Elementary differential equations and boundary value problemssite.iugaza.edu.ps/rbsalha/files/2012/02/DE.pdf · 1 ةزغ – ةيملاسلإا ةعماجلا مولعلا ةيلك

1

غزة –الجامعة اإلسالمية

كلية العلوم

قسم الرياضيات

المعادالت التفاضلية العادية

Elementary differential equations and

boundary value problems

المحاضرون

أ.د. رائد صالحة

د. فاتن أبو شوقة

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3

4

5

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الرحيمبسم هللا الرحمن

The Islamic University of Gaza

Department of Mathematics

Ordinary Differential Equations

2nd

Semester 2017-2018

Instructors: Prof. Raid Salha, Dr. Faten Abu-Shoga

Course Description: First order differential equations, Second order linear

equations, Higher order linear equations, Series solutions of second order linear

equations, The Laplace transform.

Course Objectives: At the end of this course, the student should be able

1- To understand the definition of differential equations and how to classify them.

2- To solve several types of first order differential equations. 3- To solve linear differential equations with constant coefficients. 4- To use series to solve second order linear equations with nonconstant

coefficients. 5- To use Laplace transformation in solving initial value problems.

Text Book:

Elementary Differential Equations and Boundary Value Problems

“Eight Edition”

By

W. E. Boyce and R. C. Diprima

8

Methods of Teaching: Lectures, discussions, solving selected problems.

Evaluation and Grading:

Midterm Exam 40%

Quizzes 10%

Final Exam 50%

Total 100%

Office Hours:

Saturday, Monday, Wednesday: 11-12

توزيع المادة الدراسية على أسابيع الفصل الدراسي

Week Sections to be covered Week Sections to be covered

1st week Chapter 1 9th week 4.1, 4.2

2nd week 2.1, 2.2 10th week 4.3, 4.4

3rd week 2.4, 2.6 11th week 5.1, 5.2

4th week 2.8 12th week 5.2, 5.3

5th week 3.1, 3.2 13th week 5.4, 5.5

6th week 3.2, 3.4 14th week 6.1, 6.2

7th week 3.5, 3.6 15th week General Review

8th week 3.7

Midterm Exam

16th week Final Exam

9

Definition:

The differential equation (DE) is an equation contains one or more

derivatives of unknown function.

Examples: The following are DE

1. 2 1y y

2. dp

r p kdt

3. (sin ) 3 cosxy x y y x

4. F ma (Newton’s Law)

Example: Consider the DE

(1)

10

(2)

To determine the constant c, we need a condition, as for example

(3)

Subsitute (3) into (2), we get

Then c= -50.

Definition 2:

Condition (3) is called an initial condition and the DE (1) with the initial

condition (3) is called an initial value problem (IVP).

Example 2 : Consider the IVP,

11

is the general solution.

Usingthe initial condition,

Then

Is the solution of the IVP.

1. If the unknown function depends on a single independent variable,

then only ordinary derivatives appear in the differential equation and

it is called an ordinary differential equation.

2. If the unknown function depends on several independent variables,

then only partial derivatives appear in the differential equation and it

is called a partial differential equation.

12

Examples: Classify the following differential equation

Order of the DE.

The order of a DE is the order of the highest derivative that appears in

the DE.

Example: Find the order of the following differential equations

1. 2 1y y

2.

(4)

13

.

The general linear ordinary DE of order n is

(5)

An equation that is not of the form (5) is nonlinear equation.

For example the DE

is non linear because of the term .

14

Chapter 2

First Order Differential Equations

2.1 Linear Equations; Method of Integrating Factors We will usually the general first order linear equations in the form

( ) ( ) (1)y p t y g t

where ( )p t and ( )g t are given functions of the independent variable t .

How we can solve the DE (1)?

To solve the DE (1),we multiply it by a certain function ( )t to get

( ) ( ) ( ) ( ) ( ) (2)t y t p t y t g t

such that ( ( ) ) ( ) ( ) (3)t y t g t

From equation (2) and (3), we get the following

( ) ( ) ( ) ( ) ( )t y t y t y p t t y

This implies that

( ) ( ) ( )t p t t

15

( )( )

( )

tp t

t

Now, If ( )t is positive for all t , then

ln ( ) ( )t p t dt k

By choosing the arbitrary constant k to be zero, we get that

( ) exp ( )t p t dt

Note: ( )t is positive for all t as we assumed and it is called an integrating factor.

Now using equation (3), we get that

( ) ( ) ( )t y t g t dt c

This implies that the general solution of the DE (1) is given by

1( ) ( ) (4)

( )y t g t dt c

t

Example1: Solve the initial value problem (IVP)

22 4 , (1) 2.ty y t y

16

Note: If the integral ( ) ( )t g t dt cannot be evaluated in terms of the usual elementary functional, so we leave the integral unevaluated and the general solution of the DE (1) is given by

1( ) ( ) (5)

( )o

t

t

y s g s ds ct

Where ot is some convenient lower limit of integration.


1, (0) 1.

2

ty y e y

17

H.W. Problems 1-20 Pages 39-40.

Example3(Q.8, page 39). Solve the DE

2 2 2(1 ) 4 (1 ) .t y ty t

Example4 (Q.20,page 40). Solve the (IVP)

( 1) , (ln 2) 1, 0.ty t y t y t

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2.2 Separable Equations Consider the first Order DE

( , ) (1)y f x y

If equation (1) is nonlinear, we can write it in the form

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( , ) ( , ) 0 (2)dy

M x y N x ydx

This form in equation (2) is not unique.

If M is a function of x only and N is a function of y only then equation (2) becomes

( ) ( ) 0 (3)dy

M x N ydx

This equation is called separable because it is written in the form

( ) ( ) 0 (4)M x dx N y dy

A separable equation in (4) can be solved by integrating the functions M

and N .

Example1: Solve the Differential equation

2

2.

1

xy

y

20


23 4 2, (0) 1.

2( 1)

x xy y

y

H.W. Problems 1-20 Pages 47-48.

21

Example3 (Q.11, page 48). Solve the IVP

0, (0) 1xxdx ye dy y

Example4 (Q.20,page 48). Solve the (IVP)

1

2 2 12(1 ) sin 0, (0) 1y x dy x dx y

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Homogeneous Equations

If the right side of the DE ( , )dy

f x ydx

can be expressed as a function

of the ratio y

x only, then the equation is said to be homogeneous. Such

DE can always be transformed into separable DE by a change of the dependent variable.

Example5 (Q.30, page 49). Solve the DE

4.

dy y x

dx x y

23


4 3.

2

dy y x

dx x y

24


2 23.

2

dy x y

dx xy

2.4 Differences Between Linear and Nonlinear Equations Theorem 2.4.1

If the functions p and g are continuous on an open interval :I t

containing the point 0t t , then there exist a unique function ( )y t satisfies the differential equation

25

( ) ( ) (1)y p t y g t

for each t I , and that also satisfies the initial condition

0 0( ) (2)y t y

Where 0y is an arbitrary prescribed initial value.

Note:

1. (Existence and Uniqueness) Theorem 2.4.1 states that the given (IVP) has a solution and it is unique.

2. It also states that the solution exists through any interval

I containing the initial point 0t in which p and g are continuous.

Example 1: Use Theorem 2.4.1 to find an interval in which the (IVP)

22 4 , (1) 2ty y t y

has a unique solution.

Now, we generalize Theorem 2.4.1 to the case of nonlinear DE.

Theorem 2.4.2

Let the functions f and f

y

be continuous in some rectangle

, ,t y containing the point 0 0( , )t y , Then in some interval

0 0t h t t h contained in ,t there is a unique solution ( )y t of the IVP

26

0 0( , ), ( ) (3)y f t y y t y

Example 2: Use Theorem 2.4.2 to show the (IVP)

23 4 2, (0) 1

2( 1)

x xy y

y


Example3: Find an interval in which the solution of the (IVP) exists

(ln ) cot , (2) 3.t y y t y

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Example 4: State where in the ty plane the hypothesis of Theorem2.4.2 are satisfied.

2

2

1

3

dy t

dt y y

Bernoulli Equations:

The DE of the form

( ) ( ) , 0,1 (4)ny p t y g t y n

is called a Bernoulli equation which is nonlinear.

Note:

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1. If 0,n then DE (4) has the form ( ) ( ),y p t y g t

which is a linear DE.

2. If 1,n then DE (4) has the form ( ) ( ) ,y p t y g t y

which can be rewritten in the form ( ) ( ) 0,y p t g t y

which is a linear DE.

3. If 1n or 0,n then the DE has the form

1( ) ( ), (5)n ny y p t y g t

Let 1 1(1 )

1

n n ndz dy dy dzz y n y y

dt dt dt n dt

Substitute in DE (5) becomes

1( ) ( ), (6)

1

dzp t z g t

n dt

Note that DE (6) is a linear DE, where the dependent variable is .z We solve it for Z, then we replace z by 1 ny , then we solve it for

.y

Example 5. Solve the DE

32 ( 1) 6x y x y y

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Example 6. Solve the DE 2 32 0, 0.t y ty y t

30

Discontinuous coefficients

Example 7. Solve the DE 2 ( ), (0) 0,y y g t y

Where 1, 0 1,

( )0, 1.

tg t

t

H.W. Problems 1-12, 24, 25, 28, 33.

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32

2.6 Exact Equations and Integrating Factors

Consider the DE ( , ) ( , ) 0.M x y N x y y (1)

Suppose that we identify a function ( , )x y such that

( , ) ( , )( , ), ( , )

x y x yM x y N x y

x y

(2)

and such that ( , )x y c .

Defines ( )y x implicitly as a differentiable function of x . Then

( , ) ( , )( , ) ( , )

, ( ) .

x y x y dyM x y N x y y

x y dx

dx x

dx

So the DE in (1) becomes

, ( ) 0d

x xdx

(3)

In this case DE(1) is said to be exact DE. The solution of DE(1)or the equivalent DE(3) are given implicitly by

( , )x y c (4)

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Example 1. The following DE is an exact DE

22 2 0x y xy y

Theorem 2.6.1

Let the functions , , yM N M and xN be continuous in rectangular region : , .R x y Then the DE

( , ) ( , ) 0M x y N x y y

is an exact DE in R if and only if

( , ) ( , )y xM x y N x y (5) At each point of R.

34

Example 2 : Solve the DE 2cos 2 (sin 1) 0y yy x xe x x e y

Example 3 Solve the DE 2 2(2 2 ) (2 2 ) 0.xy y x y x y

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Integrating Factors

Example 4 Determine whether the DE is exact or not

2 2(3 ) ( ) 0.xy y x xy y

Note: It is sometimes possible to convert a DE that is not exact into an exact equation by multiplying the equation by suitable integrating factor.

Consider the DE ( , ) ( , ) 0.M x y dx N x y dy (6)

Multiply DE(6) by a function ( , )x y

( , ) ( , ) ( , ) ( , ) 0.x y M x y dx x y N x y dy (7)

Assume DE (7) is an exact, then ( ) ( )y xM N (8)

This implies that

y y x xM M N N (9)

36

Note:

The most important situation in which simple integrating factors can be found when is a function of one of the variables x and .y

Case 1: If is a function of x , Then Eq(8) becomes

y x

dM N N

dx

(9)

This implies that y xM Nd

dx N

(10)

If y xM N

N

is a function of x only, then there is an integrating

factor that depends on x only and it can be found by solving DE (10) which is both linear and separable and it's solution is given by

( ) exp .y xM N

x dxN

(11)

Example 5 Find an integrating factor for the DE

2 2(3 ) ( ) 0.xy y x xy y

then solve it.

37

Case 2: If is a function of y only. How we can find an integrating factor?. See problem 23 page 100.

Example 6 Find an integrating factor for the DE

( cot 2 csc ) 0.x xe dx e y y y dy

then solve it.

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Note: If the DE is not exact, then we have the following.

1.There is no integrating factor for the DE.

2. There is one integrating factor for the DE.

3. There is more than one integrating factor for the DE.

Example 7 Use the integrating factor 1

( , ) (2 )x y xy x y

To solve the DE 2 2(3 ) ( ) 0.xy y x xy y

39

H.W. Problems 1-32 except (17, 24, 31) page 99-101.

40

41

Example 8: Solve the IVP

2(9 1) (4 ) 0, (1) 0.x y dx y x dy y

Example 9 : Find the value of b for which the given DE is exact then

solve the DE 2 2( ) 0.xy xyy e x dx bx e dy

42

Example 10 Solve the IVP

3

2 234 ( ) 3 4 0.x xdx y dy

yy y

Miscellaneous Problems (Page 131)

H.W. Problems 1-32 Page 131.

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Some special second order equations

1- Equations with the dependent variable missing

Consider the second order DE ( , )y f t y (1)

Let v y this implies v y and this transform DE(1) to the first order DE ( , )v f t v (2)

We solve DE(1) for v then we find y by integrating v .

Example1: Solve the DE

2 2 1 0, 0.t y ty t

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2- Equations with the independent variable missing

Consider the second order DE ( , )y f y y (3)

Let v y this implies ( , )dv

v f y vdt

(4)

If we think of y as an independent variable, then by the chain rule, we

have dv dv dy dv

vdt dy dt dy

This implies that DE (4) can be written as ( , )dv

v f y vdy

(5)

We solve DE (5) for v as a function of y then we replace v by dyy

dt

which gives us a separable DE with dependent variable y and

independent variable t . We solve this separable DE to find y .

Example2: Solve the DE 2( ) 0, 0.yy y t

45

Example3: Solve the DE 2 2( ) , 0.t y y t

46

Example 4:(Q 47 Page 133) Solve the DE

2( ) 2 , 0.yy y e t

H.W. Problems 36-51 Page 133.

47

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Chapter 3 Second Order Differential Equations

3.1 Homogeneous Equations with constant coefficients

A second order linear DE is written in the form

( ) ( ) ( ) ( )P t y Q t y R t y G t (1)

( ) ( ) ( )y p t y q t y g t (2)

where ( ) ( )( ) , ( )

( ) ( )

Q t R tp t q t

P t P t and ( )

( )( )

G tg t

P t .

Note: If the DE can't be written in the form (1) and (2) then it is a nonlinear DE.

An initial value problem (IVP) consists of a DE such as in (1) and (2) with a pair of initial conditions

0 0, 0 0( ) ( )y t y y t y (3)

Where 0y and 0y are given numbers.

Definition: A second order linear equation is said to be homogeneous if

the term ( )G t in DE (1) or the term ( )g t in DE (2) is zero for all t .

Otherwise the DE is called a nonhomogeneous.

49

Note: In this chapter, we will concentrate our attention on DEs in which the functions ,P Q and R in DE (1) are constants, that is in the form

0ay by cy (4)

Where ,a b and c are constants.

To solve DE (4), suppose r ty e is a solution of the DE (4).

2,r t r ty re y r e

Now substitute in DE (4), we get

2

2

0

( ) 0

r t r t r t

r t

ar e bre ce

e ar br c

Since 0r te then

2 0ar br c (5)

Equation (5) is called the Characteristic equation for the DE (4).

Assume the two roots of equation (5) are two different real roots, 1r

and 2r , 1 2r r . Then 1

1

r ty e and 2

2

r ty e are two solutions of the DE (4). The general solution of the DE (4) is given by

1 2

1 1 2 2 1 2

r t r ty c y c y c e c e (6)

50

Example 1: Solve the DE 0.y y

Example 2: Find the general solution of the DE 5 6 0.y y y

51

Example 3: Find the solution of the IVP

5 6 0, (0) 2, (0) 3.y y y y y


4 8 3 0, (0) 2, (0) 3.y y y y y

52

Example 5: Solve the IVP

5 3 0, (0) 1, (0) 0.y y y y y

Example 6. Find a DE whose general solution is

221 2

t

ty c e c e

53

3.2 Fundamental Solutions of Linear Homogeneous Equations:

Theorem 3.2.1: Consider the initial value problem ( ) ( ) ( )y p t y q t y g t , 0 0, 0 0( ) ( )y t y y t y

54

where ( ), ( )p t q t and ( )g t are continuous on an interval I that contains the point 0t . Then there is exactly one solution of this IVP, and the solution exists throughout the interval I .

Note:

1. (Existence and Uniqueness) Theorem 3.2.1 states that the given (IVP) has a solution and it is unique.

It also states that the solution exists through any interval I containing the

initial point 0t in which ,p q and g are continuous.

Example 1: Find the longest interval in which the (IVP)

2( 3 ) ( 3) 0, (1) 2, (1) 2t t y ty t y y y


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Theorem 3.2.2:

If 1y and 2y are two solutions of the DE ( ) ( ) 0y p t y q t y

Then the linear combination 1 1 2 2c y c y is also a solution for any values

of the constants 1c and 2c on an interval I that contains the point 0t .

Definition: Let 1y and 2y be two solutions of the

( ) ( ) 0y p t y q t y .

The wronskian determinate (or simply the Wronskian) of the solutions

1y and 2y is given by 1 2

1 2 1 2 2 1

1 2

( , ) .y y

W y y y y y yy y

Theorem 3.2.3: If 1y and 2y are two solutions of the DE

( ) ( ) 0y p t y q t y and the Wronskian 1 2 1 2 2 1( , )W y y y y y y

is not zero at the point 0t where the initial conditions

0 0, 0 0( ) ( )y t y y t y .

Then there are constants 1c and 2c for which 1 1 2 2y c y c y the DE and the initial conditions.

Theorem 3.2.4: If 1y and 2y are two solutions of the DE

( ) ( ) 0y p t y q t y and if there is a point 0t where the

Wronskian of 1y and 2y is nonzero, then the family of solutions

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1 1 2 2y c y c y

With arbitrary coefficients 1c and 2c includes every solution of the DE.

Definition: The solution 1 1 2 2y c y c y is called the general solution of

the DE ( ) ( ) 0y p t y q t y .

The solutions 1y and 2y with a nonzero Wronskian are said to form a fundamental set of solutions of the DE.

Example 2: Suppose 1

1

r ty e and 2

2

r ty e are two solutions of ( ) ( ) 0y p t y q t y . Show that they form a fundamental set of

solutions if 1 2r r .

57

Example 3: Show that 1

21y t and 1

2y t form a fundamental set of

solutions of 22 3 0, 0.t y ty y t

Example 4(Q.12, page 151). Find the longest interval in which the (IVP)

( 2) ( 2)(tan ) 0, (3) 1, (3) 2x y y x x y y y has a unique solution.

58

Example 5: If the Wronskian of f and g is 2 tt e and if ( )f t t find ( )g t .

Example 6: Is the functions 1y x and 2

xy xe form a fundamental

set of solutions of the DE 2 ( 2) ( 2) 0, 0.x y x x y x y x

59

60

3.3 Linear independence and Wronskian

Definition: Two functions ( )f t and ( )g t are said to be linearly dependent on an interval I if there exist two constants 1k and 2k no both zero, such that 1 2( ) ( ) 0k f t k g t (1)

For all t I . The functions ( )f t and ( )g t are said to be linearly independent on an interval I if they are not linearly dependent.

Theorem 3.3.1 If ( )f t and ( )g t are differentiable functions on an open interval I and if 0( , )( ) 0W f g t for some point 0t I then ( )f t and ( )g t are linearly independent on an interval I .

Moreover, if ( )f t and ( )g t are linearly dependent on an interval I then ( , )( ) 0W f g t for every t I .

Theorem 3.3.2 (Abel's Theorem) If 1y and 2y are two solutions of the DE ( ) ( ) 0y p t y q t y (2)

Where ( )p t and ( )q t are continuous on an open interval I , then the Wronskian 1 1( , )( )W y y t is given by

1 1( , )( ) exp ( )W y y t c p t dt (3)

Where c is a constant that depends on 1y and 2y but not on t .

Note: 1 1

0, 0,( , )( )

0, 0.

cW y y t

c

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Example 1: Assume that 1

21y t and 1

2y t are solution of the DE

22 3 0, 0.t y ty y t

Use Abel's Theorem to find the constant c in (3).

Theorem 3.3.3: Let 1y and 2y be solutions of the DE

( ) ( ) 0y p t y q t y

where ( )p t and ( )q t are continuous on an open interval I , then

1y and 2y are linearly dependent on I if and only if 1 1( , )( )W y y t is zero for all t I . Alternatively, 1y and 2y are linearly independent on I if and only if 1 1( , )( )W y y t is never zero for all t I .

62

Summary: We can summarize the facts about fundamental set of solutions, Wronskian and linear independence in the following

Let 1y and 2y be solutions of the DE ( ) ( ) 0y p t y q t y

where ( )p t and ( )q t are continuous on an open interval I . Then the following statements are equivalent.

1- The functions 1y and 2y are a fundamental set of solutions on I .

2- The functions 1y and 2y are lineary independent on I

3- 1 1 0( , )( ) 0W y y t for some 0t I .

4- 1 1( , )( ) 0W y y t for all t I .

Example 2: Determine whether the following functions are linearly

independent or linearly dependent 3( ) xf x e and 3( 1)( ) xg x e

63

Example 3: Find the Wronskian of two solutions of the given DE

Example 4(Q.21, page 158)

64

65

3.4 Complex roots of the Characteristic Equations

Consider the DE 0ay by cy (1)

where ,a b and c are constants.

The Characteristic equation of the DE(1) 2 0ar br c (2)

Suppose 2 4 0b ac . Then the roots of Eq.(2) are conjugate complex

numbers, we denote them by 1 2,r i r i (3)

Then the two solutions of DE (1), are

( ) ( )

1 2,i t i ty e y e (4)

Euler,s formula cos sintie t i t (5)

cos sintie t i t

Now, using Euler,s formula, we have

( )

1 (cos sin )i t t ti ty e e e e t i t (6)

( )

2 (cos sin )i t t ti ty e e e e t i t (7)

Note, the two solutions 1y and 2y are complex-valued functions. We

want a real–valued solutions. To get real solutions from 1y , and 2y , we use their sum and difference

66

1 2

1 2

(cos sin ) (cos sin ) 2 cos

(cos sin ) (cos sin ) 2 sin

t t t

t t t

y y e t i t e t i t e t

y y e t i t e t i t ie t

Let ( ) cos , ( ) sint tu t e t v t e t (8)

2cos sin( , )( ) 0,

cos sin sin cos

t t

t

t t t t

e t e tW u v t e

e t e t e t e t

If 0.

This implies that ( )u t and ( )v t form a fundamental set of solutions of DE(1).

From the above, we conclude that, the general solution of DE(1) is

1 2cos sint ty c e t c e t (9)

where 1c and 2c are arbitrary constants.

Example 1: Find the general solution of 0.y y y

67

Example 2: Find the general solution of 9 0.y y


16 8 145 0, (0) 2, (0) 1.y y y y y

68

69

3.5 Repeated roots; Reduction of order

Consider theDE 0ay by cy (1)

where ,a b and c are constants.

The Characteristic equation of the DE(1)

2 0ar br c (2)

Suppose 2 4 0b ac . Then Eq.(2) has two equal roots

1 22

br r

a

(3)

Then we have one solution of DE (1), 2

1

bt

ay e

To find a second solution 2y that is linearly independent with the first

solution 1y , we assume the general solution of DE(1)

2

1

2 2

22 2 2

2

( ) ( )

2

4

bt

a

b bt t

a a

b b bt t t

a a a

y v t y v t e

by v e v e

a

b by v e v e v e

a a

Substitute in DE(1), we get

70

22 2 2 2 2

2

2

2 2

1 2

4 2

( ) 0.

( ) ( ) 04 2

0

b b b b bt t t t t

a a a a a

bt

a

b b ba v e v e v e b v e v e

a a a

c v t e

b bav b b v c v

a a

v

v c t c

2 2 2 2

1 1 2 1 2( ) ( ) ( )

b b b bt t t t

a a a ay v t y v t e c t c e c te c e

2

1

bt

ay e

and 2

2

bt

ay te

2 2

1 22 2 2

( , ) 0.

2 2

b bt t

a ab

ta

b b bt t t

a a a

e teW y y e

b bte e e

a a

This implies that 1y and 2y form a fundamental set of solutions of DE(1).

Example 1: Find the general solution of the DE 4 4 0.y y y

71


10.25 0, (0) 2, (0) .

3y y y y y

Summary

Now, we can summarize the results that we have obtain for the second order linear homogeneous DEs with constant coefficients

as follows: Consider the DE 0ay by cy

Let 1r and 2r be two roots of the corresponding characteristic

equation 2 0ar br c

72

1- If 1r and 2r are different real roots 1 2r r ,then the general

solution is given by 1 2

1 2

r t r ty c e c e

2- If 1r and 2r are complex conjugate roots

1 2,r i r i ,then the general solution is

given by 1 2cos sint ty c e t c e t .

3- If 1 2r r r , then the general solution is given by

1 2

rt rty c e c te .

Reduction of order

Suppose we know one solution 1( )y t , not every where zero of the DE

( ) ( ) 0y p t y q t y (4)

To find a second solution 1( ) ( )y v t y t (5)

Then 1 1 1 1 1, 2y v y vy y v y v y vy

Substitute , ,y y y in DE(4) and colleting, we have

1 1 1 1 1(2 ) ( ) 0y v y py v y py qy v

1 1 1(2 ) 0y v y py v (6)

73

Equation (6) is a first order equation for the function v and can be solved as a linear DE or a separable DE. After finding v , then v can be obtained by an integration.

Note: This method is called the method of reduction of order because the main step is to solve DE (6) which is a first order DE

for v rather than solving DE (4) which is second order DE for y .

Example 3: Given that 1

1y t is a solution of

22 3 0, 0,t y ty y t

Find a second linearly independent solution.

74

H.W. 1-14, 23-30 Page 172-174.

75

3.6 Nonhomogeneous Equations, Method of undetermined coefficients

Consider the nonhomogeneous DE

( ) ( ) ( )y p t y q t y g t (1)

where ,p q and g are given continuous functions on an open interval I . The DE

( ) ( ) 0y p t y q t y (2)

in which ( ) 0g t and ,p q as in DE(1) is called the corresponding homogeneous DE to DE(1).

Theorem 3.6.1 If 1Y and 2Y are two solutions of the nonhomogeneous DE(1), then their difference 1 2Y Y is a solution of the corresponding homogeneous DE(2).

In addition if 1y and 2y are a fundamental set of solutions of DE(2), then

1 2 1 1 2 2( ) ( ) ( ) ( )Y t Y t c y t c y t

Where 1c and 2c are constants.

Theorem 3.6.2 The general solution of the nonhomogeneous DE(1) can be written in the form

1 1 2 2( ) ( ) ( ) ( )y t c y t c y t Y t (3)

76

Where 1y and 2y are a fundamental set of the corresponding

homogeneous DE(2), 1c and 2c are arbitrary constants and Y is some specific solution of the nonhomogeneous DE(1).

Note: The solution of the nonhomogeneous DE(1) can be obtained using the following three steps.

1. Find the complementary solution 1 1 2 2cy c y c y

which is the general solution of corresponding homogeneous DE(2).

2. Find a particular solution py of the nonhomogeneous DE(1).

3. Find the general solution of the nonhomogeneous DE(1) by adding

cy and py . From the first two steps c py y y

The method of undetermined coefficients

This method usually is used only for problems in which the homogeneous DE has constant coefficients and the nonhomogeneous terms that consist of polynomials, exponential functions, sines and cosines.

Example 1: Find a particular solution of the DE 23 4 3 .ty y y e

77

Example 2: Find a particular solution of the DE 3 4 2sin .y y y t

Example 3: Find a particular solution of the DE

23 4 4 1.y y y t

78

Note:

1 1

0 1 0 1

( )

sin cos sin cos

p

t t

n n n n

n n

g t y

e Ae

t or t A t B t

a t a t a A t A t A

Remark 1:

When ( )g t is a product of any two or all three of these types of functions, we use the product the same functions.


3 4 8 cos2 .ty y y e t

79

Remark 2:

Suppose that 1 2( ) ( ) ( )g t g t g t and suppose that 1( )pY t and

2( )pY t

are particular solutions of the DEs

1

2

( ).

( ).

ay by cy g t

ay by cy g t

respectively, then 1( )pY t +

2( )pY t is a particular solution of the DE

( ).ay by cy g t


23 4 3 2sin 8 cos2 .t ty y y e t e t

80

Example 6: Find a particular solution of the DE 3 4 2 .ty y y e

Remark 3: If the form of the particular solution that we choose is a solution of the corresponding homogeneous DE, then we multiply it

by , 1,2.st s

Example 7(Q. 14 page 184): Solve the following DE

24 3 , (0) 0, (0) 1.ty y t e y y

81

Example 8(Q. 26 page 184): Determine a sutible form of the particular solution of the following DE

22 5 3 cos2 2 cos .t ty y y te t te t

82

83

3.7 Variation of parameters

This method is more general the method of undetermined coefficients, since it can enable us to get a particular solution of an arbitrary second order linear nonhomogeneous DE.

( ) ( ) ( )y p t y q t y g t (1)

where ,p q and g are given continuous functions on an open interval I .

If ( )g t is not one of the functions as in Section 3.6 ( exponential, polynomial, sines, cosines, their product or sums) then we can't use the method of undetermined coefficients, therefore we will use the method of variation of parameters.

Suppose 1 2,y y are two linearly independent solutions of the corresponding homogeneous DE(2).

( ) ( ) 0y p t y q t y (2)

Then 1 1 2 2( ) ( )cy c y t c y t ,

where 1c and 2c are constants is the solution of DE(2).

In the method of Variation to find a particular solution of DE(1), we

replace the constants 1c and 2c by two functions 1( )u t and 2 ( )u t to get

particular solution 1 1 2 2( ) ( ) ( ) ( )py u t y t u t y t

84

where, 2 11 2

1 2 1 2

( ) ( ) ( ) ( )( ) , ( ) .

( , )( ) ( , )( )

y t g t y t g tu t dt u t dt

W y y t W y y t

The general solution of the nonhomogeneous DE(1) can be written in the

form 1 1 2 2( ) ( ) ( ) ( )py t c y t c y t y t (3)

where 1y and 2y are a fundamental set of the corresponding

homogeneous DE(2), 1c and 2c are arbitrary constants and ( )py t is some specific solution of the nonhomogeneous DE(1).


4 3csc .y y t

85

Example 2: Solve the following DE tan , 02

y y t t

Example 3: Verify that 2

1y x and 2

2 lny x x are solutions of the corresponding homogeneous DE of the following nonhomogeneous DE

2 23 4 ln , 0.x y xy y x x x

86

Example 4(Q. 29 page 192): Use the method of reduction of order to solve

the DE 2 2

12 2 4 , 0, .t y ty y t t y t

87

88

Chapter 4

Higher Order Linear Equations

4.1 General Theory of the nth Order Linear Equations

Definition 1: An nth order linear DE is an equation written in the form

( ) ( 1)

0 1( ) ( ) ( ) ( )n n

nP t y P t y P t y G t (1)

Where 0 1( ), ( ), , ( )nP t P t P t and ( )G t are continuous real-valued

functions on some interval :I t and 0 ( )P t

is nowhere zero in I . Then by dividing DE(1) by 0 ( )P t

( ) ( 1)

1( ) ( ) ( )n n

ny p t y p t y g t (2)

Definition 2:

An initial value problem (IVP) consists of DE(2) with n initial conditions

where ( 1) ( 1

0 0 0 0 0 0 0 0( ) , ( ) , ( ) , , ( )n ny t y y t y y t y y t y

(3) where 0t is a point in I .

Theorem 4.1.1:

If the functions 1 2( ), ( ), , ( )np t p t p t and ( )g t are continuous real-

valued functions on an open interval I , then there exists exactly one

89

solution ( )y t of the DE(2) that also satisfies the initial conditions

(3). This solution exists through the interval I .

Definition 3: A homogeneous equation is an equation as in (2) where the

term ( ) 0g t i.e. ( ) ( 1)

1( ) ( ) 0n n

ny p t y p t y (4)

Otherwise the DE is called a nonhomogeneous.

Theorem 4.1.2:

If the functions 1 2, , , np p p are continuous on an open interval I , if

the functions 1 2, , , ny y y are solutions of DE(4) and if

1 2( , , , ) 0nW y y y . For at least one point in I then every solution of DE can be expressed as a linear combination of the solutions

1 2, , , ny y y .

Definition 4:

A set of solutions 1 2, , , ny y y of DE (4) where the wronskian is nonzero is called a fundamental set of solutions. The solution

1 1 2 2 n ny c y c y c y is called the general solution of the DE(4).

Definition 5: The functions 1 2, , , nf f f are said to be linearly dependent

on I if there exist a set of constants 1 2, , , nk k k not all zeros, such

that 1 1 2 2 0n nk f k f k f .

90

The functions 1 2, , , nf f f are said to be linearly independent if they are not linearly dependent.

Remark: 1 2, , , ny y y is a fundamental set of solutions of DE(4) if and only if they are linearly independent

Example 1(Q5, Page 222): Determine intervals where the solutions of the

DEs exist. (4)( 1) ( 1) (tan ) 0.x y x y x y

91

Example 2(Q15, Page 222):Verify that the give functions are solutions of the DE and determine their wronskian

30, 1, , .xy y x x

92

4.2 Homogeneous Equations with Constant Coefficients

Consider the nth order linear homogeneous DE

( ) ( 1)

0 1 0n n

na y a y a y (1)

Where 0 1, , , na a a are constants.

rty e be a solution of DE(1) then 1

0 1 0n n

na r a r a (2)

is called the characteristic equation of the DE(1). Eq(2) has n roots

1 2, , , nr r r .

1- Real and unequal roots. The general solution

1 2

1 2nr t r t r t

ny c e c e c e (3)

Example 1: Find the general solution of the DE

(4) 7 6 0, (0) 1,

(0) 0, (0) 2, (0) 1

y y y y y y

y y y

93

2.Complex Roots: If the characteristic equation has complex roots, they must occur in conjugate pairs i

Example 2: Find the general solution of the DE (4) 0.y y

Repeated Roots:

If the characteristic equation has s repeated roots, 2 1, , , ,rt rt rt s rte te t e t e .

If the complex root i is repeated s times, then its conjugate i is repeated s times.

94


(4) 2 0.y y y

Example 4: Find the general solution of the DE (4) 0.y y

95

Example 5 (Q.7, Page 230): Find the roots of 1

31

Example 6: Find the general solution of the DE 5 3 0.y y y y

96

97

4.3 The Method of Undetermined Coefficients

This method as we have discussed in Section 3.6 and the main difference that we use it for higher order DE.


3 3 4 .ty y y y e


(4) 2 3sin 5cos .y y y t t

98


24 3cos .ty y t t e

99

Example 4: Without finding the constants, find a formula for a particular solution of the nonhomogeneous DE if

(a) 1 2 3 4 5sin 2 cos2t t

cy c c e c te c t c t

(i) ( ) 3 2 costg t e t t

(ii) ( ) 3 2 costg t e t t

100

(b) 2

1 2 3 4 5 6 7( )sin3 ( )cos3cy c c t c t c c t t c c t t

2( ) ( 1) sin3g t t t t

Example 5 (Q.5, Page 235): Determine the general solution of the DE

(4) 24 .ty y t e

101

102

4.4 The Method of Variation of Parameters

This method is a generalization of the method for second order DE from Section 3.7. Consider the DE

( ) ( 1)

1( ) ( ) ( )n n

ny p t y p t y g t (1)

To find a particular solution, at first we solve the corresponding

homogeneous DE ( ) ( 1)

1( ) ( ) 0n n

ny p t y p t y (2)

Let 1 1 2 2c n ny c y c y c y (3)

Be the general solution of the DE(2)

Then 1 1 2 2p n ny u y u y u y (4)

Where 1 2, , , nu u u are functions of t and

( ) ( )( ) , 1,2, ,

( )

mm

W t g tu t dt m n

W t

Where 1 2( ) ( , , , )( )nW t W y y y t

( )mW t is the determinate obtained from ( )W t by replacing the mth column by the column (0,0, ,1) .

103

Example 1(Q1, Page 240): Find the general solution of the DE

tan , 0 .y y t t

Example 2: Solve the DE .y y t

104

105

Chapter 5

Series solutions of Second Order Linear Equations

5.1 Review of Power Series

Definition: A power series 0

0

( )n

n

n

a x x

is said to converge at a point

0x if 0

0

lim ( )m

n

nm

n

a x x

exists for that 0x .

Definition: The series 0

0

( )n

n

n

a x x

is said to converge absolutely at a

point 0x if the series 0 0

0 0

| ( ) | | | | |n n

n n

n n

a x x a x x

converges.

Remark: If the series converges absolutely, then the series also converges, but the converse is not necessarily true.

Definition: A nonnegative number is called the radius of convergence of

a series 0

0

( )n

n

n

a x x

if it converges absolutely for 0| |x x and

diverges for 0| |x x .

Remark:

1- If the series converges only at 0x then 0 .

2- If the series converges for all x then .

106

3- If 0 then the interval 0| |x x is called the interval of

convergence.

Definition: If a function f is continuous and has derivatives of all order

for 0| |x x . Then the power series generated by f about 0x x given by

( )

00

0

200 0 0 0

( )( ) ( )

!

( )( ) ( )( ) ( )

2!

nn

n

f xf x x x

n

f xf x f x x x x x

is called a Taylor series expansion of f about 0x x .

If 0 0x then it is called a Maclaurin series.

Definition: A function f that has Taylor series expansion about 0x x

with radius of convergence 0 is said to be analytic at 0x x .

Examples: The Maclaurin series of

1. 0 !

nx

n

xe

n

2.

2 1

0

( 1)sin

(2 1)!

n n

n

xx

n

2. 2

0

( 1)cos

(2 )!

n n

n

xx

n

4.

01

n

n

aa x

x

107

Shift of Index of Summation

2

2

0 2

n n

n n

n n

a x a x

1

1

1 0

( 1)n n

n n

n n

na x n a x

2

2

2 0

( 1) ( 2)( 1)n n

n n

n n

n n a x n n a x

2

2 0 0

0 2

( 4)( 3) ( ) ( 2)( 1) ( )n n

n n

n n

n n a x x n n a x x

5.2 Series Solutions Near an Ordinary Point

Part I

( ) ( ) ( ) 0P x y Q x y R x y (1)

Suppose ,P Q and R are polynomials and that they have no common factors. Suppose that we want to solve DE(1) in the neighborhood of a point 0x .

Definition: A point 0x such that 0( ) 0P x is called an ordinary point.

108

Since P is continuous, it follows that there is an interval about 0x in which ( ) 0P x . In that interval, we can divide DE(1) by ( )P x to get DE(2) ( ) ( ) 0y p x y q x y (2)

Definition: A point 0x such that 0( ) 0P x is called a singular point.

Example 1: Find a series solution of the DE 0y y

Example 2: Find a series solution of the DE

0. .y xy x

109

Example 3: Find a series solution in powers of ( 1)x of the DE

0. .y xy x

Example 4 (Q 9 page 259): Find a series solution of the DE

2(1 ) 4 6 0.x y xy y

110

Example 5 (Q 15 page 259): Find a series solution of the IVP

0, (0) 2, (0) 1.y xy y y y

111

112

5.3 Series Solutions Near an Ordinary Point

Part II

Consider the DE ( ) ( ) ( ) 0P x y Q x y R x y (1)

Suppose ,P Q and R are polynomials.

In the neighborhood of a point 0x , Suppose that

0

0

( ) ( )n

n

n

y x a x x

(2)

is a solution of the DE(1).

Now, by differentiating Equation (2) m times and setting 0x x , it

follows ( )

0! ( ), 0,1,2,m

mm a x m

Since ( )y x is a solution of DE (1), we get that

( ) ( ) ( ) ( ) ( ) ( ) 0P x x Q x x R x x

From which we get that

( ) ( )( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( )

Q x R xx x x

P x P x

p x x q x x

(3)

0 0 0 0 0( ) ( ) ( ) ( ) ( )x p x x q x x

2 0 0 0 02! ( ) ( ) ( ) ( )a p x x q x x

113

To find 3a we differentiate Equation (3) and set 0x x .

Remark: To determine the 'sna we need to compute infinitely many derivatives of the functions ( )p x and ( )q x .

Example 1 (Q 2 page 265):

If ( )y x is a solution of the DE Find (4)

0 0 0( ), ( ), ( )x x x .

(sin ) (cos ) 0, (0) 0, (0) 1,y x y x y y y

Example 2: What is the radius of convergence of the Taylor series for 2 1(1 )x about 0 0x ?

114

Example 3: What is the radius of convergence of the Taylor series for 2 1( 2 2)x x about 0 0x ? about 0 1x ?

Example 4 (Page 264): Determine a lower bound for the radius of convergence of series solution of the DE

2 2(1 ) 2 4 0x y xy x y

about 0 0x ? about 0

1

2x ?

115

Example 5 (Page 264):

Can we determine a series solution about 0 0x for the DE

2(sin ) (1 ) 0y x y x y and if so, what is the radius of convergence?

Example 6 (Q7. Page 265): Determine a lower bound for the radius of

convergence of series solution of the DE 3(1 ) 4 0x y xy y

about 0 0x ? about 0 2x ?

116

5.4 Regular Singular Points

Consider the DE ( ) ( ) ( ) 0P x y Q x y R x y (1)

Definition: The singular point 0x of the DE(1) is called a regular singular point if both the functions

0

( )( )

( )

Q xx x

P x and 2

0

( )( )

( )

R xx x

P x (2)

Are analytic functions, (i. e. the functions in (2) have a convergent Taylor series expansion about 0x x ).

Remark 1: If ( ), ( )P x Q x and ( )R x are polynomials, then a singular point 0x is regular singular point if the limits

00

( )lim ( ) ,

( )x x

Q xx x

P x and

0

2

0

( )lim ( )

( )x x

R xx x

P x are both finite.

117

Remark 2: If 0x is not a regular singular point of DE(1), the it is called an irregular singular point.

Example 1: Find and classify the singular points of the DE

2(1 ) 2 ( 1) 0x y xy y .


22 ( 2) 3 ( 2) 0x x y xy x y .

118


2( ) (cos ) (sin ) 02

x y x y x y

.

119


2( 2) ( 1) 2 0x x y x y y .

5.5 Euler equations

Definition: The DE of the form 2 0x y xy y (1)

Where and are real constants is called an Euler equation.

Note: The point 0 0x is a regular singular point of the DE(1).

In any interval not including the origin, Eq(1) has a general solution of

the form 1 1 2 2y c y c y (2)

where 1y and 2y a re linearly independent

There are two cases when we solve DE(1).

Case 1, For 0x .

120

Assume ry x is a solution of DE(1).

This implies that 1 2, ( 1)r ry rx y r r x ,

Then by substituting in DE(1) we get

2 2 1( ( 1) ) ( ) 0.

( 1) 0.

r r r

r

x r r x x rx x

x r r r

This implies that 2 ( 1) 0.r r (3)

Eq(3) has two roots and they are given by

2

1 2

( 1) ( 1) 4, .

2r r

Real distinct roots: If Eq.(3) has two distinct real roots 1 2,r r , then 1 2

1 2,r r

y x y x are two solutions of DE(1) and the general solution of

DE(1) is given by 1 2

1 2

r ry c x c x

Example 1: Solve the DE 22 3 0, 0x y xy y x .

121

Equal Real roots:

If Eq.(3) has two equal real roots 1 2r r r , then 1 2, lnr ry x y x x are two solutions of DE(1) and the general solution of DE(1) is given by

1 2 lnr ry c x c x x

Example 2: Solve the DE 2 5 4 0, 0x y xy y x .

122

Complex roots:

If Eq.(3) has two complex roots 1 2,r i r i , then

1 2cos ln , sin lny x x y x x are two solutions of DE(1) and the general solution of DE(1) is given by 1 2cos ln sin lny c x x c x x

Example 3: Solve the DE 2 0, 0x y xy y x .

Case 2, For 0x . This case as the first case but we replace x by | |x .

Remark: The solution of the Euler equation of the form

2

0 0( ) ( ) 0x x y x x y y (4)

is the same as for DE(1) but we replace x by 0( )x x . Also, we can

transform DE(4) to the form of DE(1) by putting 0t x x

123

Example 4(Q.10 Page 278): Solve the DE

2( 2) 5( 2) 8 0x y x y y .

Transform the Euler equation to an equation with constant coefficients

Let , ln , 0.zx e z x x

1

zdy dy dx dy dye x

dz dx dz dx dx

dy dy

dx x dz

2 2 2

2 2 2 2 2 2

1 1 1 1 1( ) ( )

d y d y dy d y dy

dx x dz x x dz x dz x dz

Now by substituting in DE(1), we get

124

22

2 2 2

2

2

1 1 10

0

d y dy dyx x y

x dz x dz x dz

d y dy dyy

dz dz dz

2

2( 1) 0

d y dyy

dz dz (4)

Note that DE(4) has a constant coefficients.

125

Example 5(Q.25 Page 279): Transform the following DE into a DE with

constant coefficients then solve it 2 3 4 lnx y xy y x .

Example 6(Q.28 Page 279): Transform the following DE into a DE with

constant coefficients then solve it 2 4 sin(ln ).x y xy y x .

126

Chapter 6

The Laplace Transformation

6.1 Definition of The Laplace Transformation

Definition:

An improper integral over an unbounded interval is defined a limit of integrals over finite intervals, thus

( ) lim ( )A

a aAf t dt f t dt

, (1)

where A is a positive real number.

If the integral from a to A exists for each A a

and if limit as A exists then the integral is said to converge to that limiting value. Otherwise the integral is said to diverge.

Example 1: 0

cte dt

127

Example 2: 1

dt

t

Example 3: 0

, 1pt dt p

128

Definition:

Let ( )f t be given for 0t and suppose that ( )f t satisfies certain conditions. Then the Laplace transformation of ( )f t is denoted by

{ ( )}L f t or ( )F s is defined by 0

{ ( )} ( ) ( )stL f t F s e f t dt

(2)

whenever the improper integral converges.

Example 4: Find the Laplace transformation of the function ( ) 1, 0.f t t

Example 5: Find the Laplace transformation of the function ( ) , 0.atf t e t

129

Example 6: Find the Laplace transformation of the function ( ) sin , 0.f t at t

Remark:

The Laplace transformation is linear operator because if 1 2( ), ( )f t f t are

two functions whose Laplace transformation exist for 1s a and 2s a

Respectively, then for 1 2max{ , }s a a

1 1 2 2 1 1 2 2{ ( ) ( )} { ( )} { ( )}L c f t c f t c L f t c L f t

130

Example 7: Find the Laplace transformation of the function 2( ) 5 3sin 4 , 0.tf t e t t

Example 8(Q. 7, Page 313): Find the Laplace transformation of the function ( ) coshf t bt .

131

132

6.2 Solution of The Initial Value Problems

In this section, we use the Laplace transformation in solving initial value problems.

Theorem 6.2.1:

Suppose that f is continuous and f is piecewise continuous on any interval 0 t A .

Suppose further that there exist constants ,K a and M such that | ( ) | atf t Ke for t M . Then { ( )}L f t exists for s a and moreover

{ ( )} { ( )} (0)L f t sL f t f .

Theorem 6.2.2:

Suppose that the functions ( 1), , , nf f f are continuous and ( )nf is piecewise continuous on any interval 0 t A . Suppose further that

there exist constants ,K a and M such that ( 1)| ( ) | ,| ( ) | , ,| ( ) |at at n atf t Ke f t Ke f t Ke

for t M . Then ( ){ ( )}nL f t exists for s a and moreover

( ) 1 2 ( 2) ( 1){ ( )} { ( )} (0) (0) (0) (0).n n n n n nL f t s L f t s f s f sf f

133

Example 1: Use Laplace transformation to solve the IVP

2 0, (0) 1, (0) 0y y y y y


sin 2 , (0) 2, (0) 1y y t y y

134


(4) 0, (0) 0, (0) 1, (0) 0, (0) 0y y y y y y

Example 4(Q. 9, Page 322):

Find the inverse Laplace transform of the function 2

1 2

4 5

s

s s

.

135

Example 5(Q. 16, Page 322): Use Laplace transformation to solve the IVP

2 5 0, (0) 2, (0) 1y y y y y

Example 6(Q. 23, Page 322): Use Laplace transformation to solve the IVP

2 4 , (0) 2, (0) 1ty y y e y y

136

137

138

Islamic University of Gaza Differential Equations

Department of Math. Midterm exam

Second semester 22/4/2014 Time: one hour

Solve the following questions:

Q1) (a) [6Marks] Solve the following initial value problem

2

cos, (0) 1

1 2

y xy y

y

.

(b) [8 Marks] Solve the following differential equation

(tan ) sin 2 , 02

y x y x x x

.

Q2) (a) [6 Marks] Determine the interval in which the solution of the given initial value problem exists and unique

cos( 2) 3 2 , ( 1) 1

xx y x y y y

x .

(b) [10Marks] Solve the differential equation

2 2( 1) ( 2 ) 0, 0x y dx x x y dy x .

Q3)(a) [6 Marks] Find a differential equation whose general solution is

2 2

1 2cos3 sin3t ty c e t c e t .

(b) [10 Marks] If 2

1y x is one solution of the differential equation

2 2 2 0, 0x y xy y x ,

use the method of reduction of order to find a second linearly independent solution of the differential equation.

Q4) [14 Marks] Find the general solution of the differential equation

2

4 4 , 0xe

y y y xx

.

139



Summer semester 12/8/2009 Time: one hour


Q1) Find the general solution of the differential equation

2 2 2 3costy y y e t .

Q2) Solve the following differential equations:

a) 2( ) 2y ye y e y .

b)2 2 2( 3 ) 0x dy x xy y dx .

Q3) a) Solve the initial value problem

2 3(1 2 ) 0, ( 2) 1xy dy y dx y .

b) If the wronskian of the functions f and g is 43 te and if 2( ) tf t e , find ( )g t .

Q4) Use the method of reduction of the order to find a second solution of the

differential equation

2 3 4 0, 0x y xy y x , if 2

1( )y x x .

140



Summer semester 4/8/2010 Time: 75 min.


Q1) (a) [5 Marks] Solve the following differential equation

22 ty y e .

(b)[7 Marks] Solve the following differential equation

4 3

2

dy x y

dx x y

.

Q2) (a) [6 Marks] Solve the following differential equation

2 32 2 , 0x y y x y x .

(b) [6 Marks] Solve the initial value problem

3 2(2 1) 0, (0) 1y dx xy dy y

Q3) (a) [5 Marks] Suppose that 3

1( )y x x is one solution of the differential equation

2 5 9 0, 0x y x y y x .

Find a second linearly independent solution of the differential equation.

(b) [3 Marks] Suppose that 1y and 2y are two linearly independent solutions of the

differential equation

( ) ( ) 0y p t y q t y ,

and if 2

1 2( , )( ) tW y y t ct e . Find ( )p t .

Q4) [8 Marks] Use the method of variation of parameters to solve the differential equation

25 6 ty y y te .

141


Department of Math. Final exam

Second semester 22/5/2014 Time: Two hours

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Total االسم

60 12 8 8 8 8 8 8 الرقم


Q1) (a) [4 marks] Find the general solution of the differential equation

2( ) 0yy y .

(b) [4 marks] Find a function ( )f x such that the differential equation

cos (cos ) ( ) 0y x x f x y is an exact and (0) 0f .

Q2) (a)[4 marks] Solve the initial value problems

2 2 30, ( ) 0y x y xy y e .

(b) [4 marks] Solve the differential equation

3(4 8 ) 0y x y y .

Q3) Consider the differential equation

2 35 2 24 cos2x xy y y y e xe x

(a)[5 marks] Find the complementary solution, cy .

(b) [3 marks] Find a formula for the particular solution, py .

(Don't evaluate the coefficients).

142

Q4) [8 marks] Find a series solution in powers of ( 2)x for the following

differential equation (Include at least three nonzero terms in each

series) 0y xy y .

Q5)(a)[4 marks] Solve the following initial value problem

2 2 6 0, (1) 0, (1) 2x y x y y y y .

(b)[4 marks] Classify the singular points of the following differential equation

as regular or irregular

4 3( ) ( 1) 0x x y x y y .

Q6)[8 marks] Use the Laplace transformation to solve the initial value

problem

2 , (0) 1, (0) 2ty y y te y y .

Q7)(a)[ 4 marks] Is the following initial value problem has a unique solution.

Explain your answer.

23 2, (1) 1

1

xy y

y

.

(b)[4 marks] Are the two functions 3

1y x , 4

2y x form a fundamental set

of solutions of the differential equation Explain your answer.

2 6 12 0, 0x y xy y x .

(c)[4marks] Suppose ,f g and h are differential functions such that (2) 3f

and ( , )(2) 4W g h . Find ( , )(2)W f g f h .

143



Summer semester 27/8/2013 Time: Two hours

Q1 Q2 Q3 Q4 Q5 Q6 Total االسم

100 18 16 16 16 18 16 الرقم


Q1) (a) [6 Marks] Determine the interval in which the solution of the given

initial value problem exists (tan ) sin , ( ) 0y t y t y .

b) [12 Marks] Suppose that 2

1( )y x x is one solution of the differential

equation

2 2 0x y y .

Find a second linearly independent solution of the differential equation.

Q2) (a) [10 Marks] Find the general solution of the differential equation

2 24 4 , 0ty y y t e t .

(b) [6 Marks]Solve the following differential equation

2 23

2

dy y x

dx xy

.

Q3) a) [10 Marks] Consider the differential equation

(4) 3 4 ty y t e .

(i) [5 Marks] Find the solution of the corresponding homogeneous equation.

(ii) [5 Marks] Use the method of undetermined coefficients to find a suitable

form of the particular solution without evaluating the constants.

144

(b) [6 Marks] Solve the differential equation

2 4 4 0x y xy y .

Q4) [16 Marks] Find two linearly independent power series solutions at 0 0x

to the differential equation (include at least three nonzero terms in each

series)

2 0y xy y .

Q5) (a) [6 Marks] Determine a lower bound for the radius of convergence of

the series solution for the given differential equation about 0 0x

2(1 ) 3 0x y xy y .

(b) [10 Marks] Determine the singular points of the differential equation

22 ( 2) 3 ( 2) 0x x y xy x y ,

and classify them as regular or irregular .

Q6) (a) [6 Marks] Use the definition to find the Laplace transformation of the

function

2( ) tf t t e .

(b) [12 Marks] Use the Laplace transform to solve the initial value problem

2 2 cos , (0) 1, (0) 0y y y t y y

145



Summer semester 1/9/2010 Time: Two hours

Q1 Q2 Q3 Q4 Q5 Total االسم

60 12 10 11 12 15 الرقم


Q1) (a) [6Marks] Solve the following differential equation

2( ) 2 yy y e .

(b) [5 Marks] Solve the following initial value problem

sin2 , (2) 1

dy xx y y

dx x .

(c) [4 Marks] Determine the interval in which the solution of the given initial

value problem exists

( 4) 3 4 ln , (3) 1, (3) 1x x y x y y x y y .

Q2)Find the general solution of the following differential equations

(a) [6 Marks] (4) sin 2y y t .

(b) [6 Marks] 2 24 4 , 0ty y y t e t .

Q3)(a) [6 Marks] Solve the following differential equation

23( 1) 6( 1) 6 6ln( 1)x y x y y x .

(b) [5Marks] Classify the singular points of the following differential equation

as regular or irregular 2( 2 3) 4 0x x y xy y .

146

Q4)[10 Marks] Find two linearly independent power series solutions at 0 1x

to the differential equation (include at least three nonzero terms in each

series)

0y xy y .

Q5) (a) [5 Marks] Use the definition to find of the Laplace transformation of

the function

( ) cos2 , 0f t t t .

(b) [7 Marks] Use the Laplace transform to solve the initial value problem

2 2 cos , (0) 1, (0) 0y y y t y y .

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