify the given point. The function x x−x is not analytic
2
at x = 0 therefore x = 0 is a singular point. The functions x − 1 and
1 + x are analytic at x = 0 therefore x = 0 is a R.S.P., we can use the
method of Frobenius. Evaluating the derivatives of y and inserting them in
the equation, we obtain
−1
1 an
4(n + 1)(n + 2 )
an+1 =
∞
For simplicity, we may choose a0 = 1. Then
n=0
∞
−
(n + r)an x
(−1)
2n!
y1 =
n=0
n=0
+
an x
n+r
∞
an xn+r+1 = 0
+
n=0
n=0
∞
∞
(n + r)(n + r − 1)an x
n=0
∞
n n
(n + r)an xn+r+1
+
Let’s replace n by n − 1 in the second and fifth terms.
Therefore the first solution is:
∞
n+r
∞
n+r
n=0
n
an =
∞
(n + r)(n + r − 1)an x
a0
a1
a0
a2
a0
a1 = − , a 2 = −
, a3 = −
,...
3 =
5 = −
2
4!
6!
4.2. 2
4.3. 2
−
√
(−1) x
= cos x
2n!
1
If r = , the recursion relation is
2
a1 = −
√
(−1)n xn
= sin x
(2n + 1)!
y2 = x1/2
Example 7.2 Solve x2 y + (x2 − x)y + (1 + x)y = 0 around x0 = 0.
∞
an+1 =
(−1)n
(2n + 1)!
n=0
−1
−an
an =
(2n + 3)(2n + 2)
4(n + 3 )(n + 1)
2
a0
a1
a0
a2
a0
, a2 = −
= , a3 = −
= − ,...
3.2
5.4
5!
7.6
7!
n+r
n=1
∞
(n + r)an x
n=0
n+r
an x
+
n=0
(n + r − 1)an−1 xn+r
+
n+r
∞
an−1 xn+r = 0
+
n=1
[r2 − 2r + 1]a0 xr +
∞
{[(n + r)(n + r − 1) − (n + r) + 1]an + [(n + r − 1) + 1]an−1 } xn+r = 0
n=1
The indicial equation is r2 − 2r + 1 = 0 ⇒ r = 1 (double root). Therefore
this is Case 2. The recursion relation is
an = −
n+1
an−1
n2
32. 7.2. EXAMPLES
53
54
CHAPTER 7. FROBENIUS’ METHOD
Exercises
For simplicity, let a0 = 1. Then
3
3
4
2
a1 = −2, a2 = − a1 = , a3 = − a2 = −
4
2
9
3
Find two linearly independent solutions of the following differential equations in the form of series. Find solutions around the origin (use x0 = 0).
Write the solution in closed form if possible.
1) 2x2 y − xy + (1 + x)y = 0
Therefore the first solution is :
3
2
y1 = x 1 − 2x + x2 − x3 + · · ·
2
3
2) 2xy + (1 + x)y − 2y = 0
To find the second solution, we will use reduction of order. Let y2 = uy1 .
Inserting y2 in the equation, we obtain
3) (x2 + 2x)y + (3x + 1)y + y = 0
4) xy − y − 4x3 y = 0
2
2
2
x y1 u + (2x y1 − xy1 + x y1 )u = 0
Let w = u then
−2
To evaluate the integral u =
1−x+
w=
u=
1
x
⇒
w=
8) (2x2 + 2x)y − y − 4y = 0
xe−x
2
y1
9) 2x2 y + (2x2 − x)y + y = 0
1
w dx we need to find 2 . This is also a series.
y1
3
2
1 − 2x + x2 − x3 + · · ·
2
3
xe−x
=x
2
y1
7) x2 y + (x2 − x)y + y = 0
y1 1
+ − 1 dx
y1 x
ln w = −2 ln y1 + ln x − x
w=
6) 3x2 y + (−10x − 3x2 )y + (14 + 4x)y = 0
1
y
w + 2 1 − +1 w =0
y1 x
dw
=
w
1
1
= 2
2
y1
x
5) xy + y − xy = 0
−2
=
x2 x3
−
+ ···
2!
3!
1 + 3x +
1
x2
1
x2
1 + 4x + 9x2 +
1 + 4x + 9x2 +
11 2 13 3
x + x + ···
2
6
w dx = ln x + 3x +
11 2 13 3
x + x + ···
4
18
13
3
y2 = uy1 = y1 ln x + x 3x − x2 + x3 + · · ·
4
2
46 3
x + ···
3
46 3
x + ···
3
10) 4x2 y + (2x2 − 10x)y + (12 − x)y = 0
11) (x2 + 2x)y + (4x + 1)y + 2y = 0
Use Frobenius’ method to solve the following differential equations around
origin. Find the roots of the indicial equation, find the recursion relation,
and two linearly independent solutions.
12) (x2 + cx)y + [(2 + b)x + c(1 − d)]y + by = 0
(b = 0, c = 0, d is not an integer).
13) x2 y + [(1 − b − d)x + cx2 ]y + [bd + (1 − b)cx]y = 0
(c = 0, b − d is not an integer).
14) x2 y + [(1 − 2d)x + cx2 ]y + (d2 + (1 − d)cx)y = 0
(c = 0)
15) xy + [1 − d + cx2 ]y + 2cxy = 0
(c = 0, d is not an integer).
33. EXERCISES
55
∞
n=1
(−1)n xn
n! · 3 · 5 · 7 · · · (2n + 1)
1
1+
n=1
2) y = c1
(−1)n xn
n! · 1 · 3 · 5 · · · (2n − 1)
1
1 + 2x + x2
3
∞
+ c2 x
1
2
4) y = a0
n=0
1+
n=1
6
2
3) y1 = 1−x+ x2 − x3 +· · · ,
3
15
∞
n+b
an
c (n + 1 − d)
b
b(b + 1)
y1 = 1 −
x+ 2
x2 − · · ·
c(1 − d)
c (1 − d)(2 − d)
12) r = 0
∞
+ c2 x 2
(−1)n 3xn
2n n!(2n − 3)(2n − 1)(2n + 1)
3
15
35 3
y2 = x1/2 1 − x + x2 −
x + ···
4
32
128
2
x4
x6
x2
+
+
+ ···
22 (2 · 4)2 (2 · 4 · 6)2
x2
3x4
11x6
y2 = y1 ln x −
−
−
− ···
4
8 · 16 64 · 6 · 36
9
27 3
3
6) y1 = x7/3 1 + x + x2 +
x + ···
4
28
280
x2 x3
y 2 = x2 1 + x +
+
+ · · · = x2 e x
2!
3!
x
x
−
+ · · · = xe−x
2!
3!
x2
x3
y2 = xe−x ln x + xe−x x +
+
+ ···
2 · 2! 3 · 3!
1
1
1
y2 = x3/2 1 + x − x2 + x3 − · · ·
2
8
16
∞
y2 = x 1 +
n=1
(−1)n (2x)n
1 · 3 · 5 · · · (2n + 1)
∞
10) y1 = x2 e−x/2 ,
⇒
an+1 = −
y2 = x3/2 1 +
n=1
(−1)n xn
1 · 3 · 5 · · · (2n − 1)
an+1 = −
⇒
an = −
n+b+d
an
c (n + 1)
d+b
(d + b)(d + b + 1) 2
1−
x+
x − ···
c
2! c2
y 2 = xd
c
an−1
n+b−d
c2
c
x+
x2 − · · ·
1+b−d
(1 + b − d)(2 + b − d)
c
r = d ⇒ an = − an−1
n
c2 2 c3 3
y2 = xd 1 − c x + x − x + · · · = xd e−cx
2!
3!
y 1 = xb 1 −
c
an−1
n
c3
c2
1 − c x + x2 − x3 + · · ·
2!
3!
14) r = d (double root) an = −
y 1 = xd
3
7) y1 = x 1 − x +
8) y1 = 1 − 4x − 8x2 ,
r=d
2
OR y = c1 ex + c2 e−x
5) y1 = 1 +
2
⇒
13) r = b
∞
x4n
x4n+2
+ a2
,
(2n)!
(2n + 1)!
n=0
9) y1 = x1/2 e−x ,
CHAPTER 7. FROBENIUS’ METHOD
8
11) y1 = 1 − 2x + 2x2 − x3 + · · ·
5
5
35 2 105 3
1/2
1− x+
y2 = x
x −
x + ···
4
32
128
Answers
1) y = c1 x 1 +
56
y2 = xd e−cx
ecx
dx
x
y2 = xd e−cx ln x + xd e−cx cx +
15) r = 0
⇒
= xd e−cx
an+2 = −
c2 2
c3 3
x +
x + ···
2 · 2!
3 · 3!
c
an
(n + 2 − d)
c
c2
c3
x2 +
x4 −
x6 + · · ·
2−d
(2 − d)(4 − d)
(2 − d)(4 − d)(6 − d)
c
r = d ⇒ an+2 = −
an
n+2
c
c2 4
c3
y 2 = xd 1 − x2 +
x −
x6 + · · ·
2
2·4
2·4·6
y1 = 1 −
34. 58
CHAPTER 8. LAPLACE TRANSFORM I
Example 8.1 Evaluate the Laplace transform of the following functions:
a) f (t) = 1
∞
0
Chapter 8
0
1
= ,
s
s>0
b) f (t) = eat
∞
Laplace Transform I
0
c) f (t) =
Laplace transform provides an alternative method for many equations. We
first transform the differential equation to an algebraic equation, then solve
it, and then make an inverse transform. Laplace transform has a lot of
interesting properties that make these operations easy. In this chapter, we
will see the definition and the basic properties. We will also compare this
method to the method of undetermined coefficients, and see in what ways
Laplace transform is more convenient.
0
1
if
if
∞
1
∞
e−st f (t)dt
(8.1)
0
then, the inverse transform will be
f (t) = L−1 {F (s)}
(8.2)
Note that we use lowercase letters for functions and capital letters for their
transforms.
57
=
0
1
, s>a
s−a
∞
=
1
e−s
,
s
s>0
d) f (t) = t
∞
te−st dt
L {t} =
0
Using integration by parts, we obtain
L {t} = −
The Laplace transform of a function f (t) is defined as:
e−st
−s
e−st dt =
L {f } =
Definition, Existence and Inverse of Laplace
Transform
∞
0<t<1
1 t
L {t} = −t
F (s) = L {f (t)} =
e(a−s)t
a−s
eat e−st dt =
L eat =
8.1
∞
e−st
−s
e−st dt =
L {1} =
e−st
s
∞
e−st
s2
∞
∞
+
0
0
=
0
e−st
dt
s
1
, s>0
s2
The integral that defines the Laplace transform is an improper integral,
it may or may not converge. In the above examples, the transform is defined
for a certain range of s.
In practice, we can use Laplace transform on most of the functions we
encounter in differential equations. The following definitions and the theorem
answer the question Which functions have a Laplace transform?
Piecewise Continuous Functions: A function f (t) is piecewise continuous
on [a, b] if the interval can be subdivided into subintervals [ti , tj ],
a = t0 < t1 < t2 · · · < tn = b such that f (t) is continuous on each interval
and has finite one-sided limits at the endpoints (from the interior).
An example can be seen on Figure 8.1.
35. 8.2. PROPERTIES
59
60
CHAPTER 8. LAPLACE TRANSFORM I
• Transform of Derivatives
L {f }
L {f }
L f (n)
= sL {f } − f (0)
= s2 L {f } − sf (0) − f (0)
= sn L {f } − sn−1 f (0) − sn−2 f (0) − · · · − f (n−1) (0)
• Transform of Integrals
t
L
f (x) dx
=
0
F (s)
s
Example 8.2 Find the Laplace transform of sin at and cos at. Hint: Use
Euler’s formula eix = cos x + i sin x and linearity.
Figure 8.1: A piecewise continuous function
Exponential Order: f (t) is of exponential order as t → ∞ if there exist
real constants M, c, T such that |f (t)| M ect for all t T . In other words,
a function is of exponential order if it does not grow faster than ect .
Theorem 8.1: If f (t) is of exponential order and piecewise continuous on
[0, k] for all k > 0, then its Laplace transform exists for all s > c.
For example, all the polynomials have a Laplace transform. The function
t2
e does NOT have a Laplace transform.
8.2
Basic Properties of Laplace Transforms
It is difficult to evaluate the Laplace transform of each function by performing an integration. Instead of this, we use various properties of Laplace
transform.
Let L {f (t)} = F (s), then, some basic properties are: (assuming these
transforms exists)
• Linearity
L {af + bg} = aL {f } + bL {g}
• Shifting
L {eiat } − L {e−iat }
2i
1
1
1
a
L {sin at} =
−
= 2
2i s − ia s + ia
s + a2
Similarly, we can show that the transform of f (t) = cos at is
s
F (s) = 2
s + a2
1
Example 8.3 Find the inverse Laplace transform of F (s) =
.
(s + 5)2
Hint: Use shifting.
sin at =
eiat − e−iat
2i
We know that L−1
1
s2
L−1 {F (s − a)} = eat f (t)
L {sin at} =
= t. Therefore
1
(s + 5)2
L−1
= te−5t
Example 8.4 Find the Laplace transform of f (t) = t2 . Hint: Use Derivatives.
Using L {f } = sL {f } − f (0), we obtain
L {2t}
2
= 3
s
s
Example 8.5 Find the Laplace transform of f (t) = t3 . Hint: Use Integrals.
L {2t} = sL t2 − 0
⇒
L t2 =
Using the integral rule, we see that
L
L eat f (t) = F (s − a)
⇒
L {t2 }
2
= 4
s
s
6
L t3 = 4
s
t3
3
=
36. 8.3. INITIAL VALUE PROBLEMS
8.3
61
Initial Value Problems
Consider the constant-coefficient equation
y + ay + by = r(t)
(8.3)
y(0) = p, y (0) = q
62
CHAPTER 8. LAPLACE TRANSFORM I
The only disadvantage is that, sometimes finding the inverse Laplace
transform is too difficult.
We have to find roots of the polynomial s2 + as + b, which is the same as
the characteristic polynomial we would encounter if we were using method
of undetermined coefficients.
(8.4)
with initial values
Here y is a function of t (y = y(t)). We can solve it by the method of undetermined coefficients. The method of Laplace transform will be an alternative
that is more efficient in certain cases. It also works for discontinuous r(t).
Let us evaluate the Laplace transform of both sides.
Example 8.6 Solve the initial value problem
y + 4y = 0, y(0) = 5, y (0) = 3.
Let’s start by finding the transform of the equation.
L {y } + 4L {y} = 0
L {y } + aL {y } + bL {y} = L {r(t)}
(8.5)
Using L {y} = Y (s) and L {r(t)} = R(s)
s2 Y − sp − q + a(sY − p) + bY = R
(8.6)
(s2 + as + b)Y = R + (s + a)p + q
s2 Y − 5s − 3 + 4Y = 0
⇒ (s2 + 4)Y = 5s + 3
5s + 3
Y = 2
s +4
Now, we have to find the inverse transform of Y to obtain y(t).
(8.7)
Y =
y = L−1
R + (s + a)p + q
s2 + as + b
R + sp + ap + q
s2 + as + b
(8.8)
(8.9)
Y =
5s
3 2
+
+ 4 2 s2 + 4
s2
3
sin 2t
2
Note that we did not first find the general solution containing arbitrary constants. We directly found the result.
y(t) = L−1 {Y } = 5 cos 2t +
Note that this method can be generalized to higher order equations. The
advantages compared to the method of undetermined coefficients are:
Example 8.7 Solve the initial value problem
• The initial conditions are built in the solution, we don’t need to determine constants after obtaining the general solution.
• There is no distinction between homogeneous and nonhomogeneous
equations, or single and multiple roots. The same method works in all
cases the same way.
y − 4y + 3y = 1,
y(0) = 0,
y (0) = −
Transform both sides:
L {y − 4y + 3y} = L {1}
Use the derivative rule
• The function on the right hand side r(t) belongs to a wider class. For
example, it can be discontinuous.
s2 Y − s.0 +
1
1
− 4(sY − 0) + 3Y =
3
s
1
3
37. 8.3. INITIAL VALUE PROBLEMS
Isolate Y
63
3−s
1 1
− =
s 3
3s
s−3
(s − 1)(s − 3)Y = −
3s
1
1 1
1
=
−
Y =−
3s(s − 1)
3 s s−1
64
CHAPTER 8. LAPLACE TRANSFORM I
(s2 − 4s + 3)Y =
f (t)
F (s)
f (t)
F (s)
1
1
s
eat − ebt
a−b
1
(s − a)(s − b)
t
1
s2
aeat − bebt
a−b
s
(s − a)(s − b)
n!
eat sin bt
b
(s − a)2 + b2
s−a
(s − a)2 + b2
Find the inverse transform
y(t) = L−1 {Y } =
1 1 t
− e
3 3
As you can see, there’s no difference between homogeneous and nonhomogeneous equations. Laplace transform works for both types in the same way.
tn
sn+1
eat
Example 8.8 Solve the initial value problem
y + 4y + 4y = 42te−2t , y(0) = 0, y (0) = 0
1
s−a
eat cos bt
teat
1
(s − a)2
a
s 2 + a2
s
s 2 + a2
tn eat
−2t
L {y } + 4L {y } + 4L {y} = 42L te
1
s2 Y + 4sY + 4Y = 42 ·
(s + 2)2
42
(s2 + 4s + 4)Y =
(s + 2)2
42
Y =
(s + 2)4
sin at
cos at
sinh at
cosh at
n!
(s − a)n+1
a
s 2 − a2
s
s 2 − a2
42 3 −2t
te
3!
2as
(s2 + a2 )2
t sinh at
2as
(s2 − a2 )2
t cos at
s 2 − a2
(s2 + a2 )2
t cosh at
s 2 + a2
(s2 − a2 )2
sin at − at cos at
y(t) = L−1 {Y (s)} =
t sin at
2a3
(s2 + a2 )2
sin at + at cos at
2as2
(s2 + a2 )2
y(t) = 7t3 e−2t
If you try the method of undetermined coefficients on this problem, you will
appreciate the efficiency of Laplace transforms better.
Table 8.1: A Table of Laplace Transforms
38. EXERCISES
65
Exercises
3) f (t) = 2e−t cos2 t
4) f (t) = (t + 1)2 et
5) f (t) = t3 e3t
6) f (t) =
7) f (t) =
CHAPTER 8. LAPLACE TRANSFORM I
Answers
Find the Laplace transform of the following functions:
t
2) f (t) = et sin 3t
1) f (t) = cos2 2
t 0<t<a
0
a<t
66
1 0<t<a
0
a<t
t 0<t<a
8) f (t) =
1 a<t<b
0
b<t
1) F (s) =
2) F (s) =
3) F (s) =
4) F (s) =
5) F (s) =
1
s
+ 2
2s 2s + 2
3
(s − 1)2 + 9
1
s+1
+
s + 1 s2 + 2s + 5
2
2
1
+
+
(s − 1)3 (s − 1)2 s − 1
6
(s − 3)4
1 − e−as
s
1
ae−as e−as
7) F (s) = 2 −
− 2
s
s
s
−as
−as
1−e
e
− ae−as − e−bs
8) F (s) =
+
s2
s
9) f (t) = cosh 2t − 2 sinh 2t
6) F (s) =
Find the inverse Laplace transform of the following functions:
s−4
9) F (s) = 2
s −4
3
10) F (s) =
(s − 2)2
6
11) F (s) =
s(s + 4)
1
12) F (s) =
s(s2 + 9)
1
13) F (s) = 2
s (s + 1)
5s + 1
14) F (s) = 2
s +4
15) F (s) =
1
s+8
1
16) F (s) =
(s − a)n
Solve the following initial value problems using Laplace transform:
17) y − 2y + y = 0,
y(0) = 4, y (0) = −3
18) y − 2y + 2y = 0, y(0) = 0, y (0) = 1
19) y + 2y = 4t2 + 12, y(0) = 4, y (0) = 0
20) y + 6y + 9y = e−3t , y(0) = 0, y (0) = 0
10) f (t) = 3te2t
11) f (t) = (3 − 3e−4t )/2
12) f (t) = (1 − cos 3t)/9
13) f (t) = e−t + t − 1
1
14) f (t) = 5 cos 2t + sin 2t
2
−8t
15) f (t) = e
16) f (t) =
tn−1 eat
(n − 1)!
17) y(t) = 4et − 7tet
18) y(t) = et sin t
19) y(t) = 4 + 2t2
1
20) y(t) = e−3t t2
2
39. 68
CHAPTER 9. LAPLACE TRANSFORM II
Reversing the order of integration, we obtain:
∞
∞
f (x) g(t − x)e−st dt dx
=
0
x
Making the substitution u = t − x, we obtain:
∞
∞
f (x) g(u)e−su−sx dudx
L {f ∗ g} =
Chapter 9
0
∞
0
∞
0
0
Laplace Transform II
g(u)e−su du
f (x)e−sx dx
=
=F (s) G(s)
Example 9.1 Find the inverse Laplace transform of F (s) =
In this chapter, we will study more advanced properties of Laplace transform.
At the end, we will be able to find transform and inverse transform of a wider
range of functions. This will enable us to solve almost any linear constant
coefficient equation, including discontinuous inputs.
9.1
L−1
1
s2
1
s+4
= t, L−1
t
(9.1)
0
The convolution operation is commutative, in other words f ∗ g = g ∗ f
Theorem 9.1: The transform of convolution of two functions is equal to
the product of their transforms, i.e.
L {f ∗ g} = F (s) · G(s)
−1
L
{F (s) · G(s)} = f ∗ g
where L {f } = F (s) and L {g} = G(s).
Proof: Using the definitions of convolution and Laplace transform,
t
L {f ∗ g} =L
f (x) g(t − x) dx
0
∞
t
f (x) g(t − x)e−st dx dt
=
0
0
67
L−1
1
1
·
2 s+4
s
(9.2)
(9.3)
= t ∗ e−4t
xe−4(t−x) dx
0
xe4x e4x
−
4
16
−4t
t
1
e
= −
+
4 16
16
t
= e−4t
0
Example 9.2 Find the inverse Laplace transform of F (s) =
The convolution of two functions f and g is defined as
f (x)g(t − x) dx
⇒
1
.
+ 4s2
t
f (t) = t ∗ e−4t =
Convolution
h(t) = (f ∗ g)(t) =
= e−4t
s3
s
.
(s2 + 1)2
s
1
· 2
= L {cos t} · L {sin t},
+ 1) (s + 1)
we will see that f (t) = L−1 {F } = cos t ∗ sin t.
If we express F as F (s) =
(s2
t
cos(x) sin(t − x) dx
f (t) =
0
t
=
0
1
=
2
=
1
[sin(t − x + x) + sin(t − x − x)] dx
2
t
[sin(t) + sin(t − 2x)] dx
0
1
cos(t − 2x)
x sin t +
2
2
t
0
1
1
= t sin t + (cos t − cos t)
2
2
1
= t sin t
2
40. 9.2. UNIT STEP FUNCTION
9.2
69
70
CHAPTER 9. LAPLACE TRANSFORM II
Unit Step Function
∞
e−st f (t)dt
F (s) =
The Heaviside step function (or unit step function) is defined as
0
F (s) =
ua (t) = u(t − a) =
0
1
if
if
t<a
t a
(9.4)
This is a simple on off function. It is especially useful to express discontinuous inputs.
Theorem 9.2: [t−shifting] Let L {f (t)} = F (s), then
(9.5)
Proof: Using the definition,
∞
e−st f (t − a) u(t − a) dt
∞
e−st f (t − a) dt
a
∞
e−sa−sx f (x) dx
=
0
−as
=e
( where x = t − a)
F (s)
Example 9.3 Find the Laplace transform of g(t) =
0
t
if
if
t<5
t 5
We can express g(t) as g(t) = u(t − 5)f (t − 5) where f (t) = (t + 5). Then
F (s) = L {f (t)} =
9.3
In other words
L {tf (t)} = −F (s)
5
1
+
2
s
s
⇒
(9.7)
Repeating this procedure n times, we obtain:
dn
F (s)
dsn
Using the derivative formula, we find
L {t sin t} = −
=
f (t)dt
0
Example 9.4 Find the Laplace transform of f (t) = t sin t.
L {f (t − a) u(t − a)} = e−as F (s)
0
(−t)e
−st
L {tn f (t)} = (−1)n
Figure 9.1: u(t − a) and its effect on f (t)
L {f (t − a) u(t − a)} =
(9.6)
∞
L {g(t)} = e−5s
1
5
+
2
s
s
Differentiation of Transforms
If f (t) is piecewise continuous and of exponential order, then we can differentiate its Laplace transform integral.
d
ds
1
1 + s2
=
2s
(1 + s2 )2
(9.8)
41. 9.4. PARTIAL FRACTIONS EXPANSION
9.4
71
Partial Fractions Expansion
CHAPTER 9. LAPLACE TRANSFORM II
9.5
In many applications of Laplace transform, we need to expand a rational
function in partial fractions. Here, we will review this technique by examples.
2x + 1
A
B
C
=
+
+
(x − 2)(x + 3)(x − 1)
x−2 x+3 x−1
2
x + 4x − 5
B
C
A
D
+
+
=
+
(x − 2)(x − 1)3
x − 2 x − 1 (x − 1)2 (x − 1)3
x3 + 1
Dx + E
A Bx + C
+ 2
+ 2
=
2 + 4)2
x(x
x
x +4
(x + 4)2
3
A
B
x3 − 4x2 + x + 9
= x+1+ 2
=x+1+
+
x2 − 5x + 6
x − 5x + 6
x−2 x−3
• We can express any polynomial as a product of first and second order
polynomials.
• For second order polynomials in the expansion, we have to use Ax + B
(not simply a constant) in the numerator.
• If numerator’s degree is greater or equal to the denominator, we should
first divide them using polynomial division.
2
Example 9.5 Find the inverse Laplace transform of F (s) =
72
−s + 7s − 1
.
(s − 2)(s − 5)2
Applications
Now we are in a position to solve a wider class of differential equations using
Laplace transform.
Example 9.6 Solve the initial value problem
y − 6y + 8y = 2e2t , y(0) = 11, y (0) = 37
We will first find the Laplace transform of both sides, then find Y (s)
L {y } − 6L {y } + 8L {y} = L 2e2t
s2 Y − 11s − 37 − 6(sY − 11) + 8Y =
(s2 − 6s + 8)Y =
−s2 + 7s − 1 = A(s − 5)2 + B(s − 2)(s − 5) + C(s − 2)
Inserting s = 2, we see that 9 = 9A ⇒ A = 1.
Inserting s = 5, we see that 9 = 3C ⇒ C = 3.
The coefficient of s2 : A + B = −1 therefore B = −2. So
1
−s2 + 7s − 1
2
3
=
−
+
2
(s − 2)(s − 5)
s − 2 s − 5 (s − 5)2
Now we can easily find the inverse Laplace transform:
L−1 {F (s)} = e2t − 2e5t + 3te5t
2
+ 11s − 29
s−2
The factors of s2 − 6s + 8 are (s − 2) and (s − 4), so
Y =
2
11s − 29
+
(s − 2)(s − 2)(s − 4) (s − 2)(s − 4)
Y =
First, we have to express F (s) in terms of simpler fractions:
−s2 + 7s − 1
A
B
C
=
+
+
(s − 2)(s − 5)2
s − 2 s − 5 (s − 5)2
2
s−2
11s2 − 51s + 60
(s − 2)2 (s − 4)
Now we need to find the inverse Laplace transform. Using partial fractions
expansion
Y =
A
C
B
+
+
2
s − 2 (s − 2)
s−4
After some algebra we find that A = 3, B = −1, C = 8 so
Y (s) =
8
1
3
−
+
2
s − 2 (s − 2)
s−4
y(t) = L−1 {Y (s)} = 3e2t − te2t + 8e4t
42. 9.5. APPLICATIONS
73
Example 9.7 Solve the initial value problem
74
CHAPTER 9. LAPLACE TRANSFORM II
Example 9.8 Solve the initial value problem
y + y = f (t), y(0) = 0, y (0) = 3
where f (t) =
0
2 cos t
if
if
0 < t < 5π
5π < t
y + 2y + y = r(t), y(0) = 0, y (0) = 0
where r(t) =
As you can see, the input function is discontinuous, but this makes no
difference for Laplace transform.
L {y } + L {y} = L {f }
t
0
if
if
0<t<1
1<t
Once again we have a discontinuous input. This time we will use unit
step function. First, we have to express r(t) with a single formula.
r(t) = t − u(t − 1)t = t − u(t − 1)(t − 1) − u(t − 1)
2
s Y −3+Y =F
F +3
Y = 2
s +1
1
Using the fact that L {sin t} = 2
, we can obtain y(t) by convolution:
s +1
y(t) = L−1 {Y } = f (t) ∗ sin t + 3 sin t
Its Laplace transform is
R(s) = L {r(t)} =
Finding the Laplace transform of the equation, we obtain
(s2 + 2s + 1)Y = R
Using the definition of convolution,
t
f ∗ sin t =
Y =
f (x) sin(t − x) dx
0
If t < 5π, f = 0 therefore this integral is also zero. If t > 5π we have
Y =
t
f ∗ sin t =
1
e−s e−s
− 2 −
2
s
s
s
2 cos x sin(t − x) dx
s2 (s
R
(s + 1)2
1
e−s
− 2
2
+ 1)
s (s + 1)
Using partial fractions expansion
5π
Using the trigonometric identity 2 sin A cos B = sin(A + B) + sin(A − B) we
obtain
t
f ∗ sin t =
sin t + sin(t − 2x) dx
2
1
2
1
1
1
1
Y =− + 2 +
+
− e−s − + 2 +
s s
s + 1 (s + 1)2
s s
s+1
Using the fact that L−1 {e−as F (s)} = f (t − a)u(t − a), we obtain
5π
=
x sin t +
cos(t − 2x)
2
t
5π
= (t − 5π) sin t
Therefore
y(t) = −2 + t + 2e−t + te−t − u(t − 1) −1 + (t − 1) + e−(t−1)
We know that u(t − 1) = 0 for t > 1 and u(t − 1) = 1 for t > 1 so
y(t) =
y(t) =
3 sin t
(t − 5π + 3) sin t
if
if
0 < t < 5π
5π < t
−2 + t + 2e−t + te−t
(2 − e)e−t + te−t
if
if
0<t<1
1<t
43. EXERCISES
75
Exercises
1) F (s) =
3
s+3
(s2 + 4)2
8) F (s) =
2
2) F (s) =
s
s4 + 4a4
10) F (s) =
sin 4t − 4t cos 4t
128
4t sin 2t + 3 sin 2t − 6t cos 2t
7) f (t) =
16
1
t
9) f (t) = sin 2t + cos 2t
4
2
5) f (t) =
6) f (t) =
10) f (t) = e2t + 2 cos 3t +
12) y = 3 cos t + (4 + t) sin t
13) y = −25 + 8t2
Solve the following initial value problems : (where y = y(t))
1 −t
e − et cos t + 7et sin t
5
15) y = t − sin t
14) y =
11) y − y − 2y = 0, y(0) = 8, y (0) = 7
12) y + y = 2 cos t, y(0) = 3, y (0) = 4
13) y + 0.64y = 5.12t2 , y(0) = −25, y (0) = 0
1
0 < t < 2π
cos t
2π < t
16) y =
14) y − 2y + 2y = e−t , y(0) = 0, y (0) = 1
15) y + y = t, y(0) = 0, y (0) = 0
16) y + y = r(t), y(0) = 1, y (0) = 0 where r(t) =
1
0
if
if
0 < t < 2π
2π < t
5
0
if
if
17) y + y = e−2t sin t, y(0) = 0, y (0) = 0
18) y +2y +5y = r(t), y(0) = 0, y (0) = 0 where r(t) =
0<t<π
π<t
19) 4y + 4y + 17y = g(t), y(0) = 0, y (0) = 0
sin t
0
if
if
0 < t < 3π
1
7
, y(0) = , y (0) = −
50
50
3π < t
1
1
17) y = (sin t − cos t) + e−2t (sin t + cos t)
8
8
1 − e−t cos 2t + sin 2t , 0 < t < π
2
18) y =
−t π
e (e − 1) cos 2t + sin 2t
π<t
2
19) y =
1
8
t
1
e− 2 (t−x) sin 2(t − x)g(x) dx
0
1
(cos t − 7 sin t)
50
20) y =
1 −9π 3t
2
e e − e6π e−2t
50
50
1 e−t e−3t
−
+
3
2
6
8) f (t) = cosh at cos at
11) y = 3e−t + 5e2t
3s − 2s + 5
(s − 2)(s2 + 9)
12s2 − 16
(s2 + 4)3
4) f (t) = u(t − 1) cos(2t − 2)
2
s
(s2 + 4)2
20) y − y − 6y =
s2 + 2s
(s2 + 2s + 2)2
3) f (t) = u(t − 3) sin(t − 3)
Find the inverse Laplace transform transform of the following functions:
se−s
e−3s
4) F (s) = 2
3) F (s) = 2
s +1
s +4
1
1
5) F (s) = 2
6) F (s) = 3
2
(s + 16)
s + 4s2 + 3s
9) F (s) =
CHAPTER 9. LAPLACE TRANSFORM II
Answers
Find the Laplace transform transform of the following functions:
1) f (t) = te−t cos t
2) f (t) = t2 sin 2t
7) F (s) =
76
if
0 < t < 3π
if
3π < t
2
sin 3t
3
44. 78
CHAPTER 10. FOURIER ANALYSIS I
L
cos
nπx
mπx
cos
dx = 0 (m = n)
L
L
(10.3)
sin
nπx
mπx
sin
dx = 0 (m = n)
L
L
(10.4)
−L
L
Chapter 10
−L
L
cos2
Fourier Analysis I
−L
−L
nπx
mπx
sin
dx = 0 (for all m, n)
L
L
77
kπx
dx =
L
L
a0 cos
−L
∞
+
+
(10.1)
(10.2)
kπx
dx
L
L
cos
nπx
kπx
cos
dx
L
L
sin
an
nπx
kπx
cos
dx
L
L
−L
n=1
∞
∞
If possible, this expansion would be very useful in all kinds of applications.
Once we solve a question for sine and cosine functions, we will be able to
solve it for any periodic f . Here, an and bn are the coordinates of f in the
space of sine and cosine functions. But then how can we find an and bn ? The
following identities will help us:
(10.6)
Now, suppose the expansion (10.1) exists. To find ak , we will multiply
both sides by cos kπx and then integrate from −L to L.
L
L
nπx
nπx
f (x) = a0 +
an cos
+
bn sin
L
L
n=1
n=1
−L
(10.5)
2 cos A sin B = sin(A + B) − sin(A − B)
Let f (x) be a periodic function with period 2L. It is sufficient that f be
defined on [−L, L]. Is it possible to express f as a linear combination of sine
and cosine functions?
cos
−L
nπx
dx = L
L
2 sin A sin B = cos(A − B) − cos(A + B)
f (x) cos
L
sin2
2 cos A cos B = cos(A − B) + cos(A + B)
Fourier Series
∞
L
In the terminology of linear algebra, the trigonometric functions form
an orthogonal coordinate basis. We can easily prove these formulas if we
remember the following trigonometric identities:
The trigonometric functions sine and cosine are the simplest periodic functions. If we can express an arbitrary periodic function in terms of these,
many problems would be simplified. In this chapter, we will see how to
find the Fourier series of a periodic function. Fourier series is important in
many applications. We will also need them when we solve partial differential
equations.
10.1
nπx
dx =
L
L
bn
n=1
−L
(10.7)
Using the property of orthogonality, we can see that all those integrals
are zero, except the kth one. Therefore
L
f (x) cos
−L
kπx
dx = ak L
L
⇒
ak =
1
L
L
f (x) cos
−L
kπx
dx
L
(10.8)
We can apply the same procedure to find a0 and bn . In the end, we will
obtain the following formulas for a function f defined on [−L, L].
45. 10.1. FOURIER SERIES
79
a0 =
Fourier coefficients:
an =
bn =
1
2L
1
L
1
L
CHAPTER 10. FOURIER ANALYSIS I
L
f (x) dx
−L
L
f (x) cos
nπx
dx
L
f (x) sin
nπx
dx
L
−L
L
−L
(10.9)
∞
∞
Fourier series: f (x) = a0 +
80
an cos
n=1
nπx
nπx
+
bn sin
L
L
n=1
(10.10)
Example 10.1 Find the Fourier series of the periodic function
f (x) = x2 , −L x L having period= 2L.
a0 =
=
1
2L
L
x2 dx
−L
3 L
1 x
2L 3
=
−L
L2
3
Using integration by parts two times we find:
an =
=
bn =
1
L
L
x2 cos
−L
nπx
dx
L
4L2 cos nπ
n2 π 2
1
L
L
x2 sin
−L
nπx
dx = 0
L
Therefore the Fourier series is:
∞
x2 =
4L2
nπx
L2
+
(−1)n 2 2 cos
3
nπ
L
n=1
The plot of the Fourier series up to n = 1, 2 and 3 is given in Figure 10.1.
Figure 10.1: Fourier Series of f = x2 for n = 1, 2, 3
46. 10.2. CONVERGENCE OF FOURIER SERIES
10.2
81
82
CHAPTER 10. FOURIER ANALYSIS I
Convergence of Fourier Series
Like any infinite series, Fourier series is of no use if it is divergent. But
most functions that we are interested in have Fourier series that converge
and converge to the function.
Theorem 10.1: Let f be periodic with period 2L and let f and f be
piecewise continuous on the interval [−L, L]. Then the Fourier expansion of
f converges to:
• f (x) if f is continuous at x.
f (x+ ) + f (x− )
if f is discontinuous at x.
2
Example 10.2 Find the Fourier series of the periodic function
•
a
b
f (x) =
Figure 10.2: Convergence at a discontinuity
10.3
Theorem 10.2: Let f be continuous on [−L, L], f (L) = f (−L) and let f
be piecewise continuous. Then the Fourier coefficients of f satisfy:
∞
2a2 +
0
having period= 2L. Then evaluate the series at x = L.
a0 =
1
2L
a dx +
−L
1
2L
L
b dx =
0
a+b
2
L
1
L
f (x)2 dx
∞
f 2 (x) = a0 f (x) +
∞
L
f 2 (x) dx = a0
f (x) dx +
−L
∞
n=1
an cos nπx +
L
bn =
1
L
0
nπx
1
a cos
dx +
L
L
−L
0
a sin
−L
nπx
1
dx +
L
L
a L
nπx
=−
cos
L nπ
L
0
L
0
n=1
∞
L
an
f (x) cos
−L
nπx
bn
dx +
L
n=1
b sin
0
L
0
b−a
=
(1 − (−1)n )
nπ
Therefore the Fourier series is:
∞
a+b
b−a
nπx
+
[1 − (−1)n ] sin
f (x) =
2
nπ
L
n=1
a + b 2(b − a)
+
2
π
πx 1
3πx 1
5πx
+ sin
+ sin
+ ···
L
3
L
5
L
a+b
If we insert x = L in that series, we obtain f (L) =
. Thus the value at
2
discontinuity is the average of left and right limits. The summation of the
series up to n = 1, 5 and 9 is plotted on Figure 10.2.
=
f (x) sin
−L
1
1
1
= 1 + 4 + 4 + ···
4
n
2
3
n=1
2
(Hint: Use the Fourier series of f (x) = x on the interval −π < x < π)
Example 10.3 Find the sum of the series S =
nπx
dx
L
b L
nπx
−
cos
L nπ
L
−L
L
Using equation (10.9) to evaluate these integrals, we can obtain the result.
nπx
b cos
dx = 0
L
L
sin nπx .
L
nπx
nπx
+
bn f (x) sin
L
L
n=1
∞
1
an =
L
∞
n=1 bn
∞
an f (x) cos
n=1
L
(10.11)
−L
Proof: We can express f (x) as f (x) = a0 +
Now multiply both sides by f and integrate
−L
0
(a2 + b2 ) =
n
n
n=1
−L < x < 0
0<x<L
if
if
Parseval’s Identity
Evaluating the integrals in (10.9) for f (x) = x2 we obtain
π2
4(−1)n
and bn = 0 so
a0 = , a n =
3
n2
f (x) =
1
π2
1
− 4 cos x − cos 2x + cos 3x − · · ·
3
4
9
Using Parseval’s theorem, we have
sin
1
2π 4
1
+ 16 1 + 4 + 4 + · · ·
9
2
3
Therefore
1 π 4
x dx
π −π
2
= π4
5
=
nπx
d
L
47. 10.3. PARSEVAL’S IDENTITY
1
1
2 2
−
16 1 + 4 + 4 + · · · = π 4
2
3
5 9
1
1
π4
S = 1 + 4 + 4 + ··· =
2
3
90
83
84
CHAPTER 10. FOURIER ANALYSIS I
Exercises
Find the Fourier series of the periodic function f (x) defined on the given
interval
1) f (x) = x, −π < x < π
3) f (x) =
0
1
2) f (x) = x, 0 < x < 2π
−π < x < 0
0<x<π
if
if
5) f (x) = sin2 x, −π < x < π
7) f (x) =
−π/4
π/4
if
if
−1 < x < 0
0<x<1
9) f (x) = |x|, −2 < x < 2
11) f (x) =
x
1−x
if
if
4) f (x) = x2 , 0 < x < 2π
6) f (x) = x + |x|, −π < x < π
8) f (x) =
π
x
if
if
10) f (x) = | sin x|, −π < x < π
0<x<1
1<x<2
13) f (x) = ax + b, −L < x < L
15) f (x) = x3 , −π < x < π
12) f (x) =
−a
a
1
π2
1
+
+ ··· = .
9 25
8
−L < x < 0
0<x<L
16) f (x) = ex , −π < x < π
x sin ax cos ax
+
a
a2
x cos ax sin ax
+
x sin ax dx = −
a
a2
2
x sin ax 2x cos ax 2 sin ax
x2 cos ax dx =
+
−
a
a2
a3
2
x cos ax 2x sin ax 2 cos ax
x2 sin ax dx = −
+
+
a
a2
a3
x cos ax dx =
if
if
14) f (x) = 1 − x2 , −1 < x < 1
17) Using integration by parts, show that:
18) Show that 1 +
−π < x < 0
0<x<π
48. EXERCISES
85
Answers
1
1
sin 2x + sin 3x − · · ·
2
3
2) f (x) = π − 2 sin x +
1
1
sin 2x + sin 3x + · · ·
2
3
1
1
sin x + sin 3x + sin 5x + · · ·
3
5
4π 2
1
1
+ 4 cos x + cos 2x + cos 3x + · · ·
3
4
9
−4π sin x +
5) f (x) =
6) f (x) =
π
4
−
2 π
1
1
sin 2x + sin 3x + · · ·
2
3
1 1
− cos 2x
2 2
cos x +
+2 sin x −
7) f (x) = sin πx +
1
1
cos 3x +
cos 5x + · · ·
9
25
1
1
1
sin 2x + sin 3x − sin 4x + · · ·
2
3
4
1
1
sin 3πx + sin 5πx + · · ·
3
5
∞
3π
(−1)n − 1
1
+
cos nx − sin nx
2
4
πn
n
n=1
9) f (x) = 1 −
10) f (x) =
8
π2
cos
2
4
−
π π
11) f (x) = −
+
4
π2
2
π
sin
13) f (x) = b +
2aL
π
14) f (x) =
2
4
+ 2
3 π
πx 1
3πx 1
5πx
+ sin
+ sin
+ ···
L
3
L
5
L
sin
∞
n=1
3πx
1
5πx
πx 1
+ cos
+
cos
+ ···
2
9
2
25
2
cos 2nx
4n2 − 1
cos πx +
sin πx +
1
1
cos 3πx +
cos 5πx + · · ·
9
25
1
sin 3πx + · · ·
3
(−1)n+1
15) f (x) = 2
n=1
πx
1
2πx 1
3πx
− sin
+ sin
− ···
L
2
L
3
L
cos πx −
∞
∞
1 − (−1)n
1 2
1
sin nx = +
3) f (x) = +
2 n=1
nπ
2 π
8) f (x) =
CHAPTER 10. FOURIER ANALYSIS I
4a
π
12) f (x) =
1) f (x) = 2π sin x −
4) f (x) =
86
1
1
cos 2πx + cos 3πx + · · ·
4
9
(nπ)2 − 6
sin nx
n3
∞
16) f (x) =
2 sinh π 1
(−1)n
+
(cos nx − n sin nx)
π
2 n=1 1 + n2
18) Use the function in exercise 12 in Parseval’s identity
49. 88
CHAPTER 11. FOURIER ANALYSIS II
Figure 11.1: Plots of Some Even and Odd Functions
Chapter 11
Fourier Analysis II
In this chapter, we will study more advanced properties of Fourier series. We
will find the even and odd periodic extensions of a given function, we will
express the series using complex notation and finally, we will extend the idea
of Fourier series to nonperiodic functions in the form of a Fourier integral.
As you can see in Figure 11.1, an even function is symmetric with respect
to y−axis, an odd function is symmetric with respect to origin.
Half Range Extensions: Let f be a function defined on [0, L]. If we want
to expand it in terms of sine and cosine functions, we can think of it as
periodic with period 2L. Now we need to define f on the interval [−L, 0].
There are infinitely many possibilities, but for simplicity, we are interested
in making f an even or an odd function. If we define f for negative x
values as f (x) = f (−x), we obtain the even periodic extension of f , which
is represented by a Fourier cosine series. If we define f for negative x values
as f (x) = −f (−x), we obtain the odd periodic extension of f , which is
represented by a Fourier sine series.
Half-Range Cosine Expansion: (or Fourier cosine series)
∞
f (x) = a0 +
11.1
n=1
Fourier Cosine and Sine Series
If f (−x) = f (x), f is an even function. If f (−x) = −f (x), f is an odd
function. We can easily see that, for functions:
even × even = even,
odd × odd = even,
even × odd = odd
For example |x|, x2 , x4 , cos x, cos nx, cosh x are even functions. x, x3 , sin x, sin nx, sinh x
are odd functions. ex is neither even nor odd.
L
L
If f is even:
f (x) dx
f (x) dx = 2
−L
L
If f is odd:
(11.1)
0
f (x) dx = 0
(11.2)
−L
Using the above equations, we can see that in the Fourier expansion of an
even function, bn = 0, and in the expansion of an odd function, an = 0. This
will cut our work in half if we can recognize the given function as odd or
even.
87
an cos
where a0 =
1
L
nπx
, (0 < x < L)
L
L
f (x) dx,
0
an =
2
L
L
f (x) cos
0
nπx
dx
L
(11.3)
(11.4)
50. 11.1. FOURIER COSINE AND SINE SERIES
89
∞
bn sin
n=1
nπx
, (0 < x < L)
L
CHAPTER 11. FOURIER ANALYSIS II
11.2
Half-Range Sine Expansion: (or Fourier sine series)
f (x) =
90
(11.5)
Complex Fourier Series
Consider the Fourier series of f (x):
∞
∞
f (x) = a0 +
where
L
2
L
bn =
f (x) sin
0
nπx
dx
L
0
f (x) =
bn sin nx
(11.7)
n=1
Using Euler’s formula eix = cos x + i sin x we can express the sine and cosine
functions as:
0<x< π
2
π
<x<π
2
if
if
π
2
n=1
(11.6)
Example 11.1 Find the half-range cosine and sine expansions of
an cos nx +
cos nx =
einx + e−inx
,
2
sin nx =
einx − e−inx
2i
(11.8)
an + ibn
2
(11.9)
Therefore
Here, L = π, therefore
1
a0 =
π
2
π
an =
π
π
2
π
π
2
sin nx
=
n
π
2
einx +
e−inx
If we define c0 = a0 and
π
cos nx dx
2
π
an − ibn
2
an cos nx + bn sin nx =
π
π
dx =
2
4
cn =
nπ
2
sin
=−
n
an + ibn
an − ibn
, c−n =
, n = 1, 2, 3, . . .
2
2
(11.10)
We will obtain
∞
cn einx
(11.11)
f (x)e−inx dx n = 0, ±1, ±2, . . .
(11.12)
f (x) =
Therefore half-range cosine series of f is
n=−∞
∞
sin nπ
π
π
1
1
2
f (x) = −
cos nx = − cos x − cos 3x + cos 5x − · · ·
4 n=1 n
4
3
5
where
cn =
On the other hand,
2
bn =
π
π
π
2
π
sin nx dx
2
− cos nx
=
n
π
π
2
cos nπ − cos nπ
2
=
n
Therefore half-range sine series of f is
∞
f (x) =
n=1
cos nπ − cos nπ
1
1
2
sin nx = sin x − sin 2x + sin 3x + sin 5x + · · ·
n
3
5
1
2π
π
−π
For a function of period 2L we have
∞
cn einπx/L ,
f (x) =
cn =
n=−∞
1
2L
L
f (x)e−inπx/L dx
−L
Example 11.2 Find the complex Fourier series of
f (x) = x if −π < x < π and f (x + 2π) = f (x).
We have to evaluate the integral
cn =
1
2π
π
xe−inx dx
−π
(11.13)
51. 11.2. COMPLEX FOURIER SERIES
91
For n = 0 this integral is zero, so we have c0 = 0. For n = 0
cn =
−inx π
xe
−in
1
2π
π
−
−π
inπ
−π
1
2π
Therefore
CHAPTER 11. FOURIER ANALYSIS II
11.3
Fourier Integral Representation
−inx
e
dx
−in
πe−inπ + πe
−0
−in
1 einπ + e−inπ
cos nπ
=−
=−
in
2
in
i
n
= (−1)
n
=
92
In this section, we will apply the basic idea of the Fourier series to nonperiodic functions.
Consider a periodic function with period= 2L and its Fourier series. In
the limit L → ∞, the summation will be an integral, and f will be a nonperiodic function. Then we will obtain the Fourier integral representation:
∞
i
(−1)n einx ,
n
n=−∞
n=0
where
A(u) =
Note that we can obtain the real Fourier series from the complex one. If we
add nth and −nth terms we get
cos nx + i sin nx
cos(−nx) + i sin(−nx)
sin nx
i(−1)
+ i(−1)−n
= (−1)n+1
n
−n
n
B(u) =
n
∞
n+1 sin nx
(−1)
x=
n
n=1
1
π
1
π
∞
f (x) cos ux dx
(11.15)
f (x) sin ux dx
(11.16)
−∞
∞
−∞
Like the Fourier series, we have A(u) = 0 for odd functions and B(u) = 0 for
even functions.
Theorem 11.1: If f and f are piecewise continuous in every finite interval
∞
This is the real Fourier series.
|f | dx is convergent, then the Fourier integral of f converges to:
and if
−∞
Example 11.3 Find the complex Fourier series of f (x) = k
• f (x) if f is continuous at x.
π
1
ke−inx dx
2π −π
π
k e−inx
=
(n = 0)
2π −in −π
cn =
inπ
•
f (x+) + f (x−)
if f is discontinuous at x.
2
Example 11.4 Find the Fourier integral representation of
−inπ
k e −e
nπ
2i
k
=
sin nπ
nπ
=0
=
If n = 0 we have
(11.14)
0
∞
x=
[A(u) cos ux + B(u) sin ux] du
f (x) =
1
c0 =
2π
=k
f (x) =
π/2
0
if
if
|x| < 1
1 < |x|
Note that f is even therefore B(u) = 0
π
k dx
A(u) =
−π
∞
1
π
f (x) cos ux dx =
−∞
1
=
cos ux dx =
0
sin ux
u
1
π
1
=
0
1
−1
π
cos ux dx
2
sin u
u
52. 11.3. FOURIER INTEGRAL REPRESENTATION
93
Therefore, Fourier integral representation of f is
∞
f (x) =
0
sin u
cos ux du
u
94
CHAPTER 11. FOURIER ANALYSIS II
Exercises
For the following functions defined on 0 < x < L, find the half-range
cosine and half-range sine expansions:
Example 11.5 Prove the following formulas using two different methods:
eax
e cos bx dx = 2
(a cos bx + b sin bx)
a + b2
eax
eax sin bx dx = 2
(a sin bx − b cos bx)
a + b2
We can obtain the formulas using integration by parts, but this is the
long way. A better method is to express the integrals as a single complex
integral using eibx = cos bx + i sin bx, then evaluate it at one step, and then
separate the real and imaginary parts.
ax
Example 11.6 Find the Fourier integral representation of
f (x) =
−ex cos x
e−x cos x
if
if
x<0
0<x
This function is odd therefore A(u) = 0.
2 ∞ −x
1 ∞
f (x) sin ux dx =
e cos x sin ux dx
π −∞
π 0
2 ∞ −x sin(ux + x) + sin(ux − x)
e
dx
=
π 0
2
∞
1
e−x
=
[− sin(u + 1)x − (u + 1) cos(u + 1)x]
π 1 + (u + 1)2
0
∞
e−x
1
[− sin(u − 1)x − (u − 1) cos(u − 1)x]
+
π 1 + (u − 1)2
0
1
u−1
u+1
=
+
π 1 + (u + 1)2 1 + (u − 1)2
2 u3
=
π u4 + 4
B(u) =
So
f (x) =
2
π
∞
0
u3
sin ux du
+4
u4
1) f (x) =
2kx/L
2k(L − x)/L
if
if
0 < x < L/2
2) f (x) = ex
L/2 < x < L
3) f (x) = k
4) f (x) = x4
5) f (x) = cos 2x 0 < x < π
6) f (x) =
0
k
if
if
0 < x < L/2
L/2 < x < L
Find the complex Fourier series of the following functions:
7) f (x) =
0
1
if
if
−π < x < 0
0<x<π
9) f (x) = sin x
8) f (x) = x2 , −L < x < L
10) f (x) = cos 2x
Find the Fourier integral representations of the following functions:
π
π
cos x, |x| <
π − x, 0 < x < π
2
2
11) f (x) =
(f odd)
12) f (x) =
π
0,
π<x
0,
|x| >
2
13) f (x) =
e−x , 0 < x
ex , x < 0
π
0
14) f (x) =
Prove the following formulas. (Hint: Define a suitable
then find its Fourier integral representation.)
πx2 /2,
∞
2
cos ux
2
15)
1 − 2 sin u + cos u
du =
π/4,
u
u
u
0
0,
0,
x<0
∞
cos ux + u sin ux
16)
du =
π/2, x = 0
−x
1 + u2
0
πe , x > 0
if
if
0<x<1
Otherwise
function f and
0
x<1
x=1
1<x
