MAT2039 Differential Equations I - Ki??isel Sayfalarkisi.deu.edu.tr/meltem.topcuoglu/MAT2039DiffEq/mat2039...Diﬀerential Equations and Their Solutions Example 1 Consider the following

MAT2039

Differential Equations I

Department of Mathematics,Dokuz Eylul University,

35160 Buca, Izmir, Turkey.

Tuesday 3rd October, 201711:48

Differential Equations I 1/332

Table of Contents I

1 Differential Equations and Their SolutionsSolutions of Ordinary Differential Equations

2 Initial Value Problems

3 First-Order Differential EquationsSeparable EquationsExact EquationsIntegrating Factors

4 First-Order Linear Differential Equations

5 Bernoulli Equations

6 First-Order Homogeneous Differential EquationsA Special TransformationSubstitution

7 Riccati Equations

8 Clairaut EquationsDifferential Equations I 2/332

Table of Contents II

9 Applications of First-Order Differential EquationsOrthogonal TrajectoriesOblique Trajectories

10 Some Physics ProblemsProblems in CoolingProblems in MechanicsRate ProblemsMixture Problems

11 Higher-Order Linear Differential EquationsNon-Homogeneous Linear Differential Equations

12 Reduction of Order for Homogeneous Linear Differential Equations

13 Homogeneous Linear DEs with Constant CoefficientsDistinct Real RootsRepeated Real RootsComplex Conjugate Roots

14 Non-Homogeneous Linear Differential EquationsThe Method of Undetermined CoefficientsThe Method of Variation of Parameters


Table of Contents III

15 Cauchy-Euler Equations

16 Power Series Solutions

17 The Method of Frobenious (To Be Skipped)

18 The Laplace TransformConvolutionThe Inverse Laplace TransformSolving Autonomous Differential Equations with Laplace TransformApplications to Differential EquationsThe Impulse Function (To Be Skipped)

19 Systems of Linear Differential Equations (To Be Skipped)

20 Solving Linear SystemsAn Operator Method for Autonomous Systems (To Be Skipped)Solving Autonomous Differential Systems with Laplace Transform

21 Fourier SeriesFourier Cosine Series and Fourier Sine Series



Differential Equations and Their Solutions


Many important problems in engineering, physical science require thedetermination of a function satisfying an equation which contains thederivatives of the unknown function.

Definition 1 (Differential Equation)

A “differential equation” is an equation involving one or morederivatives of an unknown function.



Example 1

Consider the following equations.

1. dydx

+ y = x2.

2. y ′′ = −k2y .

3. d4y

dx4+(

d2y

dx2

)2

+ cos(x) = 0.

4. ∂2u∂x2

+ ∂2u∂y2 = 0.

Equation 1, Equation 2 and Equation 3 are called “ordinary differential

equation” (ODE) since the unknown function y depends only on a singleindependent variable x . In Equation 4, the unknown function u, dependson more than one variable (x and y) and hence the equation involvespartial derivatives. Such a differential equation is called a “partialdifferential equation” (PDE).



In this course, we consider only ordinary differential equations. Our aim isto develop methods for determining all possible unknown functions thatsatisfy a given ordinary differential equation.

Definition 2

The order of the highest derivative occurring in a differential equation iscalled the order of the differential equation. Any n-th order ordinarydifferential equation can be written in the form

F(x , y , y ′, · · · , y (n)

)= 0,

where y ′ := dydx, · · · , y (n) := d

nydxn

, y is the “dependent variable” and x isthe “independent variable”.



Definition 3

A differential equation that can be written in the form

a0(x)y(n)(x) + a1(x)y

(n−1)(x) + · · ·+ an(x)y(x) = f (x),

where a0, a1, · · · , an with a0 6= 0 and f are functions of x only, is called a“linear differential equation of order n”. Such an equation is linear iny , y ′, · · · , y (n). A differential equation that does not satisfy thisdefinition is said to be a “non-linear differential equation”.



Example 2

Consider the following equations.

1. y ′′ + x2y ′ + sin(x)y = ex .

2. xy ′′ + 11+x2

y ′ = sin(x).

3. y ′′ + x sin(y ′)− xy = x2.

4. y ′ =√

yx.

5. y ′′ +(2y ′

)2= ex .

It is easy to see that Equation 1 and Equation 2 are linear whileEquation 3, Equation 4 and Equation 5 are non-linear.



Exercises 1

Exercises

Classify the following differential equations depending on their type, orderand linearity.

1. y dydx

= x .

2. d2y

dx2− 16y = 0.

3.(

dydx

)3

= y2.

4.(

∂u∂x

)(∂u∂y

)

= xyu.

5. ∂u∂x

− ∂u∂y

= 0.


Differential Equations and Their Solutions Solutions of Ordinary Differential Equations

Solutions of Ordinary Differential Equations

Definition 4 (Solution)

A “solution” of an n-th order differential equation on an interval I is anyfunction y = y(x) that is (at least) n-times differentiable on I and thatsatisfies the differential equation identically for all x in I .



Example 3

1. Show that y = e2x is a solution of the linear equation y ′ = 2y for allx ∈ R.

2. dydx

= 12√x(y − 1) is undefined when x ≤ 0, so the solution would be

defined only for x > 0.



The solutions of a differential equation can be expressed in the followingtwo different ways.• Explicit form: y = ϕ(x). • Implicit form: Φ(x , y) = 0.

Definition 5

A function y = ϕ(x) is called an “explicit solution of the differential

equation F (x , y , y ′, · · · , y (n)) = 0 on an interval I” if it satisfies theequation for all x in the interval I when substituted for y in the equation.

Definition 6

A relation Φ(x , y) = 0 is said to be an “implicit solution of the

differential equation F (x , y , y ′, · · · , y (n)) = 0 on an interval I” if itdefines one or more explicit solutions of differential equation on I .



Example 4

The relation x2 + y2 = 4 defines an implicit solution to the non-lineardifferential equation dy

dx= − x

y. This relation defines two functions

y = ±√4− x2, so it contains two explicit solutions. Since x = ±2

correspond to y = 0, (differential equation is defined only for y 6= 0).The solutions are valid for −2 < x < 2.



Example 5

Now consider the simple differential equation d2y

dx2= 5x . We can find all

solutions of this differential equation by using two integration process.Thus,

dy

dx=

5

2x2 + c1, where c1 is an arbitrary constant,

and

y =5

6x3 + c1x + c2, where c2 is another arbitrary constant.

So the solution contains two arbitrary constants. Using appropriatevalues to these constants, we can determine all solution of the differentialequation. We call y = 5

6x3 + c1x + c2 the general solution of the

differential equation.



Definition 7 (General Solution)

Let F (x , y , y ′, · · · , y (n)) = 0 be an n-th order ordinary differentialequation. A solution of the equation, which contains n arbitraryconstants, is called a “general solution”.

Example 6

Consider the differential equation y ′′ = e−x . By two integrations withrespect to x , we find that the general solution of the equation isy = e

−x + c1x + c2 for x ∈ R, where c1 and c2 are arbitrary constants.



Remark 1

Not all differential equations have a general solution. For example,

(dy

dx

)2

+ (y − 1)2 = 0.

The only solution to this differential equation is y(x) ≡ 1, hence thedifferential equation does not have a solution containing any arbitraryconstant.



Definition 8 (Particular Solution)

A solution of a differential equation is called a “particular solution” if itdoes not contain any arbitrary constants.

Remark 2

Assigning fixed values for arbitrary constants gives a particular solution.

Example 7

For example, we see from Example 6 that y = e−x + x is a particularsolution of y ′′ = e−x .



Definition 9 (Singular Solution)

A solution of an n-th order differential equation that cannot be obtainedfrom the general solution by any choice of the n arbitrary constants iscalled a “singular solution” of the differential equation.

Example 8

Consider the differential equation(

dydx

)2

− 4y = 0. For any constant c ,

y = (x + c)2 is a solution (general solution). y = (x + 2)2 andy = (x − 5)2 are particular solutions. But the constant function y ≡ 0 isalso a solution which is not contained by the family of curvesy = (x + c)2, i.e., no choice of the constant c will yield this solution.Thus, y ≡ 0 is a singular solution.



Exercises 1.1

Exercises

1. Show that the differential equation d2y

dx2− y = 0 has the general

solution y := c1 cosh(x) + c2 sinh(x), where c1 and c2 are arbitraryconstants. Determine c1 and c2 such that the equation has theparticular solution satisfying y(0) = 0 and y ′(0) = 1.

2. Find α ∈ R such that d2y

dx2− 4dy

dx+ 4y = 0 has the general solution

y := eαx(c1x + c2), where c1 and c2 arbitrary constants.

3. Show that 2y dydx

= x

(

4 +(

dydx

)2)

has a particular solution of the

form y := x2 + 1. Find h in terms of c for which y := cx + h formsthe general solution of the equation.

4. Show that y = x dydx

+(

dydx

)2

has the general solution y := cx + c2,

where c is an arbitrary constant. Moreover, show that y := − 14x

2 is asingular solution of the equation.


Initial Value Problems


The solution of the first order differential equation

dy

dx= f (x , y)

can be characterized as y = Φ(x , c), where c is an arbitrary constant andΦ(x , c) is called the “one parameter family of solutions”.

Example 9

y = cos(x) + c is one parameter family of solutions of

y ′ = − sin(x).

Among these infinite number of solutions, there is only one solution thatsatisfies the condition y(0) = 1. We can check that y = cos(x) satisfiesthe condition y(0) = 1.

Problems of this kind are known as “initial value problems” (IVPs).



IVP can be written in the following form

dy

dx= f (x , y)

y(x0) = y0.

−→ Differential Equation

−→ Initial Condition



Theorem 1 (Existence and Uniqueness)

Let f (x , y) and ∂∂y

f (x , y) be continuous on the rectangle

R := (x , y) : |x − x0| ≤ a, |y − y0| ≤ b. Then, there is a uniquesolution of the IVP

dydx

= f (x , y)

y(x0) = y0

in some interval I : |x − x0| < h, where h > 0.



Example 10

Consider dydx

= x√y

y(x0) = y0.

Then, f (x , y) := x√y is continuous for x ∈ R and y ≥ 0, and

∂∂y

f (x , y) = x2√yis continuous for x ∈ R and y > 0. Therefore, we can

say that with x0 := 1 and y0 := 2, the IVP has a unique solution definedon some interval of the form 1− h < x < 1 + h, where h > 0. However,with x0 := 0 and y0 := 0, the IVP has two solutions y ≡ 0 and y = x4

16 ,i.e., uniqueness does not hold.



Exercises 2

Exercises

1. Show that the IVP

dydx

+ p(x)y = q(x)

y(x0) = y0,

where p, q are continuous functions and x0, y0 ∈ R, admits a uniquesolution in the interval I : |x − x0| < h for some h > 0.

2. Show that the IVP dydx

= 1 + y2

y(0) = 0

admits a unique solution in some neighborhood of 0.


First-Order Differential Equations

First-Order Differential Equations

A first-order differential equation can be defined in the following forms.

• The derivative form: dydx

= f (x , y).

• The differential form: M(x , y)dx + N(x , y)dy = 0.


First-Order Differential Equations Separable Equations

Separable Equations

Definition 10 (Separable Equations)

An equation of the form

f (y)dy

dx= g(x)

is called a “separable equation”.



Solution Technique

First, we write the equation in the form

f (y)dy = g(x)dx .

By direct integration of both sides, we get the general solution

∫ y

f (ζ)dζ

︸︷︷︸

F (y)

=

∫ x

g(η)dη

︸︷︷︸

G(x)

+c ,

where c is an arbitrary constant.



Remark 3

The differential equation

dy

dx= g(x)h(y)

is separable because it can be written as

1

h(y)dy = g(x)dx

provided that h(y) 6= 0. What happens if h(y) = 0? In that case, we getlost solutions. If these solutions are not included in the general solutionfor definite values of the arbitrary constants, then they are called“singular solutions”.



Example 11

Solve the following differential equations.

1. eydydx

= x cos(x).

2. dydx

= 2xy .

3. dydx

= −2y2x .



Exercises 3.1

Exercises

Find the solutions and domain of validity for the following differentialequations.

1.(y2 + 1

)dx + (x + 1)dy = 0.

2. ey2

ln(x)dx − ydy = 0.

3. dydx

− x2y2 = x2.

4. dydx

= cos(y) sin(x).

5. dydx

=(y2 − 1

)ln(x).


First-Order Differential Equations Exact Equations

Exact Equations

Definition 11 (Exact Equations)

Let F be a function of two real variables such that F has continuous firstpartial derivatives in a domain D. The “total differential of F” isdefined as

dF =∂F

∂xdx +

∂F

∂ydy for all (x , y) ∈ D.



Now, consider the first order differential equation in the differential form

M(x , y)dx + N(x , y)dy = 0,

which is similar to the right-hand side of dF above. Thus,

M(x , y)dx + N(x , y)dy = 0

can be written as dF = 0 if and only if

∂F

∂x= M(x , y) and

∂F

∂y= N(x , y).



Definition 12


M(x , y)dx + N(x , y)dy = 0

is said to be “exact in a domain D”, if there exists a function F suchthat

∂F

∂x= M(x , y) and

∂F

∂y= N(x , y) for all (x , y) ∈ D.

It is clear that, we need a simple test to determine whether or not a givendifferential equation is exact. This is given by the following theorem.



Theorem 2 (Test for Exactness)

Let M ,N and their first-order partial derivatives ∂M∂y

and ∂N∂x

becontinuous in a rectangular domain D. Then, the ordinary differentialequation

M(x , y)dx + N(x , y)dy = 0

is exact for all (x , y) ∈ D if and only if

∂M

∂y=∂N

∂x.



Example 12

Find the general solution of

2xeydx +(x2ey + cos(y)

)dy = 0.



Exercises 3.2

Exercises

Find the general solution of the following differential equations.

1.(sec(x + y)

)2dx +

(

2y +(sec(x + y)

)2)

dy = 0.

2.(2(x + y)− 3x2

)dx + 2(x + y)dy = 0.

3.(2xy + x cos(x) + sin(x)

)dx + x2dy = 0.

4. 1x+y2 dx +

(

3 + 2yx+y2

)

dy = 0.

5. Determine a function M such that M(x , y)dx +(2x2y − 1

)dy = 0 is

exact, and then solve it.


First-Order Differential Equations Integrating Factors

Integrating Factors

If the differential equation is not exact, it is sometimes possible totransform to an exact differential equation by multiplying some non-zerofunction.

Definition 13

A non-zero function µ = µ(x , y) is called an “integrating factor” for

M(x , y)dx + N(x , y)dy = 0

if the differential equation

µ(x , y)M(x , y)dx + µ(x , y)N(x , y)dy = 0

is exact.



Theorem 3

µ = µ(x , y) is an integrating factor for

M(x , y)dx + N(x , y)dy = 0

if and only if it is a solution of the partial differential equation

N∂µ

∂x−M

∂µ

∂y= µ

(∂M

∂y− ∂N

∂x

)

.

We see that the determination of µ requires the solution of PDE that itis usually no easier to solve the PDE. So, let us determine certain specialintegrating factors of the forms µ = µ(x) and µ = µ(y).



Theorem 4

Consider the differential equation

M(x , y)dx + N(x , y)dy = 0.

Case 1. There exists an integrating factor that depends only on x if andonly if

1

N

(∂M

∂y− ∂N

∂x

)

= f (x).

Then, an integrating factor is

µ(x) = exp

∫ x

f (η)dη

.



Case 2. There exists an integrating factor that depends only on y if andonly if

1

M

(∂M

∂y− ∂N

∂x

)

= g(y).

Then, an integrating factor is

µ(y) = exp

−∫ y

g(ζ)dζ

.



Example 13

Solve(2x2 + y

)dx + x(xy − 1)dy = 0.

Example 14

Determine the integrating factor of the form µ(x , y) = xmyn for thedifferential equation

y(3x − 2y2

)dx + x

(3x − 5y2

)dy = 0.



Exercises 3.3

Exercises

1. Solve the differential equation(x + y − 1

)ydx + (x + 2y − 2)dy = 0.

2. Solve the differential equation 3xydx +(9x2 + 4y

)dy = 0.

3. Find m, n such that µ = xmyn is an integrating factor fory2dx + x(x + y)dy = 0, and then solve it.

4. Use an integrating factor of the form µ = µ(xy) to find the generalsolution of the differential equation ydx + x

(1 + x2y3

)dy = 0.


First-Order Linear Differential Equations


Definition 14 (Linear Differential Equation)

A first-order ordinary differential equation is linear in the dependentvariable y and the independent variable x , if it can be written in the form

dy

dx+ p(x)y = q(x).



Theorem 5

The linear differential equation

dy

dx+ p(x)y = q(x).

has an integrating factor

µ(x) := exp

∫ x

p(η)dη

.

A one-parameter family of solutions for this equation is

y :=1

µ(x)

[∫ x

µ(η)q(η)dη + c

]

.

Furthermore, this one-parameter family of solutions includes all solutions.



Example 15

Solve the initial value problem

dydx

+ xy = xex2

2

y(0) = 1.



Exercises 4

Exercises

1. Find the general solution of the differential equation y ′ + 3y = x .

2. Find the general solution of y ′ − tan(x)y = sec(x).

3. Solve the differential equation y ′ + exy = ex .

4. Solve the differential equation y ′ + cot(x)y = x .

5. Depending on the values of α ∈ R, solve the differential equationdydx

+ αxy = β, where β ∈ R.

6. Obtain the general solution of the equation y2dx + (3xy − 1)dy = 0.


Bernoulli Equations

Bernoulli Equations

We now consider a special type of equation which can be reduced to alinear equation by an appropriate transformation. This is the so-calledBernoulli equations.

Definition 15 (Bernoulli Equation)


dy

dx+ p(x)y = q(x)yn

is called a “Bernoulli equation”.

If n = 0 or n = 1, then it is a linear equation.


Bernoulli Equations

Theorem 6

Suppose that n 6= 0 and n 6= 1. Then, the transformation u := y1−n

reduces the Bernoulli equation to the linear equation in u.

Example 16


dydx

+ tan(x)y = 4 sin(x)y2

y(0) = 1.


Bernoulli Equations

Remark 4

Sometimes the function q in the equation may not be continuous butmay have a jump discontinuity. In this case, the existence theorem doesnot necessarily guarantee the existence of a unique solution of the IVP.

Example 17

Find a continuous solution to the IVP

dydx

+ y = q(x)

y(0) = 0,

where

q(x) :=

1, x ≥ 0

−1, x < 0.


Bernoulli Equations

Exercises 5

Exercises

1. Find the general solution of the differential equationy ′ + 1

xy =

(1− x2

)y2.

2. Find the general solution of y ′ + 12(x+1)y =

(1− x2

)y3.

3. Solve the differential equation y ′ + tan(x)y = y2.

4. Solve the differential equation y ′ + cot(x)y = cos(x)y3.

5. Obtain the general solution of xy ′ = ey2

yey2−x2

.


First-Order Homogeneous Differential Equations


Definition 16

A first-order differential equation

M(x , y)dx + N(x , y)dy = 0

is said to be “homogeneous” if, when written in the derivative formy ′ = f (x , y), there exists a function g such that f (x , y) can be expressedin the form g( y

x).



Example 18

Check homogeneity of the following differential equations.

1.(x2 − y2

)dx + 2xydy = 0.

2.(x +

√

x2 + y2)dx − xdy = 0.

Usually, it is easy to tell by inspection whether or not a given equation ishomogeneous.

Definition 17

A function F is called “homogeneous of degree n” if

F (λx , λy) = λnF (x , y).



Now, suppose that the functions M and N in the differential equation

M(x , y)dx + N(x , y)dy = 0

are both homogeneous of the same degree n. Then,M(λx , λy) = λnM(x , y) and N(λx , λy) = λnN(x , y). Letting λ = 1

x, we

get

M(λx , λy) = M

(1

xx ,

1

xy

)

=

(1

x

)n

M(x , y),

which yields M(x , y) = xnM(1, yx). By the same way,

N(x , y) = xnN(1, yx).



Conclusion 1

If M and N in M(x , y)dx + N(x , y)dy = 0 are both homogeneousfunctions of the same degree n, then the differential equation is ahomogeneous differential equation.

Theorem 7

If M(x , y)dx + N(x , y)dy = 0 is a homogeneous equation, then thechange of variable y = ux transforms the given equation into a separableequation in u and x.

Example 19

Solve the differential equation

(x2 − 3y2

)dx + 2xydy = 0.



Exercises 6

Exercises

1. Find the general solution of y ′ = 3y−x3x−y

.

2. Find the general solution of y ′ = y(y+2x)3x2 .

3. Find the general solution of xy ′ =(1 + α ln(y) − α ln(x)

)y , where

α > 0.

4. What is the general solution y ′ = y2+(α+1)xy−αx2

x

((α+1)x−(α−1)y

) , where α > 0.


First-Order Homogeneous Differential Equations A Special Transformation

A Special Transformation

Theorem 8

Let a1, b1, c1, a2, b2 and c2 be constants and consider the equation

(a1x + b1y + c1)dx + (a2x + b2y + c2)dy = 0.

Case 1. If ∆ := a1b2 − a2b1 6= 0, then the transform u := x − h andv := y − k, where (h, k) is the solution of the simultaneousalgebraic system

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

reduces the given equation to the homogeneous equation

(a1u + b1v)du + (a2u + b2v)dv = 0.

Case 2. If ∆ = 0, then z := a1x + b1y reduces the equation to separableequation in z and x.



Example 20

Solve the following differential equations.

1. (x − 2y + 1)dx + (4x − 3y − 6)dy = 0.

2. (x + 2y + 3)dx + (2x + 4y − 1)dy = 0.



Exercises 6.1

Exercises

1. Find the general solution of(x − y + α− β)dx + (x + y − α− β)dy = 0, where α, β ∈ R.

2. Find the general solution of(2x + 2αy − 1)dx +

(2αx + 2α2y − β

)dy = 0, where α, β ∈ R such

that α 6= 0.

3. Solve the differential equation 2(y +α)2dx − (y + x + α+ β)2dy = 0,where α, β ∈ R.


First-Order Homogeneous Differential Equations Substitution

Substitution

In the preceding sections, we have used some standard substitutions tosolve the given differential equation. There are some differential equationswhich look different from those, and we can solve by using some specialsubstitutions. But they do not have a general rule. Here is an example.

Example 21


(1 + 3x sin(y)

)dx − x2 cos(y)dy = 0

by letting u = sin(y).


First-Order Homogeneous Differential Equations Substitution

Exercises 6.2

Exercises

Solve the following differential equations by making suitable substitutions.

1. y ′ =(tan(x + y)

)2.

2. y ′ = (αx + βy + γ)2, where α, β, γ ∈ R.

3. y ′ = (αx+βy)2+γ

αx+βy− α

β, where α, β, γ ∈ R with β 6= 0.

4. y ′ = (x+yn)m−1nyn−1 , where n,m ∈ N.

5. nyn−1y ′ = yn + x − 1, where n ∈ N.


Riccati Equations

Riccati Equations

Definition 18


dy

dx= p(x)y2 + q(x)y + r(x)

is called a “Riccati equation”.


Riccati Equations

In many cases, depending on p, q and r , the solution of this equationcannot be expressed in terms of elementary functions.

Theorem 9

Suppose that y1 is a solution of the Riccati equation. Then, thetransformation y := y1 +

1ureduces the Riccati equation to the linear

equation in u.

Therefore, knowing a particular solution enables us to find the generalsolution.


Riccati Equations

Example 22


dy

dx= y2 − 2xy + 2,

which has a particular solution y1(x) := 2x .


Riccati Equations

Exercises 7

Exercises

1. Assume that y1 is a solution of the Riccati differential equationy ′ = p(x)y2 + q(x)y + r(x). Then, show that the substitutiony := u + y1 transforms the equation into a Bernoulli equation in u.

2. Show that the substitution y := u′

putransforms the Riccati equation

y ′ = p(x)y2 + q(x)y + r(x) into a second-order linear differentialequation provided that p is differentiable.

3. Find α and β such that y ′ = y2 − yx+ α

x2has a particular solution of

the form y1 := xβ . Then, find the general solution of the equation.

4. Find the general solution of the differential equationy ′ = (y − αx + β)2 + α, where α, β ∈ R, if it has a particular solutionin the form of a straight line.


Clairaut Equations

Clairaut Equations

Definition 19

An equation, which is of the form

y = xp + f (p),

where p := dydx

and f is a given function, is called a “Clairaut equation”.


Clairaut Equations

Solution Technique

First, differentiating the given equation with respect to x and we simplyto obtain

(x + f ′(p)

)dp

dx= 0.

Assuming x + f ′(p) 6= 0, we divide through by this factor, and solve theresulting equation to obtain p = c , where c is an arbitrary constant. Thisgives us the one-parameter family of solutions by substituting intodifferential equation

y = cx + f (c).

Next, assuming x + f ′(p) = 0 leads to an extra solution, which is not amember of the one-parameter family of solutions. Such a solution isusually called a “singular solution”.


Clairaut Equations

Example 23


y = xp + p2.


Clairaut Equations

Exercises 8

Exercises

1. Solve the Crairaut differential equation y = xp + p2 + 1.

2. Solve the Crairaut differential equation y = xp + e−p + 1.

3. Equations of the form y = xf (p) + g(p) are called “Lagrangedifferential equation”, and are solved by using the same steps forsolving Clairaut equations. Solve the Lagrange equation2y(p − 1)− x

(p2 + 1

)= 0.

4. Solve the Lagrange equation y = −(x + 1)p(ln(p)− 1

)+ 1.


Applications of First-Order Differential Equations

Applications of First-Order Differential Equations

In this chapter, we consider several applied problems that can beformulated mathematically in terms of first order differential equations.


Applications of First-Order Differential Equations Orthogonal Trajectories

Orthogonal Trajectories

Definition 20

Let Φ(x , y , c) = 0 be a given one-parameter family of curves in thexy -plane. A curve, which intersects the curves of the given family atright-angles, is called an “orthogonal trajectory” of the given family.



Example 24

Consider the family of circles x2 + y2 = c2 with center of the origin andradius of c . Each straight line through the origin y = kx is an orthogonaltrajectory of the family of circles. Conversely, each circle of the familyx2 + y2 = c2 is an orthogonal trajectory of the family of straight linesy = kx . Actually, they are orthogonal trajectories of each other.

Figure: Red circles are intersected perpendicularly by blue lines.



Solution Technique

Now, we proceed to find the orthogonal trajectories of the family ofcurves Φ(x , y , c) = 0. We obtain the differential equation of the givenfamily, by differentiating the given equation implicitly with respect to xand then eliminating the parameter c by using the given and the derivedequation.

dΦ =∂Φ

∂xdx +

∂Φ

∂ydy = 0,

which yieldsdy

dx= −

∂Φ∂x∂Φ∂y

=: f (x , y).



Thus, the curve of the given family, which passes through the point(x , y), has the slope f (x , y) there. Therefore, m1 := f (x , y) is the slopeof the curve at (x , y), and m2 := − 1

f (x,y) is the slope of the orthogonal

trajectories of the curve. Since m1m2 = −1, they intersect each other atright angles. Thus, the differential equation of the family of theorthogonal trajectories is

dy

dx= − 1

f (x , y)=

∂Φ∂y

∂Φ∂x

.

A one-parameter family Ψ(x , y , d) = 0 of solutions of the differentialequation dy

dx= − 1

f (x,y) represents the family of the orthogonal trajectories

of the original family of curves.



Example 25

Find the orthogonal trajectories of the family of circles x2 + y2 = c2.

Example 26

Find the orthogonal trajectories of the family of parabolas y = kx2.

Figure: Red parabolas are intersected perpendicularly by blue ellipses.



Exercises 9.1

Exercises

1. Find the orthogonal trajectories of the family of curves 3x2 − y2 = c .

2. Find the orthogonal trajectories of the family of curves y = cex .

3. Find the orthogonal trajectories of the family of curves y = cx2.

4. Obtain the orthogonal trajectories of the family of curves y3 = cx2.

5. Obtain the orthogonal trajectories of the family of curves y = ex

ex+c.

6. Show that the orthogonal trajectories of the family of curvesy2 = 2cx + c2 is itself.


Applications of First-Order Differential Equations Oblique Trajectories

Oblique Trajectories

Definition 21

Let Φ(x , y , c) = 0 be a one-parameter family of curves. A curve, whichintersects the curves of the given family at a constant angle α 6= π

2 , iscalled an “oblique trajectory” of the given family.

ΩΘ

Α

FHx,y,c0L=0

YHx,y,d0L=0



Let dydx

= f (x , y) be the differential equation of the given family ofcurves. Then,

tan(θ) =dy

dx= f (x , y) or θ = arctan

(dy

dx

)

,

which gives the slope of the curve (x , y). Noting that the slope of theoblique trajectory is tan(ω) and ω = θ + α, we find that

tan(ω) = tan(θ + α) =tan(θ) + tan(α)

1− tan(θ) tan(α)

=f (x , y) + tan(α)

1− f (x , y) tan(α).



Thus, the differential equation of the oblique trajectory family is

dy

dx=

f (x , y) + tan(α)

1− f (x , y) tan(α)

Example 27

Find a family of oblique trajectories that intersects the family of straightlines y = kx at the angle of α = π

4 .



Exercises 9.2

Exercises

1. Find the family of curves which intersects each of the curves y = ce−x

at an angle π4 .

2. Find the family of curves which intersects each of the curves y = cx2

at an angle of π3 .

3. Find α if yα = cx intersects 12 ln

(

1 +(yx

)2)

+ ln |x | = arctan(yx

)+ d

at an angle of π4 at each point of intersection.

4. Show that there is no family of curves, which cuts itself at an angle ofθ, where θ ∈ (−π

2 , 0) ∪ (0, π2 ).


Some Physics Problems

Some Physics Problems

In this section, we will obtain mathematical models for some simple kindsof physics problems.


Some Physics Problems Problems in Cooling

Problems in Cooling

Newton’s law of cooling that the time rate of change of temperature of abody is proportional to the temperature difference between the body andits surrounding medium. Let T denote the temperature of the body andlet Tm denote the temperature of the surrounding medium. Then, thetime rate of change of temperature of the body is dT

dtand Newton’s law

of cooling can be formulated as

dT

dt= −k(T − Tm),

where k is a positive constant of proportionality.



Example 28

A thermometer, which has been showing 26 C inside a house, is placedoutside where temperature is 10 C. Three minutes later it is found thatthe thermometer reading is 12 C. We wish to predict the thermometerreading at any time.

Let T = T (t) be the temperature of the thermometer at the time t, weare given that

T (0) = 26 and T (3) = 12, Tm := 10.



According to Newton’s law, the rate of change of temperature dTdt

isproportional to the temperature difference (T − 10). Since thetemperature is decreasing, we choose (−k) as the constant ofproportionality.Thus, T = T (t) is to be determined from the equation

dTdt

= −k(T − 10)

T (0) = 26 and T (3) = 12.

Finally, we will find that T (t) = 16(12

)t+ 10.



Exercises 10.1

Exercises

1. Suppose that an object initially having a temperature of 20 C isplaced in a large temperature controlled room of 80 C and one hourlater the object has a temperature of 35 C. What will its temperaturebe after three hours?

2. A 200 F cup of tea is left in a 65 F room. At the initial time, the teais cooling at 5 F per minute. Write an initial-value problem modelingthe temperature T = T (t) of the tea. Find the heat of the tea after 5minutes.

3. A room is kept at a constant temperature of 68 F. A pie is taken outof a 350 F oven and placed on the counter. If the pie has reduced intemperature to 150 F in 45 minutes, when will the pie reach 80 F?


Some Physics Problems Problems in Mechanics

Problems in Mechanics

Newton’s second law states that the time rate of change of momentumof a body is proportional to the resulting force of acting on the body andis in the direction of this resultant force. In the mathematical language,we have

d

dt(mv) = F and F = ma,

where m is the mass of the body, v is the velocity, F is the force and a isthe acceleration.



The Motion of Falling Objects

An object fall through the air towards the earth. Assuming that the onlyforces acting on the object are gravity and air resistance, determine thevelocity of the object as a function of time.

Here, m is the mass of the object, g is the gravitational constant and k isthe drag coefficient.



Since F = ma, we have mg − kv = m dvdt

with an initial velocity ofv(0) = v0. Then, we get

dv

mg − kv=

dt

m,

which is an equation of separable type. Solving this, we get

v(t) =1

k

(

mg − e−k( t

m+c)

)

,

where c is an arbitrary constant. Using the initial condition v(0) = v0,we obtain

v(t) =

(

v0 −mg

k

)

e− kt

m +mg

k.



Exercises 10.2

Exercises

1. A parachutist bails out of an airplane at 10 000 ft, falls freely for20 seconds, then opens her parachute. The air resistance isproportional to speed. Assume for the drag coefficient, k1 = 0.15without a parachute and k2 = 1.5 with a parachute.

a. Find the IVP to model the problem during the period of free fall.b. Find the IVP to model the problem after the parachute opens.c. Find the total amount of time for the parachutist to land on.

2. A steel ball weighting 1 kg is dropped from 793m without anyvelocity. During the fall, the air resistance is equal to 1

8v ,where v isthe velocity of the ball in meters per second.

a. Compute the limiting velocity.b. Find the approximate time it takes for the ball to hit the ground if

g = 9.8m/s2.


Some Physics Problems Rate Problems

Rate Problems

The rate at which radioactive nuclei decay is known in general to beproportional to the number of such nuclei that are present in a givensample.

Example 29

Half the original number of radioactive Polonium-210 nuclei undergoesdisintegration in a period of 138 years.

Question 1. What percentage of the original Polonium-210 nuclei willremain after 414 years?

Question 2. In how many years will only one-tenth of the originalquantity remain?



Let x = x(t) be the amount of Polonium-210 nuclei present after years.Then, dx

dtrepresents the rate at which the nuclei decay. so,

dx

dt= −kx ,

where is a constant of proportionality. The constant is negative since x isdecreasing. Let x(0) = x0 be the initial amount. Solving the equation,we get x(t) = ce−kt , where c is an arbitrary constant. Let x(0) = x0 bethe initial amount. Then, we see that x(t) = x0e

−kt .



Denoting by x(t) the amount of Polonium-210 at the time t, we obtainat the end of the first period

x(138) = x0e−138k =

1

2x0

from which we get k := ln(2)138 . Thus,

x(t) = x0

(1

2

) t138

.

Answer 1. x(414) = x0(12

) 414138 = 1

8x0, which corresponds to 12.5% of theoriginal amount.

Answer 2. Solving x(T ) = 110x0, we lead to the equation

x0(12

) T138 = 1

10x0. Hence, T := 138 ln(10)ln(2) , which is

approximately 458 years and 5 months.



Exercises 10.3

Exercises

1. During a chemical reaction, substance A is converted into substanceB at a rate that is proportional to the square of the amount of A.Initially, 50 grams of A is present, and after 2 hours only 10 grams ofA remain unconverted. How much of A is present after 4 hours?

2. A chemical substance A transforms into another substance in achemical reaction. After 1 hour, 50 grams of A remain while after 3hours 25 grams of A remain.

a. How many grams of A were present in the beginning?b. How many grams of A will remain after 5 hours?c. After how many hours there will be 2 grams of A?



Exercises 10.3

3. A certain colony of bacteria grows at a rate proportional to thenumber of bacteria present.

a. If the number of the bacteria doubles every 10 hours, how longwill it take for the colony to triple in size?

b. If the number of bacteria in the culture is 5 million at the end of 6hours and 8 million at the end of 9 hours, how many bacteria werepresent initially?


Some Physics Problems Mixture Problems

Mixture Problems

A substance S is allowed to flow in to a certain mixture in a container ata certain rate and the mixture is kept uniform stirring. Further, in a suchsituation, this uniform mixture simultaneously flows out of the containerat another rate. Determine the quantity of the substance S present inthe mixture at the time t.



Let x = x(t) represent the amount of S present at time t. Then, dxdt

denotes the rate of change of x with respect to t. If rin denotes the rateat which S enters the mixture and rout the rate at which it leaves, thenwe have

dx

dt= rin − rout.

Example 30

Consider a large tank holding 1000 lt of water into which a brine solutionof salt begins to flow at a constant rate of 6 lt/min. The solution insidethe tank is kept well stirred and is flowing out of the tank at a rate of6 lt/min. If the concentration of salt in the brine entering tank is 1 kg/lt,determine when the concentration of salt in the tank will reach 1

2 kg/lt.



6 ltmin1 kglt

6 ltmin

xHtL

Note that x(t) is the mass of the salt in the tank at any time t and

rin := (6 lt/min)(1 kg/lt) = 6 kg/min

while

rout := (6 lt/min)

(x(t)

1000kg/lt

)

=3

500x(t) kg/min.



Hence, we arrive at the IVP

dxdt

= 6− 3500x

x(0) = 0,

which is a linear differential equation, whose solution is

x(t) = 1000(

1− e− 3t

500

)

.



Thus, the concentration of salt in the tank at any time t is

x(t)

1000= 1− e

− 3t500 kg/lt.

To determine when the concentration of the salt is 12 kg/lt, we solve

x(T ) = 12 , or equivalently

1− e− 3T

500 =1

2,

which gives T := 5003 ln(2) ≈ 116min. We observe that the mass of salt

in the tank steadily increases and has the limiting value

limt→∞

x(t) = limt→∞

1000(

1− e− 3t

500

)

= 1000 kg.



Exercises 10.4

Exercises

1. A 1500 gallon tank initially contains 600 gallons of water with 5 lbs ofsalt dissolved in it. Water enters the tank at a rate of 9 gal/hr and thewater entering the tank has a salt concentration of lbs/gal. If a wellmixed solution leaves the tank at a rate of 6 gal/hr, how much saltexists in the tank after one day?

2. Consider a tank with 200 liters of salt-water solution, 30 grams ofwhich is salt. Pouring into the tank is a brine solution at a rate of4 lt/min and with a concentration of 1 gr/lt. The “well-mixed”solution pours out at a rate of 5 lt/min. Find the amount of salt attime t.

3. The air in a room whose volume is 1000m3 tests 0.15% CO2. Startingat t = 0, outside air testing 0.05% CO2 is admitted at the rate of500m3/min.

a. What is the percentage of CO2 in the air in the room after 3min?b. When does the air in the room test 0.1% CO2?


Higher-Order Linear Differential Equations


A linear ordinary differential equation of order n in the dependent variabley and the independent variable x is an equation of the form

a0(x)dny

dxn+ a1(x)

dn−1y

dxn−1+ · · ·+ an(x)y = f (x),

where a0 is not zero. We shall assume that a0, a1, · · · , an and f arecontinuous real valued functions on an interval [a, b] and that a0(x) 6= 0for all x ∈ [a, b].



Basic Theory

Let C∞ be the set of all infinitely many differentiable functions definedon the closed interval [a, b] in R. Consider the mapping D : C∞ → C∞,then Dn : C∞ → C∞, where n ∈ N. Explicitly, we have D0y = y ,Dy = y ′ and so on. If a0, a1, · · · , an are functions of class C∞, then themapping L : C∞ → C∞ defined by

L[y ](x) :=a0(x)[Dny ](x) + a1(x)[D

n−1y ](x) + · · ·+ an(x)y(x)

=a0(x)y(n)(x) + a1(x)y

(n−1)(x) + · · ·+ an(x)y(x)

is a linear mapping in y = y(x).



Definition 22

The mapping L is called the “n-th order linear differential operator”,provided that a0 is never zero. Denoting by L := L(D), we can representthe equation in the form

L(D)y = f .

The right-hand member f is called the “non-homogeneous term”. If fis identically zero, then the equation reduces to L(D)y = 0 or explicitly

a0(x)dny

dxn+ a1(x)

dn−1y

dxn−1+ · · ·+ an(x)y = 0,

which is called the “homogeneous linear differential equation of

order n”.



The IVP for n-th order linear differential equation is given by

a0(x)y(n) + a1(x)y

(n−1) + · · ·+ an(x)y = f (x)

y(x0) =y0

y ′(x0) =y1...

y (n−1)(x0) =yn−1,

where x0 ∈ [a, b] and y0, y1, · · · , yn−1 ∈ R.



Theorem 10 (Existence and Uniqueness)

Let a0, a1, · · · , an and f be continuous on an interval I such thata0(x) 6= 0 for all x ∈ I . Then, for any x0 ∈ I , the IVP

a0(x)y(n) + a1(x)y

(n−1) + · · ·+ an(x)y = f (x)

y(x0) =y0

y ′(x0) =y1...

y (n−1)(x0) =yn−1

admits a unique solution, which exists on the entire interval I .



Corollary 1 (The Trivial Solution)

Let ϕ be a solution of the n-th order homogeneous linear differentialequation L(D)y = 0 such thatϕ(x0) = 0, ϕ′(x0) = 0, · · · , ϕ(n−1)(x0) = 0 for some x0 ∈ I . Then,ϕ(x) ≡ 0 for x ∈ I .

Example 31


y ′′ + 3xy ′ + x3y = ex

y(1) =2

y ′(1) =5

has a unique solution on the entire interval (−∞,∞).



Definition 23

Let f1, f2, · · · , fn be given functions and c1, c2, · · · , cn be given constants.Then, the expression

c1f1 + c2f2 + · · ·+ cnfn

is called a “linear combination of f1, f2, · · · , fn”.

Theorem 11 (Superposition Principle)

Let ϕ1, ϕ2, · · · , ϕn be solutions of the homogeneous linear differentialequation L(D)y = 0. Then, any linear combination of these solution isalso a solution of L(D)y = 0.



Definition 24 (Linear Dependence)

The functions f1, f2, · · · , fn are called “linearly dependent on I”provided that there exists constants c1, c2, · · · , cn, not all zero, such that

c1f1(x) + c2f2(x) + · · ·+ cnfn(x) = 0 for all x ∈ I .

f1, f2, · · · , fn are called “linearly independent on I” if

c1f1(x) + c2f2(x) + · · ·+ cnfn(x) = 0 for all x ∈ I

impliesc1 = c2 = · · · = cn = 0.



Example 32

1. f1(x) := x and f2(x) := x2 are linearly independent on [0, 1]. Indeed,c1x + c2x

2 = 0 for all x ∈ [0, 1] implies c1 = c2 = 0.

2. f1(x) := sin(x) and f2(x) := 3 sin(x) are linearly dependent on [0, π].



Definition 25 (Wronskian)

Let f1, f2, · · · , fn be (n − 1)-times differentiable on I . The determinant

W [f1, f2, · · · , fn] :=

∣∣∣∣∣∣∣∣∣

f1 f2 · · · fnf ′1 f ′2 · · · f ′n...

.... . .

...

f(n−1)1 f

(n−1)2 · · · f

(n−1)n

∣∣∣∣∣∣∣∣∣

is called the “Wronskian of the functions f1, f2, · · · , fn”, and its valueat x ∈ I denoted by W [f1, f2, · · · , fn](x).



Theorem 12 (Linear Independence)

Let ϕ1, ϕ2, · · · , ϕn be solutions of the n-th order homogeneous lineardifferential equation L(D)y = 0. Then, ϕ1, ϕ2, · · · , ϕn are linearlydependent on I if and only if

W [ϕ1, ϕ2, · · · , ϕn](x) ≡ 0 for x ∈ I .

Theorem 13

Let ϕ1, ϕ2, · · · , ϕn be solutions of L(D)y = 0, and denote by W theirWronskian. Then, either W (x) ≡ 0 for x ∈ I or W (x) 6= 0 for all x ∈ I .



Corollary 2

Let ϕ1, ϕ2, · · · , ϕn be solutions of L(D)y = 0. Then, ϕ1, ϕ2, · · · , ϕn arelinearly independent if and only if W (x) 6= 0 for all x ∈ I .

Example 33

Clearly, cos and sin are linearly independent on (−∞,∞).



Example 34

Let f1(x) := x |x | and f2(x) := x2 for [−1, 1]. Recall that |x | = x sgn(x)for x ∈ R. Then, we compute that

W [f1, f2](x) =

∣∣∣∣

x |x | x2

2|x | 2x

∣∣∣∣= 2x2|x | − 2x2|x | ≡ 0 for x ∈ [−1, 1].

However, f1 and f2 are linearly independent on [−1, 1], i.e., there doesnot exist constants c1 and c2 with c21 + c22 6= 0 such thatc1x |x |+ c2x

2 = 0 for all x ∈ [−1, 1]. Indeed, if x ≥ 0, c1x2 + c2x

2 = 0for all x ∈ [−1, 1] implies c1 = −c2, and if x ≤ 0 −c1x

2 + c2x2 = 0 for

all x ∈ [−1, 1] implies c1 = c2. This is a contradiction.



Remark 5

Exercise 34 does not conflict with Corollary 2 since f1 and f2 definedtherein cannot be solutions of a second-order homogeneous lineardifferential equation defined on [−1, 1].

Theorem 14

The n-th order homogenous linear differential equation L(D)y = 0 has atotal of n linearly independent solutions ϕ1, ϕ2, · · · , ϕn. Furthermore,any other solution ϕ of L(D)y = 0 can be written as a linear combinationof these n solutions, i.e., there exists suitable constants, not all zero,such that

ϕ = c1ϕ1 + c2ϕ2 + · · ·+ cnϕn.



Definition 26 (Fundamental Solutions)

Let ϕ1, ϕ2, · · · , ϕn be linearly independent solutions of the n-th orderhomogeneous linear differential equation L(D)y = 0 on the interval I .Then, the set ϕ1, ϕ2, · · · , ϕn is called a “set of fundamental

solutions of L(D)y = 0 on I”.

Definition 27 (General Solution)

Let ϕ1, ϕ2, · · · , ϕn be a set of fundamental solutions of L(D)y = 0 onI . Then, the linear combination

ϕ = c1ϕ1 + c2ϕ2 + · · ·+ cnϕn on I

is called the “general solution of L(D)y = 0 on I”.



Exercises 11

Exercises

1. Check linear independence of the following functions.

a. 1, x , x2 and (x + 1)2.b. cos(x), sin(x) and sin(2x).

c. π,(tan(x)

)2and sec2(x).

2. For x ∈ R, consider the functions eαx and eβx , where α, β ∈ R.

a. Show that they are solutions of y ′′ − (α+ β)y ′ + αβy = 0.b. Show that they are linearly independent if and only if α 6= β.

3. For x > 0, consider the functions 1, 1xand x2.

a. Determine α, β, γ ∈ R such that they are solutions of

x3y ′′′ + αx2y ′′ + βxy ′ + γy = 0 for x > 0.

b. Show these solutions that they are linearly independent.c. Write the general solution of the equation.


Higher-Order Linear Differential Equations Non-Homogeneous Linear Differential Equations

Non-Homogeneous Linear Differential Equations

In this section, we will consider

L(D)y = f (x).

Theorem 15

Let ϕ be any solution of the non-homogeneous linear differential equation

L(D)y = f (x),

and ψ be any solution of the associated homogeneous equationL(D)y = 0. Then, the sum (ϕ+ ψ) is solution of the non-homogeneousequation L(D)y = f (x).



Definition 28

Consider the n-th order non-homogeneous linear differential equation

L(D)y = f (x) (A)

and the associated homogeneous equation

L(D)y = 0. (B)

1. The general solution of (B) is called “complementary solution of

(A)”. We will denote this solution by ϕc.

2. Any solution of (A) without arbitrary constants is called a “particularsolution of (A)”, which we will denote by ϕp.

3. The sum (ϕc + ϕp) is called the “general solution of (A)”.



Example 35

Consider the equationy ′′ + y = x

whose associated homogeneous equation y ′′ + y = 0 has thecomplementary solution ϕc(x) := c1 cos(x) + c2 sin(x). Moreover, aparticular solution of the original equation is ϕp(x) := x . Therefore, thegeneral solution of the original equation isϕ(x) := ϕc(x) + ϕp(x) = c1 cos(x) + c2 sin(x) + x .



Example 36

Consider the equation

yy ′′ +(y ′)2

= 0 for x ∈ [0, 1],

which has the solutions ϕ1(x) :≡ 1 and ϕ2(x) :=√x for x ∈ [0, 1].

Unfortunately, for any constants c1 and c2, the functionϕ(x) := c1 + c2

√x is not a solution of the given equation. Because it is

not linear.



Theorem 16

Consider the differential equation L(D)y = f , wheref := c1f1 + c2f2 + · · ·+ cmfm. Let L(D)ϕ1 = f1, L(D)ϕ2 = f2, · · · ,L(D)ϕm = fm. Then, ϕ defined by ϕ := c1ϕ1 + c2ϕ2 + · · ·+ cmϕm is asolution of L(D)y = f .



Example 37

Consider the equations

y ′′ − 5y ′ + 6y =1

y ′′ − 5y ′ + 6y =x

y ′′ − 5y ′ + 6y =ex ,

whose respective particular solutions are ϕ1(x) :=16 , ϕ2(x) :=

x6 + 5

36and ϕ3(x) :=

12e

x . Then, ϕ(x) := ex + 6x + 11 is a solution of

y ′′ − 5y ′ + 6y = 2ex + 36(x + 1).



Exercises 11.1

Exercises

1. Consider the differential equation

(x + 1)y ′′ + xy ′ − y = (x + 1)2 for x > −1. (A)

Show that two solutions of the homogeneous equation associated with(A) are ϕ1 := x and ϕ2 := e−x , and that ϕp := x2 − x + 1 is asolution of (A). Write the general solution of (A).

2. Consider the differential equation

y ′′ − tan(x)y ′ −(sec(x)

)2y = sec(x) tan(x) for x ∈ (−π

2 ,π2 ). (B)

Verify that ϕ1 := sec(x) and ϕ2 := tan(x) are solutions of thehomogeneous equation associated with (B), and ϕp := x sec(x) is asolution of (B). What is the general solution of (B)?


Reduction of Order for Homogeneous Linear Differential Equations

Reduction of Order for Homogeneous Linear Differential

Equations

If one solution of a second order homogeneous linear differential equationis known, then a second linearly independent solutions and hence afundamental set of solution can be determined.Actually, a more general result holds.



Theorem 17

Let ϕp be a non-trivial solution of the n-th order homogeneous lineardifferential equation

a0(x)dny

dxn+ a1(x)

dn−1y

dxn−1+ · · ·+ an(x)y = 0. (A)

Then, the transformation y = ϕpz reduces (A) to an (n − 1)-st orderhomogeneous linear differential equation in the dependent variablew := dz

dx.

Remark 6

In order to apply this procedure to the second-order homogeneous lineardifferential equation, one solution of the given equation must be known.



Example 38

Given that ϕp(x) := x is a solution of

(x2 + 1

)y ′′ − 2xy ′ + 2y = 0

find a linearly independent solution by reduction of order.

Example 39

Given that ϕp(x) := x is a solution of

y ′′ − 2xy ′ + 2y = 0

find the second linearly independent solution.



Exercises 12

Exercises

1. Obtain the general solution of

y ′′ − 4x

2x − 1y ′ +

4

2x − 1y = 0 for x > 1

after finding an exponential solution (i.e., ϕp := eλx).

2. Find the general solution of the equation y ′′ + r(x)y ′ + r ′(x)y = f (x),where r is a differentiable function and f is a continuous function.

3. Find a monomial type solution for

y ′′′ − 6

xy ′′ +

1

x2(x2 + 18

)y ′ − 2

x3(x2 + 12

)y = 0 for x > 0.

Then, find its general solution after reduction of order process.



Exercises 12

4. Reduce the order of the equation

y ′′′ + a1(x)y′′ + a2(x)y

′ + a3(x)y = 0

if ϕp is one of its solutions.

5. Find a third-order differential equation whose reduced equation is

w ′′ + a1(x)w′ + a2(x)w = 0.

6. Use Leibnitz’s formula for the n-th derivative of a product, i.e.,(fg)(n) =

∑nk=0

(nk

)f (k)g (n−k), to obtain the reduced equation of

dny

dxn+ a1(x)

dn−1y

dxn−1+ · · ·+ an(x)y = 0,

where a1, a2, · · · , an are continuous functions.


Homogeneous Linear DEs with Constant Coefficients

Homogeneous Linear Differential Equations with Constant

Coefficients

In this section, we consider the special case of the n-th orderhomogeneous linear differential equation whose coefficients are realconstants, i.e.,

a0dny

dxn+ a1

dn−1y

dxn−1+ · · ·+ any = 0, (1)

where a0, a1, · · · , an are real constants with a0 6= 0. Such equations arealso called “autonomous” since the coefficients are independent of x .



Next, we will find how to obtain the general solution of (1) explicitly.Let us now consider a function such that its higher-order derivatives areconstant multipliers of itself, i.e.,

dky

dxk= cy for all x .

The exponential function eλx , where λ is a constant, is such a function.

Indeed, we havedk

dxkeλx = λkeλx for all x .



Assuming that ϕ(x) := eλx , where λ is to be determined later, is asolution of (1), we getϕ′(x) = λeλx , ϕ′′(x) = λ2eλx , · · · , ϕ(n)(x) = λneλx . Substituting theseinto (1) yields that

a0λneλx + a1λ

n−1eλx + · · ·+ ane

λx = 0

or equivalently

(a0λ

n + a1λn−1 + · · ·+ an

)eλx = 0.

Since eλx 6= 0, we obtain the polynomial equation in the unknown λ as

a0λn + a1λ

n−1 + · · ·+ an = 0. (2)

(2) is called the “characteristic equation associated with (1)”.



If ϕ(x) := eλx is a solution of (1), then λ must satisfy the characteristicequation (2). Therefore, in order to solve (1), we write the characteristicequation (2) and solve it for λ.There are three following possible cases for roots of the characteristicequation (2) to be considered.

• Real and Distinct.

• Real and Repeated.

• Complex Conjugate.


Homogeneous Linear DEs with Constant Coefficients Distinct Real Roots

Distinct Real Roots

Theorem 18

Consider the n-th order homogeneous autonomous linear differentialequation L(D)y = 0. Let λ0 be a real root of the associatedcharacteristic polynomial. Then, λ0 contributes the general solution with

c1eλ0x ,

where c1 is an arbitrary constant.

Example 40

Find the general solutions of the following equations.

1. y ′′ − 3y ′ + 2y = 0.

2. y ′′′ − 4y ′′ + y ′ + 6y = 0 (λ1 := −1, λ2 := 2 and λ3 := 3).


Homogeneous Linear DEs with Constant Coefficients Distinct Real Roots

Exercises 13.1

Exercises

1. Find the general solution of y ′′′ − 5y ′′ − 4y ′ + 20y = 0.

2. Find the general solution of y iv + y ′′′ − 7y ′′ − y ′ + 6y = 0.

3. What is the general solution of yv − 5y ′′′ + 4y ′ = 0?


Homogeneous Linear DEs with Constant Coefficients Repeated Real Roots

Repeated Real Roots

Let λ0 be a root of multiplicity m. We might expect that m linearlyindependent solutions of L(D)y = 0 are

eλ0x , xeλ0x , · · · , xm−1

eλ0x .

Note that in this case the characteristic equation will be of the form

(λ− λ0)mP(λ) = 0,

where P is a polynomial such that P(λ0) 6= 0.



Then,L(D)eλx = (λ− λ0)

mP(λ)eλx , (3)

which yields 0 by substituting λ0 for λ. That is, eλ0x is a solution ofL(D)y = 0. To find the other solutions, we take the i-th partialderivative of (3) with respect to λ. Then,

∂ i

∂λiL(D)eλx =

∂ i

∂λi(λ− λ0)

mP(λ)eλx for i = 1, 2, · · · ,m− 1.

Since, for i = 1, 2, · · · ,m − 1, the right-hand side still includes the factor(λ− λ0), we get

∂ i

∂λiL(D)eλx

∣∣∣∣λ=λ0

= 0 for i = 1, 2, · · · ,m − 1.



As eλx has continuous partial derivatives of all orders with respect to λand x , we have

∂ i

∂λi∂ j

∂x jeλx =

∂ j

∂x j∂ i

∂λieλx ,

and thus

L(D)∂ i

∂λieλx

︸︷︷︸

x ieλx

∣∣∣∣λ=λ0

= 0 for i = 1, 2, · · · ,m− 1

showing thatxeλ0x , x2eλ0x , · · · , xm−1

eλ0x

are solutions of L(D)y = 0.



Theorem 19

Consider the n-th order homogeneous autonomous linear differentialequation L(D)y = 0. Let λ0 be a real root of the associatedcharacteristic polynomial of multiplicity m. Then, λ0 contributes thegeneral solution with

eλ0x

(c1 + c2x + · · ·+ cmx

m−1),

where c1, c2, · · · , cm are arbitrary constants.



Exercises 13.2

Exercises

1. Find the general solution on y ′′ − 6y ′ + 9y = 0.

2. Find the general solution of y ′′′ − 4y ′′ + 5y ′ − 2y = 0.

3. Obtain the general solution of y iv + 5y ′′′ + 6y ′′ − 4y ′ − 8y = 0.

4. Obtain the general solution of y (n) = 0.


Homogeneous Linear DEs with Constant Coefficients Complex Conjugate Roots

Complex Conjugate Roots

Suppose that the characteristic equation has complex roots. Letλ1 := α+ iβ be a complex root. Since the characteristic polynomial hasonly real coefficients, we see that λ2 := λ1 = α− iβ is also a root of thecharacteristic polynomial. Then, e(α+iβ)x and e(α−iβ)x are solutions ofL(D)y = 0, and thus for any constants d1 and d2, we see thatϕ(x) := d1e

(α+iβ)x + d2e(α−iβ)x is also a solution.



Using the Euler’s formula, we see that

ϕ(x) =d1eαx[cos(βx) + i sin(βx)

]+ d2e

αx[cos(βx) − i sin(βx)

]

=eαx[(d1 + d2)︸︷︷︸

c1

cos(βx) + i(d1 − d2)︸︷︷︸

c2

sin(βx)].

Thus,eαx cos(βx) and e

αx sin(βx)

define two linearly independent solutions for L(D)y = 0.



Theorem 20

Consider the n-th order homogeneous autonomous linear equationL(D)y = 0.

1. Let the characteristic equation have the complex conjugate roots(α± iβ). Then, its contribution to the general solution is

eαx(c1 cos(βx) + c2 sin(βx)

),

where c1 and c2 are arbitrary constants.

2. If the roots (α± iβ) are of multiplicity m with 2m ≤ n. Then, itscontribution to the general solution is

eαx[(c1 + c2x + · · ·+ cmx

m−1)cos(βx)

+(cm+1 + cm+2x + · · ·+ c2mx

m−1)sin(βx)

],

where c1, c2, · · · , c2m are arbitrary constants.



Example 41

Find the general solutions of the following equations.

1. y ′′ + y = 0.

2. y ′′ − 6y ′ + 25y = 0.

3. y (4) − 4y ′′′ + 14y ′′ − 20y ′ + 25y = 0 (λ1,2,3,4 := 1± 2i).

Conclusion 2

By Theorem 14, we obtain a total of n roots for the associatedcharacteristic polynomial of L(D)y = 0 to form its general solution.



Exercises 13.3

Exercises

1. Find the general solution of y ′′ − 4y ′ + 13y = 0.

2. Find the general solution of y iv + 4y ′′′ + 8y ′′ + 8y ′ + 4y = 0.

3. What is the general solution of

y (6) − 6yv + 18y iv − 32y ′′′ + 36y ′′ − 24y ′ + 8y = 0?



Exercises 13

Exercises

1. Find the general solution of y iv − 2y ′′′ − 2y ′′ + 8y = 0.

2. Find the general solution of

y (7) − 11y (6) + 46yv − 78y iv − 19y ′′′ + 225y ′′ − 108y ′ − 216y = 0.


y (7) − 2y (6) + 9yv − 68y iv + 76y ′′′ − 240y ′′ + 900y ′ = 0.

4. Find the differential equation of least possible order for which xe−2x ,sin(2x) and x2e2x cos(x) are included in its general solution.




In this section, we consider the special case of the n-th ordernon-homogeneous linear differential equation whose coefficients are realvalued functions, i.e.,

a0(x)dny

dxn+ a1(x)

dn−1y

dxn−1+ · · ·+ an(x)y = f (x),

where a0, a1, · · · , an : I → R are continuous functions with a0 6= 0 for allx in some interval I .


Non-Homogeneous Linear Differential Equations The Method of Undetermined Coefficients

The Method of Undetermined Coefficients

We now consider the non-homogeneous differential equation withconstant coefficients of the form

a0dny

dxn+ a1

dn−1y

dxn−1+ · · ·+ any = f (x),

where a0, a1, · · · , an are real constants with a0 6= 0 and f is a function,which is a solution of some homogeneous autonomous linear differentialequation.Let us recall that the general solution ϕ := ϕc + ϕp, where Lϕc = 0 andLϕp = f . We learned how to find ϕc in the previous section. Now, weconsider methods of determining the particular solutions.



Definition 29

A function is said to be of “UC class” if it is defined by one of thefollowing properties.

1. xm, where m ∈ N0.

2. eλx , where λ ∈ R with λ 6= 0.

3. cos(λx) or sin(λx), where λ 6= 0.

4. Finite products and sums of the functions satisfying the propertiesabove.

Example 42

For instance, e2x(cos(x) + sin(x)

), x3e4x and x5 sin(x) are of UC class.

Remark 7

The method of undetermined coefficients applies, when thenon-homogeneous function f in the differential equation is a finite linearcombination of UC functions.



Definition 30

Consider a UC function f . The set of functions consisting of f itself andall linearly independent UC functions of which the successive derivativesof f are either constant multiples or linear combinations will be called asthe “UC set of f ”.

Example 43

Obtain the UC set for each of the following functions.

1. f (x) := x3.

2. f (x) := sin(2x).

3. f (x) := x2 cos(x).



Solution Procedure

Let ϕ0 be a particular solution of L(D)y = f , wheref := c1f1 + c2f2 + · · ·+ cmfm such that c1, c2, · · · , cm are real constantsand f1, f2, · · · , fm are of UC class. Further, suppose that ϕc is obtained,i.e., L(D)ϕc = 0.

Step 1. For each UC function f1, f2, · · · , fm, form the corresponding UCset S1, S2, · · · , Sm.

Step 2. Omit Si if Si ⊂ Sj holds for some i , j = 1, 2, · · · ,m.

Step 3. Suppose one of these sets, say Si includes some terms of ϕc.Then, multiply Si by xm, where m is the multiplicity of the rootof the characteristic polynomial generating that term, and checkagain Step 2.

Step 4. Form a linear combination of all elements of these sets withunknown constant coefficients.

Step 5. Determine these unknown coefficients by substituting the linearcombination into the differential equation such that it identicallysatisfies the equation.



Example 44

Find the general solution of the equation

y ′′ − 2y ′ − 3y = 4ex − 10 sin(x).

The general solution is

ϕ(x) := c1e−x + c2e

3x − ex − cos(x) + 2 sin(x).

Example 45


y ′′ − 3y ′ + 2y = 4x2 + ex(2x + 1) + 8e3x .


ϕ(x) :=c1ex + c2e

2x

+ 2x2 + 6x + 7− exx(x + 3) + 4e3x .



Example 46


y (4) + y ′′ = 12x2 − 2 cos(x) + 4 sin(x).


ϕ(x) :=c1x + c2 + c3 cos(x) + c4 sin(x)

+ x4 − 12x2 + 2x cos(x) + x sin(x).



Exercises 14.1

Exercises

1. Find the general solution of the following differential equations.

a. y ′′′ + y ′′ = 2ex + 18e−3x .b. y ′′′ + y ′′ = cos(x)− sin(x).c. y ′′′ + y ′′ = e−x .d. y ′′′ + y ′′ = 20x3 − 12x2.e. y ′′′ + y ′′ = e

−x + 6x .

2. Find the general solution of the following differential equations.

a. y ′′′ + 4y ′ = 24x2.b. y ′′′ + 4y ′ = 3 cos(x) + 6 sin(x).c. y ′′′ + 4y ′ = e

x(7 cos(2x) + 6 sin(2x)

).

d. y ′′′ + 4y ′ = x cos(2x)− sin(2x).e. y ′′′ + 4y ′ = 64 cos(x) sin(x).

f. y ′′′ + 4y ′ =(cos(x)

)2.

3. Find the general solution of y (n) = xα for x > 0, where α ∈ R.


Non-Homogeneous Linear Differential Equations The Method of Variation of Parameters

The Method of Variation of Parameters

This method also applies in the case when the coefficients are functionsof x once the complementary solution is known. We shall develop thismethod in connection with the second-order linear differential equationswith variable coefficients of the form

a0(x)d2y

dx2+ a1(x)

dy

dx+ a2(x)y = f (x)

where the coefficients and the function on the right-hand side satisfy theprevious properties. Suppose that ϕ1 and ϕ2 are linearly independentsolutions of the corresponding homogeneous equation, then thecomplementary solution is ϕc(x) := c1ϕ1(x) + c2ϕ2(x), where c1 and c2are arbitrary constants.



The procedure in the method of variation of parameter is to replace thearbitrary constants c1 and c2 by the respective functions u1 and u2, whichwill be determined so that the resulting function

ϕp(x) := u1(x)ϕ1(x) + u2(x)ϕ2(x)

will be a particular solution of the equation.



Theorem 21

Assume that a0, a1, a2 and f are continuous on the interval I such thata0(x) 6= 0 for all x ∈ I , and consider the differential equation

a0(x)d2y

dx2+ a1(x)

dy

dx+ a2(x)y = f (x). (A)

Let ϕ1 and ϕ2 be linearly independent solutions of the correspondinghomogeneous equation. Then, a particular solution of (A) is given by

ϕp(x) := −ϕ1(x)

∫ x ϕ2(η)f (η)

a0(η)W (η)dη + ϕ2(x)

∫ x ϕ1(η)f (η)

a0(η)W (η)dη,

where W denotes the Wronskian of ϕ1 and ϕ2.



Example 47

By using the method of variation of parameters, find a particular solutionof the equation

y ′′ + y = tan(x).



Exercises 14.2

Exercises

1. Find the general solution of y ′′ − 2y ′ + y = ex

x2+1 .

2. For α ∈ R, obtain the general solution of y ′′ − 2y ′ + y = xαex .

3. Find the general solution of y ′′′ + 4y ′ = 40x4.

4. Find the general solution of y ′′′ − y ′′ + y ′ − y = cos(αx) for α ∈ R.


Cauchy-Euler Equations


Definition 31 (Cauchy-Euler Equation)


a0xn d

ny

dxn+ a1x

n−1 dn−1y

dxn−1+ · · ·+ any = f (x),

where a0, a1, · · · , an are constants with a0 6= 0 and f is a continuousfunction, is called a “Cauchy-Euler differential equation of order n”.

Note that a Cauchy-Euler differential equation is linear differentialequation with variable coefficients. The leading coefficient a0x

n becomeszero at x = 0. Thus, we will consider such equations in any interval notincluding the origin, which we will assume without loss of generality thatthis interval is (0,∞).



Theorem 22

The transformation x = et transforms

a0xn d

ny

dxn+ a1x

n−1 dn−1y

dxn−1+ · · ·+ any = f (x)

into a linear differential equation with constant coefficients, i.e.,

b0dny

dtn+ b1

dn−1y

dtn−1+ · · ·+ bny = g(t),

where b0, b1, · · · , bn are constants with b0 6= 0 and g is a continuousfunction.

Remark 8

Equations of the form

a0(x − x0)n d

ny

dxn+ a1(x − x0)

n−1 dn−1y

dxn−1+ · · ·+ any = f (x) for x > x0.

can be treated similarly by using the transformation x = et + x0.



Example 48

Show that the general solution of the Cauchy-Euler equation

x3y ′′′ − 4x2y ′′ + 8xy ′ − 8y = 32 ln(x)− 64 for x > 0

isϕ(x) := c1x + c2x

2 + c3x4 − 4 ln(x + 2) + 1,

where c1, c2 and c3 are arbitrary constants.

Example 49


(x + 2)2y ′′ − 2(x + 2)y ′ − 4y = 4 ln(x + 2) + 3 for x > −2

isϕ(x) :=

c1x + 2

+ c2(x + 2)4 − ln(x + 2),




Example 50


(2x − 3)2y ′′ − 6(2x − 3)y ′ + 12y = 4x − 6 for x >3

2

is

ϕ(x) := c1(2x − 3) + c2(2x − 3)3 − 1

4(2x − 3) ln(2x − 3),




Exercises 15

Exercises

1. Let a1, a2, · · · , an ∈ R and considerxny (n) + a1x

n−1y (n−1) + · · ·+ any = 0 for x > 0. Obtain thecharacteristic equation by substituting y = xλ. Recall that for theequation transformed by x = et , we seek solutions of the formy = eλt , which yields y = xλ.

2. Obtain the general solution of x2y ′′ + 5xy ′ + 4y = 0 for x > 0.

3. Find the general solution of x2y ′′ + xy ′ + y =(

sec(ln(x)

))3

for

x > 0.

4. Let α ∈ R, and find the general solution of x2y ′′ − 6y = xα for x > 0.

5. Solve x3y ′′′ − 3x2y ′′ + 7xy ′ − 8y = x2 ln(x) for x > 0.

6. Solve (2x + 1)2y ′′ − 2(2x + 1)y ′ − 12y(x) = 8x for x > 0.


Power Series Solutions


Consider the second-order homogeneous linear equation

a0(x)y′′ + a1(x)y

′ + a2(x)y = 0 for x ∈ I ,

where a0, a1, a2 are continuous functions with a0(x) 6= 0 for all x ∈ I , andsuppose that this equation has no solution, which is expressible as a finitelinear combination of known elementary functions. Let us assume that ithas a solution, which can be expressed in the form of an infinite series

ϕ(x) := c0 + c1(x − x0) + · · ·+ cn(x − x0)n + · · · =

∞∑

k=0

ck(x − x0)k ,

where c0, c1, · · · are constants.This expression is called a power series centered at x0. Thus, we haveassumed that the differential equation has power series solution of theabove form. In this chapter, we investigate the existence of such typessolutions and we will try to find it, i.e., determine its coefficients.



Normalized Form

This kind of equations are in general treated in the following form

y ′′ +a1(x)

a0(x)︸︷︷︸

p1(x)

y ′ +a2(x)

a0(x)︸︷︷︸

p2(x)

y = 0,

which is called the “normalized form”.

Definition 32 (Analytic Function)

A function f is said to be “analytic at x0” provided that its Taylor seriesabout x0

∞∑

k=0

f (k)(x0)

k!(x − x0)

k for |x − x0| < ρ

for some ρ > 0, exists and converges to f .



Definition 33 (Ordinary Point and Singular Point)

The point x0 is called an “ordinary point” for the differential equation

a0(x)y′′ + a1(x)y

′ + a2(x)y = 0 (A)

provided that both p1 :=a1a0

and p2 :=a2a0

are analytic at x0. If at leastone of these functions is not analytic at x0, then x0 is called a “singularpoint of (A)”.

Example 51

For the differential equation

y ′′ + xy ′ +(x2 − x + 2

)y = 0,

we have p1(x) := x and p2(x) := x2 − x + 2, which are analytic on theentire interval. So all points are ordinary points for this equation.



Example 52


(x − 1)y ′′ + xy ′ +1

xy = 0

has the normalized form

y ′′ +x

x − 1y ′ +

1

x(x − 1)y = 0.

Hence, p1(x) :=x

x−1 and p2(x) :=1

x(x−1) . Clearly, p1 is analytic for all

points except x = 1, and p2 is analytic for all points except x = 0, 1.Thus, x = 0 and x = 1 are singular points, and all other points areordinary.



Theorem 23

Let the point x0 be an ordinary point for the differential equation

a0(x)y′′ + a1(x)y

′ + a2(x)y = 0. (A)

Then, (A) has two linearly independent analytic solutions of the form

∞∑

k=0

ck(x − x0)k ,

and they converge in some interval I : |x − x0| < ρ. Moreover, the radiusof convergence of any solution is at least as large as the distance from x0to the nearest singular point (real or complex valued) of the equation (A).



Solution Procedure

Now, we will explain how we proceed to find these solutions. In otherword, how do we determine the coefficients of the series

ϕ(x) :=

∞∑

k=0

ck(x − x0)k

so that this expression does satisfy the equation? Differentiation yelds

ϕ′(x) =

∞∑

k=0

kck(x − x0)k−1 =

∞∑

k=1

kck(x − x0)k−1

ϕ′′(x) =

∞∑

k=0

k(k − 1)ck(x − x0)k−2 =

∞∑

k=2

k(k − 1)ck(x − x0)k−2

for |x − x0| ≤ ρ0, where ρ0 < ρ.



Substituting these into the equation yields

a0(x)∞∑

k=2

k(k − 1)ck(x − x0)k−2 + a1(x)

∞∑

k=1

kck(x − x0)k−1

+a2(x)

∞∑

k=0

ck(x − x0)k = 0,

which can be rearranged as

b0 + b1(x − x0) + · · ·+ bn(x − x0)n + · · · = 0,

where for i = 0, 1, · · · , the coefficient bi involves cj for j = 0, 1, · · · , i . Inorder to solve this equation, we set b0 = b1 = · · · = 0. This leads to aninfinite number of conditions, which must be satisfied by the coefficientsc0, c1, · · · .



Example 53

Find the power series solution of the differential equation

(x2 − 1)y ′′ + 4xy ′ + 2y = 0.

Example 54


(x2 − 1

)y ′′ + 6xy ′ + 4y = 0.

Note that x = 0 is an ordinary point, and x = ±1 are singular points.The series solution of this equation is

ϕ(x) := c0

∞∑

k=0

(k + 1)x2k + c1

∞∑

k=0

(2k + 3)x2k+1 for |x | < 1,




Example 55 (Shifting an Equation)


(x2 − 1

)y ′′ + 3xy ′ + xy = 0.

Note that x = 2 is an ordinary point. Then, solutions of the equation areof the form

ϕ(x) :=

∞∑

k=0

ck(x − 2)k for |x − 2| < 1.

Now, let t := x − 2, then dt = dx , which yields

ψ(t) :=

∞∑

k=0

cktk for |t| < 1,

and the equation turns out to be

(t2 + 4t + 3

)y + 3(t + 2)y + (t + 2)y = 0,

where y := dydt

and y := d2y

dt2.



Example 56 (Legendre Equation)

The general Legendre differential equation of order ν is

(1− x2

)y ′′ − 2xy ′ + ν(ν + 1)y = 0

whose general solution is

ϕ(x) :=c0

∞∑

k=0

(−1)k

(2k)!

(k−1∏

ℓ=0

(ν − 2ℓ)

)( k∏

ℓ=1

(ν + (2ℓ− 1)

))

x2k

︸︷︷︸

pν (x)

+ c1

∞∑

k=0

(−1)k

(2k + 1)!

(k−1∏

ℓ=0

(ν − (2ℓ+ 1)

))( k∏

ℓ=1

(ν + 2ℓ

))

x2k+1

︸︷︷︸

qν (x)

.

We can show that pν(x) and qν(x) converge for |x | < 1, and are linearlyindependent.



If ν = n, where n is an integer, then either pn or qn terminates. Hence,we get polynomial solutions for Legendre equations of integer order. Inthis case, the coefficients c0 and c1 can be chosen so that thesepolynomial solutions attain the value 1 at x = 1. These polynomials arecalled as the “Legendre polynomials”.



Exercises 16

Exercises

1. Find the general solution of y ′′ + 1x+1y

′ = 0 for x > 0.

2. Find the general solution of the so-called “Airy’s differential

equation” y ′′ − xy = 0.

3. Find the general solution of y ′′ + xy ′ + y = 0.


y ′′ + xy ′ +(x2 + 1

)y = 0.

5. Obtain the general solution of

(x4 + 3

)y ′′ − 4x3y ′ + 6

(x2 + 1

)y = 6x

(x2 + 3

).



6. After making the substitution y := u√xfor the dependent variable,

solve

x2y ′′ + xy ′ +

(

x2 − 1

4

)

y = 0 for x > 0,

which is known as the “Bessel equation of order 12”.

7. Apply the series solution technique for the first-order differentialequation y ′ + p(x)y = 0, where p(x) :=

∑∞k=0 pkx

k for |x | < ρ, whereρ > 0.


The Method of Frobenious (To Be Skipped)

The Method of Frobenious

Consider the homogeneous linear differential equation

a0(x)y′′ + a1(x)y

′ + a2(x)y = 0, (A)

where a0, a1, a2 are continuous functions. If x0 is a singular point of (A),then we cannot obtain solutions of power series form. In this case, welook for a solution of the form

|x − x0|r∞∑

k=0

ck(x − x0)k ,

where r is a (real or complex) constant to be determined later.



Definition 34 (Regular Singular Point and Irregular Singular Point)

Let x0 not be a regular point of (A). x0 is called a “regular singularpoint for (A)” provided that

(x − x0)p1(x) and (x − x0)2p2(x),

where p1 :=a1a0

and p2 :=a2a0, are both analytic at x0. Otherwise, x0 is

called an “irregular singular point”.



Example 57

Consider the equation

2x2y ′′ − xy ′ + (x − 5)y = 0,

whose normalized form is

y ′′ − 1

2xy ′ +

x − 5

x2y = 0.

Note that − 12x and x−5

x2are not analytic at 0. Since x

(− 1

2x

)= − 1

2 and

x2 x−5x2

= (x − 5) are analytic at 0, 0 is a regular singular point for theequation.



Example 58

The equation

x2(x − 2)2y ′′ + 2(x − 2)y ′ + (x + 5)y = 0

has the normalized form

y ′′ +2(x − 2)

x2(x − 2)2y ′ +

x + 5

x2(x − 2)2y = 0.

Clearly, 0 and 2 are the only singular points. For the singular point 0, we

see that x 2(x−2)x2(x−2)2 = 2

x(x−2) and x2 x+5x2(x−2)2 = x+1

(x−2)2 . However, the

former one is not analytic at 0, and thus 0 is an irregular singular point.

For the singular point 2, we see that (x − 2) 2(x−2)x2(x−2)2 = 2

x2and

(x − 2)2 x+5x2(x−2)2 = x+1

x2both of which are analytic at 2. Therefore, 2 is a

regular singular point.



Theorem 24

Let x0 be a regular singular point for (A). Then, there exists at least onenon-trivial solution of the form

|x − x0|r∞∑

k=0

ck(x − x0)k ,

where r is a fixed (real or complex) number. Moreover, this solution isexists and converges in a deleted interval 0 < |x − x0| < ρ.



Solution Procedure

Next, we proceed to determine the number r and the coefficientsc0, c1, · · · .Step 1. Let x0 be a regular singular point. Without loss of generality, we

let 0 < x − x0 < ρ, the case where −ρ < x − x0 < 0 can betreated similarly. Assume that

ϕ(x) := |x − x0|r∞∑

k=0

ck (x − x0)k ,

where c0 6= 0, is a solution. Equivalently, we have in this case,

ϕ(x) =

∞∑

k=0

ck(x − x0)k+r for 0 < x − x0 < ρ.



Step 2. Taking the derivatives, we have

ϕ′(x) =

∞∑

k=0

(k + r)ck(x − x0)k+r−1

ϕ′′(x) =

∞∑

k=0

(k + r)(k + r − 1)ck(x − x0)k+r−2

for 0 < x − x0 ≤ ρ0, where ρ0 < ρ.

Step 3. Substituting these into the equation and making arrangements,we obtain

b0(x − x0)r + b1(x − x0)

r+1 + · · ·+ bn(x − x0)n+r + · · · = 0,

where for i = 0, 1, · · · , the coefficient bi involves r and cj forj = 0, 1, · · · , i .



Step 4. In order to solve this equation, we set b0 = b1 = · · · = 0.

Step 5. From b0 = 0, we get the indicial equationr2 +

(p1(0)− 1

)r + p2(0) = 0. Denote by r1 and r2, the roots of

this equation such that Re(r1) ≥ Re(r2), where Re gives thereal part of a complex number.

Step 6. We now equate to zero the coefficients b1, b2, · · · , which leadsus to an infinite number of conditions on the coefficientsc0, c1, · · · involving r .



Step 7. Substituting r1 obtained in Step 5 and solving the coefficientsc0, c1, · · · recursively gives us a solution of the desired form.

Step 8. Step 7 can be repeated for r2 instead of r1 to obtain anothersolution. However, if (r1 − r2) is a nonnegative integer, then thissolution is in general linearly dependent with the former one.



Example 59

By using the method of Frobenious, find the general solution of theequation

2x2y ′′ − xy ′ + (x − 5)y = 0.

Clearly, 0 is a regular singular point. Let

ϕ(x) :=

∞∑

k=0

ckxk+r for 0 < x < ρ,

where c0 6= 0. Substituting this into the equation yields

0 =2x2∞∑

k=0

(k + r)(k + r − 1)ckxk+r−2 − x

∞∑

k=0

(k + r)ckxk+r−1

+ (x − 5)

∞∑

k=0

ckxk+r



=2

∞∑

k=0

(k + r)(k + r − 1)ckxk+r −

∞∑

k=0

(k + r)ckxk+r

+

∞∑

k=0

ckxk+r+1 − 5

∞∑

k=0

ckxk+r

=

∞∑

k=0

[2(k + r)(k + r − 1)− (k + r)− 5]ckxk+r +

∞∑

k=1

ck−1xk+r

=

∞∑

k=0

(k + r + 1)(2(k + r)− 5

)ckx

k+r +

∞∑

k=1

ck−1xk+r

=(r + 1)(2r − 5)x r

+∞∑

k=1

[(k + r + 1)

(2(k + r)− 5

)ck + ck−1

]xk+r .



Solving the indicial equation (r + 1)(2r − 5) = 0, we get r1 :=52 and

r2 := −1. It follows that the recurrence relation is

(k + r + 1)(2(k + r)− 5

)ck + ck−1 = 0 for k = 1, 2, · · ·

from which we get

ck = (−1)k( k∏

ℓ=1

1

(ℓ + r + 1)(2(ℓ+ r)− 5

)

)

c0 for k = 1, 2, · · · .



• For the first root r1 =52 , we get a solution

x52

∞∑

k=0

(−1)k

k!

( k∏

ℓ=1

1

(2ℓ+ 7)

)

xk ,

where we let c0 := 1 for simplicity.

• For the second root r2 = −1, we get the other solution

1

x

∞∑

k=0

(−1)k

k!

( k∏

ℓ=1

1

(2ℓ− 7)

)

xk ,

where we let c0 := 1 for simplicity.



Thus, the general solution of the equation is

ϕ(x) :=c1x52

∞∑

k=0

(−1)k

k!

( k∏

ℓ=1

1

(2ℓ+ 7)

)

xk

+ c21

x

∞∑

k=0

(−1)k

k!

( k∏

ℓ=1

1

(2ℓ− 7)

)

xk ,

where c1 and c2 are arbitrary constants. We can show that this solutionconverges on the entire interval except 0.

In general, it is not an easy task to obtain the coefficients explicitly fromthe recurrence relation. However, in the above example, we dealt with atwo-term recurrence relation, which can be solved without big effort.



Theorem 25

Suppose that x0 is a regular singular point of

a0(x)y′′ + a1(x)y

′ + a2(x)y = 0, (A)

and let r1 and r2 be the roots of the corresponding indicial equation suchthat Re(r1) ≥ Re(r2). The first solution of (A) has the form

ϕ1(x) := |x − x0|r1∞∑

k=0

ck(x − x0)k , where c0 6= 0.



Case 1. If ν := r1 − r2 6∈ N0, then the second linearly independentsolution is of the form

ϕ2(x) := |x − x0|r2∞∑

k=0

dk(x − x0)k , where d0 6= 0.

Case 2. If ν ∈ N then the second linearly independent solution is of theform

ϕ2(x) := |x − x0|r2∞∑

k=0

dk(x − x0)k + cϕ1(x) ln |x − x0|,

where d0 6= 0 and the constant c may or may not be zero.



Case 3. If ν = 0, then the second linearly independent solution has is ofthe form

ϕ2(x) := |x−x0|r2∞∑

k=0

dk(x−x0)k+ϕ1(x) ln |x−x0|, where d0 6= 0.

The solutions ϕ1 and ϕ2 are valid in some deleted intervalI : 0 < |x − x0| < ρ.

Remark 9

In Case 2 of Theorem 25, the constant c can be explicitly determined bythe formula

c :=dν

dxν

((x − x0)

r1

ϕ1(x)

)2∣∣∣∣x→x0

.



Example 60

Use the method of Frobenius to find the solution of differential equation

x2y ′′ − xy ′ +

(

x2 +5

4

)

y = 0 for 0 < x < ρ,

where ρ > 0. Clearly, x0 = 0 is a regular singular point for the equation.Substituting

∞∑

k=0

ckxk+r , where c0 6= 0,

into the equation yields that

0 =

∞∑

k=0

(k + r)(k + r − 1)ckxk+r −

∞∑

k=0

(k + r)ckxk+r

−∞∑

k=0

ckxk+r+2 − 5

4

∞∑

k=0

ckxk+r .



Rearranging the terms gives us

0 =1

4

∞∑

k=0

(2(k + r) + 1

)(2(k + r)− 5

)ckx

k+r −∞∑

k=2

ck−2xk+r

=1

4(2r + 1)(2r − 5)c0x

r +1

4(2r + 3)(2r − 3)c1x

r+1

+

∞∑

k=0

(1

4

(2(k + r) + 1

)(2(k + r)− 5

)ck − ck−2

)

xk+r .

Thus, the indicial equation is (2r + 1)(2r − 5) = 0 from which we getr1 :=

52 and r2 := − 1

2 . Note that ν := r1 − r2 = 3, which is a positiveinteger.



Equating to zero the coefficient of the lowest power of x , we have

(2r + 3)(2r − 3)c1 = 0(2(k + r) + 1

)(2(k + r)− 5

)ck − 4ck−2 = 0, k = 2, 3, · · · .



• Let us proceed with the first root r1 =52 . We see that the coefficient

of c1 is not zero, and thus c1 = 0. Using this in the latter equation,we see that c3 = c5 = · · · = 0 too. For the even indices, we see that

ck =1

k(k + 3)ck−2 for k = 2, 4, · · ·

or explicitly

c2k =6(k + 1)

(2k + 3)!c0 for k = 1, 2, · · · .

Thus, with c0 := 1, the first solution is

ϕ1(x) := x52

∞∑

k=0

6(k + 1)

(2k + 3)!x2k .



• By Case 2 of Theorem 25, we seek for the second solution, which isof the form

x12

∞∑

k=0

dkxk + cϕ1(x) ln(x).

A simple examination of Remark 9 implies that c must be zero.

Indeed,(

x52

ϕ1(x)

)2

is an analytic even function. So, its odd-order

(third-order) derivative will yield an odd function, and any oddfunction passes through the origin proving that c = 0. Thus, thesecond linearly independent solution is of the form

ϕ2(x) :=1

x12

∞∑

k=0

dkxk .



Repeating the same procedure for this solution, we see that

d1 = 0

d3 = arbitrary

dk = 1(k−3)k dk−2, k = 2, 4, 5, 6, · · · .

Letting d3 := 0 for simplicity, we see that d1 = d3 = · · · = 0. On theother hand, we compute that

d2k = − (2k − 1)

(2k)!d0 for k = 1, 2, · · · .

Thus, with d0 := −1, the second solution is

ϕ2(x) :=1

x12

∞∑

k=0

(2k − 1)

(2k)!x2k .



Therefore, the general solution of the equation is

ϕ(x) := c1x52

∞∑

k=0

6(k + 1)

(2k + 3)!x2k + c2

1

x12

∞∑

k=0

(2k − 1)

(2k)!x2k ,




Example 61

Let us now solvex2y ′′ + (x − 3)xy ′ + 3y = 0

at the regular singular point 0. Substituting

∞∑

k=0

ckxk+r , where c0 6= 0,

into the equation gives us

0 =

∞∑

k=0

(k + r − 3)(k + r − 1)ckxk+r +

∞∑

k=1

(k + r − 1)ck−1xk+r

=(r − 3)(r − 1)c0xr +

∞∑

k=1

(k + r − 1)((k + r − 3)ck + ck−1

)xk+r .

Thus, the indicial roots are r1 := 3 and r2 := 1 of which difference is apositive integer.



Hence, we obtain the recurrence relation

ck = − 1

k + r − 3ck−1 for k = 1, 2, · · · .

• For the first root r1 = 3, we get

ck = − 1

kck−1 for k = 1, 2, · · ·

or explicitly

ck =(−1)k

k!c0 for k = 1, 2, · · · .

Thus, with c0 := 1, the first solution is

ϕ1(x) := x3∞∑

k=0

(−1)k

k!xk = x3e−x .



• By Case 2 of Theorem 25, we seek for the second solution, which isof the form

x

∞∑

k=0

dkxk + cx3e−x ln(x),

where d0 6= 0. Substituting this into the equation, we get

0 =

∞∑

k=1

k((k − 2)dk + dk−1

)xk+1 − cx3(x − 2)e−x

=

∞∑

k=1

k((k − 2)dk + dk−1

)xk+1 + c

∞∑

k=2

(−1)kk

(k − 2)!xk+1

=(d0 − d1)x2 +

∞∑

k=2

k

(

(k − 2)dk + dk−1 + c(−1)kk

(k − 2)!

)

xk+1.



Thus, we have

d1 = d0

c = −d1

d2 = arbitrary

dk = − 1k−2dk−1 − c (−1)k

(k−2)(k−2)! , k = 3, 4, · · · .

Applying induction to this two-term recurrence relation, we find

dk =(−1)k

(k − 2)!

(

d2 + d0

k∑

ℓ=3

1

ℓ − 2

)

for k = 2, 3, · · · .

We let d2 := 0, otherwise it contributes to the first solution since

x

∞∑

k=0

(−1)k

(k − 2)!xk = x3e−x .



Hence, we obtain

ϕ2(x) := x

∞∑

k=0

(−1)k

(k − 2)!

( k∑

ℓ=3

1

ℓ− 2

)

xk − x3e−x ln(x),

where we let d0 := 1 (c := −1) for simplicity.

Therefore, the general solution of the equation is

ϕ(x) := c1x3e−x + c2

[

x

∞∑

k=0

(−1)k

(k − 2)!

( k∑

ℓ=3

1

ℓ− 2

)

xk − x3e−x ln(x)

]

,




Remark 10

Once a solution of the equation is obtained, the second solution can befound by using reduction of order explained in Section 12.



Exercises 17

Exercises

1.

2.

3.

4.


The Laplace Transform


Definition 35 (The Laplace Transform)

Let f be a function of t, which is defined for t ≥ 0. The Laplacetransform of f , which we will denote by Lf (t) or F (s), is defined by

Lf (t) :=

∫ ∞

0

e−st f (t)dt,

whenever this integral converges.

In general, the parameter s is complex but for our discussion, we will onlyconsider real values for s.



Example 62

We have the following list of the Laplace transform of some functions.

1. L1 = 1sfor s > 0.

2. Lt = 1s2

for s > 0.

3. Leαt = 1s−α

for s > α.

4. Lcos(αt) = ss2+α2 for s > 0.

5. Lsin(αt) = αs2+α2 for s > 0.



Theorem 26

Suppose that the following conditions hold.

Property 1. f is a piecewise continuous on [0,T ) for any T > 0 (i.e., fis continuous on this interval except for a finite number ofpoints where it has jump discontinuities).

Property 2. There exist constants K , α, h > 0 such that |f (t)| ≤ Keαt

for all t > h.

Then, the Laplace transform F (s) = Lf (t) exists for s > α.

Remark 11

Functions satisfying Property 1 and Property 2 of Theorem 26 are called“piecewise continuous and of exponential order α”.



Proof of Theorem 26

We know that if f is piecewise continuous on the interval [a, b), then∫ b

af (t)dt exists. Thus, splitting the integral in two parts, we get

∫ ∞

0

e−st f (t)dt =

∫ h

0

e−st f (t)dt

︸︷︷︸

I1

+

∫ ∞

h

e−st f (t)dt

︸︷︷︸

I2

.

We know that I1 exists.



For I2, we have

|I2| ≤∫ ∞

h

|e−st f (t)|dt =∫ ∞

h

e−st |f (t)|dt

≤K

∫ ∞

h

e−st

eαtdt = K

∫ ∞

h

e(α−s)t

dt.

Note that

∫ ∞

h

e(α−s)t

dt = limR→∞

∫ R

h

e(α−s)t

dt

=1

α− slim

R→∞

(

e(α−s)R − e

(α−s)h)

,

which converges if (α− s) < 0. Using this in |I2| completes the proof.



In this chapter, we will deal with the functions which satisfy theconditions of this theorem.

Theorem 27 (Linearity)

The Laplace transform is a linear operator. More precisely, F (s) andG(s) denote the Laplace transform of the functions f and g, respectively.Assume that F (s) and G(s) exist for s > α and s > β, respectively.Then,

Laf (t) + bg(t) = aF (s) + bG(s) for s > maxα, β,

where a, b ∈ R.



Example 63

Consider that

L(

cos(αt))2

=L1 + cos(2αt)

2

=1

2L1+ 1

2Lcos(2αt)

=1

2

1

s+

1

2

s

s2 + (2α)2

for s > 0.



Theorem 28 (First Translation Theorem)

Let F (s) be the Laplace transform of f , which exists for s > α. Then,

Leλt f (t)

= F (s − λ) for s > α+ λ.



Proof of Theorem 28

By the definition, we have

Leλt f (t)

=

∫ ∞

0

e−st

eλt f (t)dt =

∫ ∞

0

e−(s−λ)t f (t)dt

=F (s − λ)

for s > α+ λ.



Example 64

Considering Example 62, we can compute the following transforms.

1. Le3tt = 1(s−3)2 for s > 3.

2. Le5t cos(2t) = s−5(s−5)2+4 for s > 5.



Theorem 29 (Derivative of the Transform)

Let F (s) be the Laplace transform of f , which exists for s > α. Then, fors > α, we have

Ltnf (t)

= (−1)n

dn

dsnF (s) for n = 0, 1, · · · .



Proof of Theorem 29

Differentiating both sides of F (s) = Lf (t) with respect to s, we have

F ′(s) =

∫ ∞

0

e−st(−t)f (t)dt = −Ltf (t)

F ′′(s) =

∫ ∞

0

e−st(−t)2f (t)dt = L

t2f (t)

...

F (n)(s) =

∫ ∞

0

e−st(−t)nf (t)dt = (−1)nL

tnf (t)

.

This completes the proof.



Remark 12

It should be noted here that F (s) converges uniformly for s ≥ β > α,and thus reversal of integral and the derivative applied in the proof ofTheorem 29 is permissible.



Example 65

By using Example 62, we have for s > 0 that

Lt2 sin(2t)

=(−1)2

d2

ds2Lsin(2t) =

d2

ds22

s2 + 4

=4(3s2 − 4

)

(s2 + 4

)3 .

Example 66

We see for s > 0 that

Ltn =Ltn · 1 = (−1)ndn

dsnL1

=(−1)ndn

dsn1

s=

n!

sn+1.



Theorem 30 (Transform of the Derivative)

Let f be continuous on [0,∞) and of exponential order α, and that f ′ ispiecewise continuous on [0,∞). Then,

Lf ′(t) = sLf (t) − f (0) for s > α.



Proof of Theorem 30

Integrating by parts, we get for R > 0 that

∫ R

0

e−st f ′(t)dt =

∫ R

0

e−st f ′(t)dt = e

−st f (t)∣∣∣

R

0+ s

∫ R

0

e−st f (t)dt

=e−sR f (R)− f (0) + s

∫ R

0

e−st f (t)dt.

Since f is of exponential order α, we see for some K > 0 that

∣∣e

−sR f (R)∣∣ ≤ Ke

−(s−α)R for all large R and s > α

from which we get f (R)e−sR → 0 as R → ∞ if s > α. Thus, by lettingR → ∞, we find that

Lf ′(t) = sLf (t) − f (0).



Using induction, we can extend the next theorem to higher-orderderivatives of f .

Corollary 3

Let f , f ′, · · · , f (n−1) be continuous on [0,∞) and of exponential order α,and that f (n) is piecewise continuous on [0,∞). Then,

Lf (n)(t) = snLf (t) −n−1∑

k=0

s(n−1)−k f (k)(0) for s > α.



Theorem 31 (Second Translation Theorem)

Let F (s) = Lf (t) exist for s ≥ α. For λ ≥ 0, define g by

g(t) :=

f (t − λ), t ≥ λ

0, λ > t ≥ 0.

Then,Lg(t) = e

−λsF (s) for s > α.



Example 67

Let us compute Lg(t) for

g(t) :=

sin(t), t ≥ π2

0, π2 > t ≥ 0.

Since sin(t) = cos(t − π

2

), we have for s > 0 that

Lg(t) = e−π

2 sLcos(t) =se−

π

2 s

s2 + 1.



Exercises 18

Exercises

1. Compute the Laplace transform of the following functions.

a. f (t) := 3e−5t + e3t + 2t2 − 1.b. f (t) := 2 sinh(5t) + 3 cos(2t).

c. f (t) := t32 .

2. Compute the Laplace transform of the following functions.

a. f (t) :=

1, t ≥ 1

0, 0 ≤ t < 1.b. f (t) :=

0, t ≥ π

sin(t), 0 ≤ t ≤ π.

c. f (t) :=

et , t ≥ 1

e, 0 ≤ t < 1.d. f (t) :=

1, 1 ≤ t ≤ 2

0, otherwise.



3. Compute the Laplace transform of f (t) := 1t

(1− e−t

)for t > 0.

4. Show that the Laplace transform of f (t) := eet

does not exist.

5. Let f be a piecewise continuous and of exponential order α, anddefine g(t) :=

∫ t

0 f (η)dη for t ≥ 0.

a. Prove that g is also of exponential order α.b. Compute the Laplace transform of the function g .

6. Let f be a piecewise continuous and of exponential order α. Provethat F (s) → 0 as s → ∞.


The Laplace Transform Convolution

Convolution

Definition 36 (Convolution)

Let f and g be two functions, which are piecewise continuous on everyfinite interval, and of the same exponential order. The function denotedby (f ∗ g) and defined as

(f ∗ g)(t) :=∫ t

0

f (η)g(t − η)dη for t ≥ 0

is called the “convolution of the functions f and g”.



Let us change the variable of the integration by means of transformationζ := t − η, then we have

(f ∗ g)(t) =∫ t

0

f (η)g(t − η)dη =

∫ 0

t

f (t − ζ)g(ζ)(−dζ)

=

∫ t

0

g(ζ)f (t − ζ)dζ = (g ∗ f )(t),

which proves that convolution is commutative.



The following properties of convolution can be easily justified.

Property 1. f ∗ g = g ∗ f (commutative)

Property 2. f ∗ (g + h) = f ∗ g + f ∗ h (distributive)

Property 3. f ∗ (g ∗ h) = (f ∗ g) ∗ h (associative)

Property 4. f ∗ 0 = 0 ∗ f = 0.



Remark 13

However, it should be noted that f ∗ 1 6= f .

Example 68

For instance,

(sin ∗1)(t) =∫ t

0

sin(η)dη = 1− cos(t) 6= sin(t).



Theorem 32 (Convolution Theorem)

Let f and g be two functions, which are piecewise continuous on everyfinite interval, and of exponential order α. Then,

L(f ∗ g)(t) = Lf (t) · Lg(t) for s > α.



Example 69

Show that the solution of the initial value problem

y ′′(t) + 4y(t) = f (t)

y(0) = 2 and y ′(0) = −6

is

ϕ(t) := 2 cos(2t)− 3 sin(2t) +1

2

∫ t

0

sin(2(t − η)

)f (η)dη.



Exercises 18.1

Exercises

1. Prove distributive and associative properties of convolution.

2. Compute the convolution of the following functions.

a. f (t) := cos(αt) and g(t) := sin(βt), where α, β ∈ R.b. f (t) := tn and g(t) := eαt , where n ∈ N and α ∈ R.c. f (t) := cosh(t) and g(t) := sinh(t).

3. Show that if f and g are piecewise continuous and of someexponential, then f ∗ g is also of some exponential order.


The Laplace Transform The Inverse Laplace Transform

The Inverse Laplace Transform

Given a function f , which is defined on [0,∞), then we define its Laplacetransform by F (s) := Lf (t). Now, we consider the inverse problem.Given a function F (s), can we find a function f such that

Lf (t) = F (s)

holds? We introduce the notation L−1F (s) to denote the function fsuch that

Lf (t) = F (s),

and we call the operator L−1 as the “inverse Laplace transform”. Thisinverse transformation is unique provided that f is continuous with theLaplace transform F (s), i.e., there does not exist any other continuousfunction having the same Laplace transform.



For F (s) := Lf (t) and G(s) := Lg(t), the following properties ofthe inverse Laplace transform can be justified easily.

Property 1. L−1aF (s) + bG(s) = af (t) + bg(t) (linearity)

Property 2. L−1F (s − λ) = eλt f (t)

Property 3. L−1

dn

dsnF (s)

= (−1)ntnf (t)

Property 4. L−1F (s)G(s) = (f ∗ g)(t)



Example 70

Show that

L−1

6

(s − 2)4

= t3e2t .

Example 71

Show that

L−1

1

s2 + 6s + 13

=1

2e−3t sin(2t).



Example 72

Show by using partial fractions that

L−1

1

s(s2 + 1

)

= 1− cos(t).

Example 73

By using convolution, we have

L−1

1

s(s2 + 1

)

= L−1

1

s2 + 1· 1s

= (sin ∗1)(t) = 1− cos(t).



Example 74

By using convolution, we have

L−1

s

(s − 3)(s2 + 1

)

=L−1

s

s2 + 1· 1

s − 3

=(cos ∗e3·

)(t)

=

∫ t

0

cos(η)e3(t−η)dη

=1

10

(3e3t − 3 cos(t) + sin(t)

).

Example 75

Solve the integral equation

y(t) +

∫ t

0

et−ηy(η)dη = sin(t)

by using the Laplace transform.



Exercises 18.2

Exercises

1. Compute the inverses of the following Laplace transforms.

a. F (s) := 1(s−1)(s+3) .

b. F (s) := 1s3(s2+9) .

c. F (s) := s+1s2−9 +

4s−3 .

d. F (s) := s(s2+1)3 .

2. Does the inverse Laplace transform of F (s) := 1 exist as a piecewisecontinuous function of some exponential?


The Laplace Transform Solving Autonomous Differential Equations with Laplace Transform

Solving Autonomous Differential Equations with Laplace

Transform

Consider the n-th order linear differential equation with constantcoefficients

a0dny

dtn+ a1

dn−1y

dtn−1+ · · ·+ any = f (t)

with the initial conditions

y(0) = y0, y′(0) = y1, · · · , y (n−1)(0) = yn−1.

Note that this is an IVP and has a unique solution if f is continuous.



We now take the Laplace transform of both sides of the IVP and get

a0Ly (n)

+ a1L

y (n−1)

+ · · ·+ anLy = Lf (t)

or equivalently

a0[snLy(t) − sn−1y(0)− sn−2y ′(0)− · · · − y (n−1)(0)

]

+ a1[sn−1Ly(t) − sn−2y(0)− sn−3y ′(0)− · · · − y (n−2)(0)

]

+ · · ·+ anLy(t) = Lf (t),



Denote by Y and F the respective Laplace transforms of y and f , we get

[a0s

n + a1sn−1 + · · ·+ an

]Y (s)

−[a0s

n−1 + a1sn−2 + · · ·+ an−1

]y0

−[a0s

n−2 + a1sn−3 + · · ·+ an−2

]y1

− · · · − anyn−1 = F (s).

Once F (s) is known, we can find Y (s) by solving this algebraic equationand then find y by taking the inverse Laplace transform, i.e,y(t) := L−1Y (s).



Example 76

Show by using the Laplace transform that the first-order initial valueproblem

y ′ − 2y = 3e5t

y(0) = 3

has the solutionϕ(t) := 2e2t + e

5t .



Example 77

Show by using the Laplace transform that the second-order initial valueproblem

y ′′ + y = 8e−2t sin(t)

y(0) =0

y ′(0) =0

has the solution

ϕ(t) :=(e−2t − 1

)cos(t) +

(e−2t + 1

)sin(t).



Exercises 18.3

Exercises

1. Determine the solutions of the following initial-value problems.

a.

y ′′ + 6y ′ + 5y = 0

y(0) = 4 and y ′(0) = 0.b.

y ′′ − y ′ − 6y = 30

y(0) = 0 and y ′(0) = 15.

2. Determine the solutions of the following boundary-value problems.

a.

y ′′ + 4y ′ + 4y = 0

y(0) = e2 and y(1) = −e

2.b.

y ′′ − 2y ′ + 5y = 0

y(0) = 1 and y(1) = 0.

c.

y ′′ + y = sin(t)

y(0) = 1 and y(π2

)= 1.

d.

y ′′ + y = sin(t)

y(0) = π2 and y(π) = 0.

Hint. First let y ′(0) = y1 and solve the problem as an IVP. At theend, determine the value of y1 for which the terminal condition holds.


The Laplace Transform Applications to Differential Equations

Example 78


y ′′ + 2ty ′ − 4y = 1

y(0) = 0, y ′(0) = 0.

Hint. Use the conclusion of Exercise 6 in Section 18.



Example 79


ty ′′ − 2y ′ + ty = 0

y(0) = 1, y ′(0) = 0.

Hint. L−1

1(s2+1)2

= 12

(sin(t)− t cos(t)

).



Example 80


ty ′′ − y ′ = 3t2

y(0) = 0.

What can be said about a solution satisfying y ′(0) = 1?



Example 81


ty ′′ + (t + 2)y ′ + y = 2

y ′(0) = 1.

Hint. Exercise 3 of Section 18.What can be said about a solution satisfying y(0) = 1?



Exercises 18.4

Exercises

1.

2.

3.

4.


The Laplace Transform The Impulse Function (To Be Skipped)

The Impulse Function

Consider the function

fε(t) :=

1ε, ε > t > 0

0, t ≥ ε and t = 0,

whose graphic is given below.

¶

t

1¶

y



We compute that∫ ∞

0

fε(t)dt =

∫ ε

0

1

εdt ≡ 1

and similarly,

Lfε(t) =1

ε

∫ ε

0

e−st

dt =1− e−εs

εs.

Letting ε→ 0+, the magnitude of fε over (0, ε) increases without limitand its Laplace transform tends to unity. Indeed, by using L’Hopital’srule, we get

limε→0

1− e−εs

εs= lim

ε→0

se−εs

s= 1.



That is, the transform of fε approaches to the unity as ε→ 0+. If t andfε(t) represents time and force at that time, respectively, then

∫ ε

0 fε(t)dtrepresents the impulse of the force acting over the time interval (0, ε).We have unit impulse when an infinite force acting over a zero timeinterval. The limiting form of fε is frequently called the “unit impulse

function”. We will denote it by δ = δ(t), and this is also called as“Dirac Delta function”. Therefore,

limε→0+

fε(t) = δ(t) and Lδ(t) = 1.



Exercises 18.5

Exercises

1.

2.

3.

4.


Systems of Linear Differential Equations (To Be Skipped)

Systems of Linear Differential Equations

Let I be an open interval in R. We define an m × n matrix valuedfunction AAA = AAA(x) on I by

AAA(x) :=

a11(x) a12(x) · · · a1n(x)a21(x) a22(x) · · · a2n(x)

......

. . ....

am1(x) am2(x) · · · amn(x)

m×n

,

where aij is real valued functions on I for each (i , j). We say that AAA isk-times (k ≥ 0) “continuously differentiable” on I , if each aij isk-times continuously differentiable on I . That is, all components havek-th order derivatives and are continuous.



We denote by the k-th order derivative of AAA by

AAA(k)(x) :=

a(k)11 (x) a

(k)12 (x) · · · a

(k)1n (x)

a(k)21 (x) a

(k)22 (x) · · · a

(k)2n (x)

......

. . ....

a(k)m1(x) a

(k)m2(x) · · · a

(k)mn(x)

m×n

for k ∈ N,

and AAA(0)(x) := AAA(x).



Let BBB be a matrix of the form

BBB(x) :=

b11(x) b12(x) · · · b1n(x)b21(x) b22(x) · · · b2n(x)

......

. . ....

bm1(x) bm2(x) · · · bmn(x)

m×n

,

then we say AAA = BBB on I if aij(x) = bij(x) for all x ∈ I , i = 1, 2, · · · ,mand j = 1, 2, · · · , n.



The Sum: CCC := AAA+BBB is the m × n matrix-valued function on Iwhose entry located at (i , j) is defined by cij(x) := aij(x) + bij(x) forx ∈ I .

The Scalar Multiple: DDD := f AAA, where f : I → R, is the m × nmatrix-valued function on I whose entry located at (i , j) is definedby dij(x) := f (x)aij(x) for x ∈ I .

The Null Matrix: The matrix whose all entries are 0 is called the“null matrix” and it satisfies AAA+000 = 000 +AAA = AAA for all AAA.



Derivative of Matrices: For k-times differentiable matrices AAA and BBBof the same size, we have for ℓ = 0, 1, · · · , k and a scalar c that

(AAA+BBB)(ℓ)(x) =AAA(ℓ)(x) +BBB(ℓ)(x)

(cAAA)(ℓ)(x) =cAAA(ℓ)(x).

The Matrix Product: Let AAA and BBB be matrices of the sizes m × nand n× r , respectively. Then, CCC := AAABBB has the size m × r and isdefined to have

cij(x) :=

n∑

k=1

aik(x)bkj(x) for x ∈ I

at its cij entry located at (i , j). It should be noted that the matrixproduct is not commutative, i.e., AAABBB 6= BBBAAA in general. Also,[AAA+BBB]CCC = AAACCC +BBBCCC if AAA and BBB can be multiplied by CCC .



Definition 37 (Systems of Differential Equations)

Let AAA0,AAA1, · · · ,AAAm be r × n matrix-valued functions on I , and let fff becolumn of size r vector-valued function on I . Then, an equation in theform

AAA0(x)yyy(m) +AAA1(x)yyy

(m−1) + · · ·+AAAm(x)yyy = fff (x),

where yyy is a column vector of the form

yyy(x) :=(y1(x), y2(x), · · · , yn(x)

)T, is called a “system of r linear

differential equations of order m in n unknowns y1, y2, · · · , yn on I”provided that AAA0 is never equal to 000r×n on I .If fff = (0, 0, · · · , 0

︸︷︷︸

r-times

)T on I , then the system is called “homogeneous”

otherwise it is called “non-homogeneous”. If AAA0,AAA1, · · · ,AAAm arematrices of real numbers, then the system is called a “autonomous” or a“system of r linear differential equations of order m in n unknowns

y1, y2, · · · , yn with constant coefficients”.



Definition 38 (Solution)

Let AAA0,AAA1, · · · ,AAAm be r × n matrix-valued continuous functions on I ,and let fff be column of size r vector-valued continuous function on I . Bya “solution” of the system


(m−1) + · · ·+AAAm(x)yyy = fff (x), (A)

we mean a vector-valued function yyy = ϕϕϕ, where

ϕϕϕ(x) =(ϕ1(x), ϕ2(x), · · · , ϕn(x)

)T, which is m-times differentiable

and satisfies (A) identically on I .



Example 82

Show that ϕϕϕ(x) :=

(e2x

e2x

)

is a solution of the system

y ′1 − 2y2 =0

y ′2 − y1 − y2 =0.

Note that this system can be put in the matrix form

(1 00 1

)

ϕϕϕ′ −(

0 21 1

)

ϕϕϕ =

(00

)

.



Example 83

Find the general solution of the system

y ′1 − y2 =0

y2 =f (x),

where f is a continuous function. Substituting the latter equation intothe former one, we get

y ′1 = y2 = f (x) =⇒ y1 =

∫ x

f (η)dη + c ,

where c is an arbitrary constant. Thus, the general solution is

ϕϕϕ(x) :=

( ∫ xf (η)dη + cf (x)

)

.



Theorem 33

Let a0, a1, · · · , am and b be continuous functions on I , and consider them-th order linear differential equation

a0(x)y(m) + a1(x)y

(m−1) + · · ·+ am(x)y = b(x). (A)

Then, (A) can be transformed into a system of m linear differentialequations of first order in m unknowns on I . Moreover, there is aone-to-one correspondence between the solutions of (A) and thetransformed system.



Example 84

Transform the equation

y ′′′ + 2xy ′′ − 4y ′ + y = ex

into a system of linear differential equations.



Definition 39

Let fff 1, fff 2, · · · , fff m be of the same size vector-valued continuousfunctions, and c1, c2, . . . , cm be real numbers. Then, the expression

c1fff 1 + c2fff 2 + · · ·+ cmfff m

is called a “linear combination of fff 1, fff 2, · · · , fff m”.



Definition 40

Let fff 1, fff 2, · · · , fff m be of the same size vector-valued continuous functionson I . Then, fff 1, fff 2, · · · , fff m are called “linearly dependent on I” if thereexist constants c1, c2, · · · , cm, not all zero, such that

c1fff 1(x) + c2fff 2(x) + · · ·+ cmfff m(x) ≡ 000 for x ∈ I .

Otherwise, they are called “linearly independent on I”.

Example 85

Check whether or not fff 1(x) :=

(ex

ex

)

and fff 2(x) :=

(−e−x

e−x

)

are

linearly dependent.



Theorem 34

Let AAA0,AAA1, · · · ,AAAm be r × n matrix-valued continuous functions on I .Then, any linear combination of the solutions of the homogeneousdifferential system


(m−1) + · · ·+AAAm(x)yyy = 000

is also a solution of the system on I .



Theorem 35

Suppose that AAA0,AAA1, · · · ,AAAm are r × n matrix-valued continuousfunctions on I , and fff is a column of size r vector-valued continuousfunction on I . Let ψψψ be a solution of the non-homogeneous differentialsystem


(m−1) + · · ·+AAAm(x)yyy = fff (x) (A)

and ϕϕϕ be a solution of the associated homogeneous differential system


(m−1) + · · ·+AAAm(x)yyy = 000. (B)

Then, (ϕϕϕ+ψψψ) is a solution of (B).



Definition 41

1. The general solution of (B) (which contains a total of n arbitraryconstants) is called “complementary solution of (A)”. We willdenote this solution by ϕc .

2. Any solution of (A) without arbitrary constants is called a “particularsolution of (A)”, which we will denote by ϕp .

3. The sum (ϕc + ϕp) is called the “general solution of (A)”.



Example 86

Find the general solution of yyy ′ − yyy =(ex , ex , · · · , ex︸︷︷︸

r -times

)T. That is,

y ′i − yi = ex for i = 1, 2, · · · , r , or in the operator form (D− 1)yi = ex

for i = 1, 2, · · · , r .



Theorem 36

Let AAA be a n × n matrix-valued continuous function on I and fff be acolumn of size n vector-valued continuous function on I . For any yyy 0

column vector of size n of real numbers and x0 ∈ I , the initial-valueproblem

yyy ′ +AAA(x)yyy = fff (x)

yyy(x0) = yyy0

has a unique solution defined on I .



Theorem 37

Let AAA1,AAA2, · · · ,AAAm be a n × n matrix-valued continuous functions on Iand fff be a column of size n vector-valued continuous function on I . Forany yyy0,yyy1, · · · ,yyym−1 column vectors of size n of real numbers andx0 ∈ I , the IVP

yyy (m) +AAA1(x)yyy(m−1) + · · ·+AAAm(x)yyy = fff (x)

yyy(x0) =yyy0

yyy ′(x0) =yyy1

...

yyy (m−1)(x0) =yyym−1

has a unique solution defined on I .



Theorem 38

Let AAA be a n × n matrix-valued continuous functions on I . Then, thehomogeneous system of n linear differential equations of first order in nunknowns

yyy ′ +AAA(x)yyy = fff (x) (A)

always possesses n solutions, which are linearly independent on I .Further, if ϕϕϕ1,ϕϕϕ2, · · · ,ϕϕϕn are n linearly independent solutions of (A) onI , then every solution of (A) can be expressed as a linear combinationc1fff 1 + c2fff 2 + · · ·+ cnfff n of these solutions by a proper choice of theconstants c1, c2, · · · , cn.



Example 87

Find the general solution of the system

y ′1 − y2 =0

y ′2 − y1 =0.



Definition 42

Let fff 1, fff 2, · · · , fff n be column of size n vector-valued continuous functionson I . Then, the “Wronskian of the functions fff 1, fff 2, · · · , fff n” is definedby

W [fff 1, fff 2, · · · , fff n] := det(fff 1, fff 2, · · · , fff n

),

and W [fff 1, fff 2, · · · , fff n](x) denotes its value at x ∈ I .



Theorem 39

Let AAA be a n × n matrix-valued continuous function on I , and letϕϕϕ1,ϕϕϕ2, · · · ,ϕϕϕn be solutions of the homogeneous differential system

yyy ′ +AAA(x)yyy = 000.

Then, the solutions ϕϕϕ1,ϕϕϕ2, · · · ,ϕϕϕn are linearly independent if and only ifW [ϕϕϕ1,ϕϕϕ2, · · · ,ϕϕϕn](x) 6= 0 for all x ∈ I .



Exercises 19

Exercises

1.

2.

3.

4.


Solving Linear Systems

Solving Linear Systems

In this section, we will discuss how to find solutions of linear systems withconstant coefficients.


Solving Linear Systems An Operator Method for Autonomous Systems (To Be Skipped)

An Operator Method for Autonomous Systems

The system of n linear equations of order with constant coefficients inunknown functions ϕ1, ϕ2, · · · , ϕn given by

AAA0yyy(m) +AAA1yyy

(m−1) + · · ·+AAAmyyy = fff (x)

can be written in the form

L11(D) L12(D) · · · L1n(D)L21(D) L22(D) · · · L2n(D)

......

. . ....

Ln1(D) Ln2(D) · · · Lnn(D)

︸︷︷︸

LLL(D)

ϕ1

ϕ2

...ϕn

︸︷︷︸

yyy

=

f1(x)f2(x)...

fn(x)

︸︷︷︸

fff (x)

,

where for i , j = 1, 2, . . . , n, Lij is a linear operator of order at most mwith constant coefficients.



If det[LLL(D)] 6= 0, then by Cramer’s rule

det[LLL(D)]ϕi =

n∑

j=1

(−1)i+j det[LLLij(D)]fj (x) for i = 1, 2, · · · , n,

where LLLij is the matrix obtained by deleting the i-th row and j-th columnof LLL.By using the operator method, we can solve the system of differentialequations for each ϕi , i = 1, 2, · · · , n.

Remark 14

The number of arbitrary constants in the general solution of the system isequal to the degree of the characteristic polynomial of LLL, i.e., the order ofthe differential operator det[LLL(D)].



Example 88

Solve the differential system

y ′1 + 2y2 =e

3x

y ′2 + y1 + y2 =e

−x .

It follows from

det[LLL(λ)] =

∣∣∣∣

λ 21 λ+ 1

∣∣∣∣= λ(λ+ 1)− 2

=λ2 + λ− 2

that the number of arbitrary constants will be 2. The solution is

ϕ1(x) :=− 2c1ex + c2e

−2x +2

5e3x + e

−x ,

ϕ2(x) :=c1ex + c2e

−2x − 1

10e3x +

1

2e−x .



Example 89

Show that the general solution of the system

y ′′1 + y ′

2 − y1 + y2 =− 1

y ′1 + y ′

2 − y1 =x2

isϕ1(x) :=(c1 + c2x)e

x + c3e−x − x2 − 4x − 6,

ϕ2(x) :=− c2ex − 2c3xe

x − x2 − 2x − 3.



Example 90

Find the general solution of the differential system

y ′′′1 − y ′′

2 + y ′1 − y2 =0

y ′2 − y1 =0.



Exercises 20.1

Exercises

1.

2.

3.

4.


Solving Linear Systems Solving Autonomous Differential Systems with Laplace Transform

Solving Autonomous Differential Systems with Laplace

Transform

The system of n linear equations of order with constant coefficients inunknown functions ϕϕϕ1,ϕϕϕ2, · · · ,ϕϕϕn given by

AAA0yyy(m) +AAA1yyy

(m−1) + · · ·+AAAmyyy = fff (x),

where AAA0,AAA1, · · · ,AAAm are n × n constant matrices of real numbers.



Solution Procedure

Step 1. First, setup an arbitrary initial condition

yyy(0) = yyy0, yyy′(0) = yyy1, · · · , yyy (m−1)(0) = yyym−1.

Step 2. Take the Laplace transform of the each equation in the systemto obtain a linear algebraic system of equations in the nunknowns Y1(s) := Ly1(x), Y2(s) := Ly2(x), · · · ,Yn(s) := Lyn(x).

Step 3. Solve the linear system of algebraic equations in order todetermine Y1(s),Y2(s), · · · ,Yn(s).

Step 4. Finally, use the inverse Laplace transform to find the solutionϕ1(x) := L−1Y1, ϕ2(x) := L−1Y2, · · · ,ϕn(x) := L−1Yn, and form the vector solution

ϕϕϕ(x) :=(ϕ1(x), ϕ2(x), · · · , ϕn(x)

)T.



Example 91

Show by using the Laplace transform that the solution of the differentialsystem

y ′1 =6y1 − 3y2 + 7ex

y ′2 =2y1 + y2 + 4ex

withy1(0) =− 1

y2(0) =0

isϕ1 :=− 2ex + e

3x

ϕ2 :=− ex + e

3x .



Example 92


y ′1 =3y1 − 3y2 + 9

y ′2 =− 6y1 − 36x

withy1(0) =3

y2(0) =10

isϕ1(x) :=2e−3x − 6x + 1

ϕ2(x) :=4e−3x − 6x + 6.



Example 93


y ′1 =− y2 + e

−x

y ′2 =y1 + 3e−x

withy1(0) =0

y2(0) =1

isϕ1(x) :=2 cos(x)− 2 sin(x)− 2e−x

ϕ2(x) :=2 cos(x) + 2 sin(x)− e−x .



Exercises 20.1

Exercises

1. Find the general solution of the system

y ′1 = −3y1 + 2y2

y ′2 = 3y1 − 4y2.

2. Find the solution of the initial value system

y ′1 = −5y1 + 3y2 + 3et

y ′2 = −3y1 + y2

withy1(0) =1

y2(0) =− 1.

3. Find the solution of the system

y ′1 = 3y1 − 4y2

y ′2 = y1 − y2

withy1(0) =1

y2(1) =e.




y ′1 = y1 + 2y2 + 1

y ′2 = −y1 − y2 + 1

y ′3 = y1 − y3

with

y1(0) =1

y2(0) =0

y3(0) =1.


y ′1 = 3y1 − y2 − y3

y ′2 = y1 + y2 − y3

y ′3 = y1 − y2 + y3

with

y1(0) =− 1

y2(0) =1

y3(0) =0.


y ′1 = 3y1 + 4y2 − 2y3 − et

y ′2 = 2y1 + y2 − 4y3 + 1

y ′3 = y1 + 2y2

with

y1(0) =0

y2(0) =0

y3(0) =0.


Fourier Series

Fourier Series

Definition 43 (Periodicity)

A function f is said to be “periodic with a period of T”, iff (x) = f (x +T ) for all x ∈ R. If T is the smallest of such values, then itis called as the “fundamental period”.

Example 94

For instance, f (x) := sin(nx) is periodic with the fundamental period 2πn.


Fourier Series

Definition 44 (Fourier Series)

A series of the form

a02

+

∞∑

k=1

ak cos

(kπ

Lx

)

+ bk sin

(kπ

Lx

)

(A)

is called as a “Fourier series with period 2L”.

Remark 15

If the series (A) converges, then it defines a function, which is2L-periodic.


Fourier Series

Let f be a periodic function with a period of 2L. We assume that anexpansion as in (A) is possible for f , i.e.,f (x) = a0

2 +∑∞

k=1

ak cos

(kπLx)+ bk sin

(kπLx)

. Now, we will try to findhow the coefficients an and bn are defined.Integrating (A) over the interval [−L, L] term by term (if permissible), weget

∫ L

−L

f (η)dη =a02

∫ L

−L

1dη

+∞∑

k=1

ak

∫ L

−L

cos

(kπ

Lη

)

dη

+ bk

∫ L

−L

sin

(kπ

Lη

)

dη

.


Fourier Series

Since we have for n = 1, 2, · · · that

∫ L

−L

cos

(nπ

Lη

)

dη =L

nπsin

(nπ

Lη

)∣∣∣∣

η=L

η=−L

=L

nπ[sin(nπ) − sin(−nπ)]

=0

and similarly∫ L

−L

sin

(nπ

Lη

)

dη = 0.

Thus, we find that

∫ L

−L

f (η)dη =a02

∫ L

−L

1dη = a0L

or simply

a0 =1

L

∫ L

−L

f (η)dη.


Fourier Series

For determining an and bn, we need the following lemma.

Lemma 1

Let m and n be positive integers. Then, the following properties hold.

(i)

∫ L

−L

cos

(mπ

Lη

)

cos

(nπ

Lη

)

dη =

L, m = n

0, m 6= n.

(ii)

∫ L

−L

sin

(mπ

Lη

)

sin

(nπ

Lη

)

dη =

L, m = n

0, m 6= n.

(iii)

∫ L

−L

cos

(mπ

Lη

)

sin

(nπ

Lη

)

dη = 0.


Fourier Series

Proof of Lemma 1

(i) • First, let m = n. Then, we compute that

∫ L

−L

[

cos

(mπ

Lη

)]2

dη =1

2

∫ L

−L

[

1 + cos

(2mπ

Lη

)]

dη

=1

2

[

η +L

2mπsin

(2mπ

Lη

)]η=L

η=−L

=L.


Fourier Series

• Next, let m 6= n, then

∫ L

−L

cos

(mπ

Lη

)

cos

(nπ

Lη

)

dη

=1

2

∫ L

−L

[

cos

((m + n)π

Lη

)

+ cos

((m − n)π

Lη

)]

dη

=1

2

[L

(m + n)πsin

((m + n)π

Lη

)

+L

(m − n)πsin

((m − n)π

Lη

)]η=L

η=−L

=0.

Proofs of (ii) and (iii) are left as exercises.


Fourier Series

Now, let us return to find an and bn in the Fourier series. To find an, wemultiply (A) by cos

(nπLx)and integrate over [−L, L], and get

∫ L

−L

f (η) cos

(nπ

Lη

)

dη =a02

∫ L

−L

cos

(nπ

Lη

)

dη

︸︷︷︸

0

+

∞∑

k=1

ak

∫ L

−L

cos

(kπ

Lη

)

cos

(nπ

Lη

)

dη

+ bk

∫ L

−L

sin

(kπ

Lη

)

cos

(nπ

Lη

)

dη

︸︷︷︸

0

.


Fourier Series

It follows from Lemma 1 (i) that

∫ L

−L

f (η) cos

(nπ

Lη

)

dη = anL,

which gives us

an =1

L

∫ L

−L

f (η) cos

(nπ

Lη

)

dη for n = 1, 2, · · · .

Similarly, multiplying (A) by sin(nπLx)and integrate over [−L, L], we get

bn =1

L

∫ L

−L

f (η) sin

(nπ

Lη

)

dη for n = 1, 2, · · · .


Fourier Series

Theorem 40

Let f be a 2L-periodic function. Suppose that f and f ′ are piecewisecontinuous. Then, the fourier series of f ,

a02

+

∞∑

k=1

ak cos

(kπ

Lx

)

+ bk sin

(kπ

Lx

)

,

where

an :=1

L

∫ L

−L

f (η) cos

(nπ

Lη

)

dη for n = 0, 1, · · ·

and

bn :=1

L

∫ L

−L

f (η) sin

(nπ

Lη

)

dη for n = 1, 2, · · · ,

converges to f (x) if x is a point of continuity. Furthermore, if f has ajump discontinuity at x, then it converges to

f (x−) + f (x+)

2.


Fourier Series

Example 95

Let

f (x) :=

1, 2n ≤ x < 2n+ 1

0, 2n− 1 ≤ x < 2nfor n ∈ Z


−3 −2 −1 1 2 3

1

x

y

Figure: The graphic of f over the interval [−3, 3].

Clearly, f is periodic with a period of 2, i.e., L = 1.


Fourier Series

We compute that

a0 :=1

1

∫ 1

−1

f (η)dη = 1,

an :=1

1

∫ 1

−1

f (η) cos(nπη)dη = 0 for n = 1, 2, · · · ,

bn :=1

1

∫ 1

−1

f (η) sin(nπη)dη =1− (−1)n

nπfor n = 1, 2, · · · .

Hence, the Fourier series is

1

2+

1

π

∞∑

k=1

1− (−1)k

ksin(kπx)

or equivalently

1

2+

2

π

∞∑

k=1

1

2k − 1sin

((2k − 1)πx

).


Fourier Series

Clearly, f is discontinuous at integers and f (n+)−f (n−)2 = 1

2 for n ∈ Z. ByTheorem 40, we have

1

2+

2

π

∞∑

k=1

1

2k − 1sin

((2k−1)πx

)=

1, 2n < x < 2n+ 112 , x = n

0, 2n − 1 < x < 2n

for n ∈ Z.

Below, the graphic of this function is given.

−3 −2 −1 1 2 3

1

x

y

Figure: The graphic of the Fourier series of f over the interval [−3, 3].


Fourier Series

Example 96

Find the Fourier series of the 4-periodic function defined by

f (x) :=

0, −2 < x < −1

1, −1 ≤ x ≤ 0

x , 0 < x ≤ 1

0, 1 < x < 2


−6 −5 −4 −3 −2 −1 1 2 3 4 5 6

1

x

y



Fourier Series

We compute for n = 1, 2, · · · that

a0 :=1

2

∫ 2

−2

f (η)dη =3

4,

an :=1

2

∫ 2

−2

f (η) cos

(nπ

2η

)

dη

=2

nπsin

(nπ

2

)

+2

(nπ)2

[

cos

(nπ

2

)

− 1

]

,

bn :=1

2

∫ 2

−2

f (η) sin

(nπ

2η

)

dη

=− 1

nπ+

2

(nπ)2sin

(nπ

2

)

.


Fourier Series

Therefore, the Fourier series is for f is

3

8+

2

π

∞∑

k=1

1

k

[

sin

(kπ

2

)

+1

kπ

[

cos

(kπ

2

)

− 1

]]

cos

(kπ

2x

)

− 1

π

∞∑

k=1

1

k

[

1− 2

kπsin

(kπ

2

)]

sin

(kπ

2x

)

.

Moreover, f is discontinuous at x = −1, 0, 1, and that

f (−1−) + f (−1+)

2=0 + 1

2=

1

2,

f (0−) + f (0+)

2=1 + 0

2=

1

2,

f (1−) + f (1+)

2=1 + 0

2=

1

2.


Fourier Series

Hence, the series converges to the function

0, −2 ≤ x < −112 , x = −1

1, −1 < x < 012 , x = 0

x , 0 < x < 112 , x = 1

0, 1 < x ≤ 2


−6 −5 −4 −3 −2 −1 1 2 3 4 5 6

1

x

y



Fourier Series

Example 97

Find the Fourier series of the 2π-periodic function

f (x) := x for − π < x < π


−3π −2π −π π 2π 3π

−π

π

x

y

Figure: The graphic of f over the interval [−3π, 3π].


Fourier Series

We can compute that

a0 :=1

π

∫ π

−π

ηdη = 0,

an :=1

π

∫ π

−π

η cos(nη)dη = 0,

bn :=1

π

∫ π

−π

η sin(nη)dη = 2(−1)n+1

n.


Fourier Series

Thus, the corresponding Fourier series is

2

∞∑

k=1

(−1)k+1

ksin(kx),

which converges to

x − 2nπ, (2n− 1)π < x < (2n+ 1)π

0, x = (2n − 1)πfor n ∈ Z.

Below, the graphic of the function above is plotted.

−3π −2π −π π 2π 3π

−π

π

x

y

Figure: The graphic of f over the interval [−3π, 3π].


Fourier Series

Exercises 21

Exercises

1. Find the Fourier series of the functions given below and draw thegraphic of the function to which the series converges.a. f (x) := max0, x for |x | < π and f (x + 2π) = f (x) for x ∈ R.

b. f (x) := sgn(x) for |x | < π and f (x + 2π) = f (x) for x ∈ R.c. f (x) := x3 for |x | < 1 and f (x + 2) = f (x) for x ∈ R.d. f (x) := e

x for |x | < 2 and f (x + 4) = f (x) for x ∈ R.

e. f (x) :=(sin(x)

)3for |x | ≤ π and f (x + 2π) = f (x) for x ∈ R.

f. f (x) := | sin(x)| for |x | ≤ π and f (x + 2π) = f (x) for x ∈ R.

g. f (x) :=

0, −π < x ≤ 0

cos(x), 0 < x < πand f (x + 2π) = f (x) for x ∈ R.

h. f (x) :=

2, |x | < 1

1, 1 < |x | < 2and f (x + 4) = f (x) for x ∈ R.


Fourier Series

3. a. Show that

x2 ∼ 1

3+

4

π2

∞∑

k=1

(−1)k

k2cos(kπx) for |x | ≤ 1.

b. By substituting a specific value for x , show that

∞∑

k=1

1

k2=π2

6.

4. a. Use Exercise 1 (a) to show that

∞∑

k=1

(−1)k−1

2k − 1=π

4.

b. Use Exercise 1 (b) to show that

∞∑

k=1

1

(2k − 1)2=π2

8.


Fourier Series Fourier Cosine Series and Fourier Sine Series

Fourier Cosine Series and Fourier Sine Series

Definition 45 (Even Function and Odd Function)

1. A function f is said to be “even” if f (−x) = f (x) for all x ∈ R.

2. A function f is said to be “odd” if f (−x) = −f (x) for all x ∈ R.

Property 1. If both f and g are even, then f ± g is even and fg is even.

Property 2. If both f and g are odd, then f ± g is odd and fg is even.

Property 3. If f is even and g is odd, then fg is odd.

Property 4. If f is even, then∫ x

−xf (η)dη = 2

∫ x

0 f (η)dη.

Property 5. If f is odd, then∫ x

−xf (η)dη = 0.

Now, we will apply these properties to Fourier series.



1. Let f be even. Then,

a0 :=1

L

∫ L

−L

f (η)︸︷︷︸

even

dη =2

L

∫ L

0

f (η)dη,

an :=1

L

∫ L

−L

f (η)︸︷︷︸

even

cos

(nπ

Lη

)

︸︷︷︸

even︸︷︷︸

even

dη =2

L

∫ L

0

f (η) cos

(nπ

Lη

)

dη,

bn :=1

L

∫ L

−L

f (η)︸︷︷︸

even

sin

(nπ

Lη

)

︸︷︷︸

odd︸︷︷︸

odd

dη = 0.



2. Let f be odd. Then,

a0 :=1

L

∫ L

−L

f (η)︸︷︷︸

odd

dη = 0,

an :=1

L

∫ L

−L

f (η)︸︷︷︸

odd

cos

(nπ

Lη

)

︸︷︷︸

even︸︷︷︸

odd

dη = 0,

bn :=1

L

∫ L

−L

f (η)︸︷︷︸

odd

sin

(nπ

Lη

)

︸︷︷︸

odd︸︷︷︸

even

dη =2

L

∫ L

0

f (η) sin

(nπ

Lη

)

dη.



Accordingly, by extending a function f defined on (0, L) evenly, we getseries with no sine terms; by extending it oddly we get no cosine terms.These special cases of Fourier series are called as “Fourier Cosineseries” and “Fourier Sine series”, respectively.



Theorem 41

Let f be a given function with both f and f ′ piecewise continuous on(0, L).

1. The Fourier Cosine series of f is a02 +

∑∞k=1 ak cos

(kπLx), where

an := 2L

∫ L

0f (η) cos

(nπLη)dη.

2. The Fourier Sine series of f is∑∞

k=1 bk sin(kπLx), where

bn := 2L

∫ L

0f (η) sin

(nπLη)dη.

Both of these series converge to f (x) if x is a point of continuity. If x is

a point of jump discontinuity, then these series converge to f (x−)+f (x+)2 .

The Fourier Cosine series converge to f (0+) and f (L−) at the left-handand right-hand ends, respectively; but the Fourier Sine series converge to0 at the end points 0 and L.



Example 98

Computing the Fourier Cosine series for f (x) :≡ 1 for 0 < x < 1, we find

a0 :=2

L

∫ L

0

f (η)dη = 2

∫ 1

0

1dη,

an :=2

L

∫ L

0

f (η) cos(nπη)dη = 2

∫ 1

0

cos(nπη)dη

=0.

Hence, f (x) = a02 = 1.



Example 99

Let us compute the Fourier Sine series for f (x) :≡ 1 for 0 < x < 1. Wefind that

bn :=2

∫ 1

0

sin(nπη)dη = − 2

nπcos(nπη)

∣∣η=1

η=0

=21− (−1)n

nπ,

which yields

4

π

∞∑

k=1

1

2k − 1sin

((2k − 1)πx

).



-2 -1 1 2

-1

1

(a) First 10 terms of the series

-2 -1 1 2

-1

1

(b) First 50 terms of the series

Figure: Plot of some partial sums of the Fourier Sine series for f (x) :≡ 1.



Example 100

For the Fourier series of f (x) := |x | for −π < x < π, which is an evenfunction, we see that bn = 0 for n = 1, 2, · · · . On the other hand, wehave

a0 :=2

π

∫ π

0

ηdη = π

an :=2

π

∫ π

0

η cos(nη)dη =2

π

[1

n2cos(nη) +

1

nη sin(nη)

]η=π

η=0

=− 2

π

1− (−1)n

n2.

Thus, the corresponding Fourier series is

f (x) =π

2− 4

π

∞∑

k=1

1

(2k − 1)2cos

((2k − 1)x

)for x ∈ R.



-3 Π -2 Π -Π 0 Π 2 Π 3 Π

Π

2

Π


-3 Π -2 Π -Π 0 Π 2 Π 3 Π

Π

2

Π


Figure: Plot of some partial sums of the Fourier series for | · |.



Example 101

Let us find the Fourier Cosine and Fourier Sine expansions for thefunction f (x) := cos(x) for 0 < x < π

2 .

• For the Fourier Cosine series, we compute that

a0 :=4

π

∫ π

2

0

cos(η)dη =4

π

an :=4

π

∫ π

2

0

cos(η) cos(2nη)dη = − 4

π

(−1)n

4n2 − 1.

Thus, we have the Fourier Cosine series

2

π− 4

π

∞∑

k=1

(−1)k

4k2 − 1cos(2kx) for 0 < x <

π

2.



-3 Π

2-Π -

Π

2

Π

2Π

3 Π

2

1

2

1


-3 Π

2-Π -

Π

2

Π

2Π

3 Π

2

1

2

1


Figure: Plot of some partial sums of the Fourier Cosine series for cos.



• For the Fourier Sine series, we compute that

bn :=4

π

∫ π

2

0

cos(η) sin(2nη)dη =4

π

2n

4n2 − 1.

Thus, we have the Fourier Sine series

8

π

∞∑

k=1

k

4k2 − 1sin(2kx) for 0 < x <

π

2.



-3 Π

2-Π -

Π

2

Π

2Π

3 Π

2

-1

1


-3 Π

2-Π -

Π

2

Π

2Π

3 Π

2

-1

1


Figure: Plot of some partial sums of the Fourier Sine series for cos.



Exercises 21.1

Exercises

1.

2.

3.

4.


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MAT2039 Differential Equations I - Ki??isel Sayfalarkisi.deu.edu.tr/meltem.topcuoglu/MAT2039DiffEq/mat2039...Diﬀerential Equations and Their Solutions Example 1 Consider the following