diff equation 4 2011 fall high order theory

8/3/2019 Diff Equation 4 2011 Fall HIGH Order Theory

http://slidepdf.com/reader/full/diff-equation-4-2011-fall-high-order-theory 1/56

S.l.dr.ing.mat. Alina BogoiDifferential

Equations

POLITEHNICA University of BucharestFaculty of Aerospace Engineering

CHAPTER 2

First-Order Differential

Equations (cont’d)



Diff_Eq_4_2011 2

Outline• Exact equations

• Integrating Factors

• Homogenous equations

• First Order Ordinary Differential equation

• First Order Non-linear Differential equation

– Bernoulli equation

– Riccati equation

– Clairaut equation



Diff_Eq_4_2011 3

A first-order DE of the form a1

( x)(dy/dx) +

a0

( x) y = g ( x) (1)

is said to be a linear equation in y.When g ( x) = 0, (1) is said to be homogeneous;

otherwise it is nonhomogeneous.

DEFINITIONLinear Equations

First Order Ordinary Differential

equation



Diff_Eq_4_2011 4

• we have the general solution:

dx x f eece y y ydx x P dx x P dx x P

pc ∫ ∫ ∫ +∫ =+=−−

)()()()(

• We call is an integrating factor and we should only memorize

this to solve problems.

∫ dx x P

e)(

Integrating Factor



Diff_Eq_4_2011 5

BERNOULLI’S EQUATION

The differential equation

where n is any real number, is called Bernoulli’sequation. For n = 0 and n = 1, the equation is

linear and can be solved by methods of the lastsection.

,)()( n y x f y x P dx

dy=+



Diff_Eq_4_2011 6

SOLVING A BERNOULLI

EQUATION

1. Rewrite the equation as

2. Use the substitution w = y1 − n, n ≠ 0, n ≠ 1. NOTE: dw/dx = (1 − n) y−n

dy/dx.

3. This substitution turns the equation in Step 1into a linear equation which can be solved by

the method of the last section.

).()(

1

x f y x P dx

dy

y

nn

=+

−−



Diff_Eq_4_2011 7

How to transform the Riccati equation to a linear one ?

Somehow we get one solution, , of a Riccati

equation, then the change of variables

transforms the Riccati equation to a linear one.

)()()(2'

x R y xQ y x P y ++=

)( x

z xS y

1)( +=

RICATTI’S EQUATION



Diff_Eq_4_2011 8

z xQ

z xS x P

z x P z'

z

1)(

1)()(2

1)(

122

++=−

)())()()(2( x P z xQ xS x P z' −=++

[2 ( ) ( ) ( )]

i ntegrating factor

( )P x S x Q x dx

f x e+∫ =

)(dx)()(

)(

1)(

x f

C x f x P

x f x z +−= ∫

Riccati Equation



Diff_Eq_4_2011 9

CLAIRAUT’S EQUATION

The nonlinear differential equation y = xy′ + F( y′)

is called Clairaut’s equation. Its solution is thefamily of straight lines y = cx + F (c), where c is

an arbitrary constant.

dx

dy p = ( ) y x p F p= +



Diff_Eq_4_2011 10

PARAMETRIC SOLUTION TO

CLAIRAUT’S EQUATION

2

1 22

1 1 2 1 1

2 1 1 1

( ) 0

0

put into the origional eq. ( )

( ) ( ) general solution

dy dp dF dp dp dF p p x xdx dx dp dx dx dp

dp d y y c x c

dx dxdy

p c c x c c x F cdx

c F c y c x F c

= = + + ⇒ + =

= = ⇒ = +

⇒ = = ⇒ + = +

⇒ = ⇒ = +

0 ( , ) 0 eliminate in the origional ODEdF

x G x p pdp

+ = ⇒ =



Diff_Eq_4_2011 11

Clairaut’s equation

)( p F px y+

2

2 2

2 2

22

Ex:

2 ( 2 ) 0

(1) ( ) ( ( )) general solution

(2) 2 0

2 2 4

4 0 singular solution4

y px p

dy dp dp dp p x p p x p

dx dx dx dx

F p p y x F c cx c

x x x x p p y

x

y x y

= +

⇒ = + + = ⇒ + =

= ⇒ = + = +

+ = ⇒ = − ⇒ = − +

= − ⇒ + =

'( ) 0 x F p+ =( ) y c x F c= +



Diff_Eq_4_2011 12

Non-linear 2nd O.D.E.

• p substitution.

•

the second derivative of y is replaced by the first derivative

of p thus eliminating y

• completely and producing a first O.D.E. in p and x.

dx

dy p =

2

2

dx

yd

dx

dp

=

A. Equations where the dependent variables does

not occur explicitly



Diff_Eq_4_2011 13

Solve axdx

dy xdx

yd =+2

2

Let dx

dy p =

ax xpdx

dp=+

⎟ ⎠

⎞⎜⎝

⎛ 2

2

1exp xintegral factor

Example



Diff_Eq_4_2011 14

Solve axdx

dy xdx

yd =+2

2

Let dx

dy p = 2

2

dx

yd

dx

dp

=and therefore

ax xpdx

dp=+

⎟ ⎠

⎞⎜⎝

⎛ 2

2

1exp xintegral factor

222

2

1

2

1

2

1 x x x

axe xpeedx

dp=+

22

2

1

2

1

)(

x x

axe pedx

d =

21exp2

ax C x dx⎛ ⎞= + −⎜ ⎟⎝ ⎠∫

Example

dx

dy xC a p =−+= )

2

1exp( 2



Diff_Eq_4_2011 15

They are solved by differentiation followed by the p

substitution.

When the p substitution is made in this case, the second

derivative of y is replaced as

Let

dx

dy p =

dy

dp p

dx

dy

dy

dp

dx

dp

dx

yd ===∴

2

2

Non-linear 2nd O.D.E.

B. Equations where the independent variables doesnot occur explicitly



Diff_Eq_4_2011 16

Solve 2

2

2

)(1dx

dy

dx

yd y =+

Let dx

dy p = and therefore

21 p

dy

dp yp =+

Separating the variables

dy

dp

pdx

yd

=2

2

dy y

dp p

p 1

12=

−

)1ln(2

1lnln 2 −=+ pa y

)1( 22 +== yadx

dy p

( )

)sinh(1

)(sinh1

)1(

1

22

caxa

y

baya

x

ya

dy x

+=

+=

+=

−

∫

Example



S.l.dr.ing.mat. Alina

Bogoi

Differential

Equations

POLITEHNICA University of BucharestFaculty of Aerospace Engineering

CHAPTER 3

Higher Order Linear

Equations



Diff_Eq_4_2011 18

Outline• Nth-order Linear Homogeneous

Equations with ConstantCoefficients

• Nth-order Linear Nonhomogeneous

Equations:

• Method of Undetermined

Coefficients• The method of variation of

parameters

• Cauchy-Euler Equation

Higher Order Linear Equations



Diff_Eq_4_2011 19

A set of f 1

( x), f 2

( x), …, f n

( x) is linearly dependent on

an interval I , if there exists constants c1 , c2 , …, cn , not all zero, such that c1

f 1

( x) +

c2

f 2

( x) + …

+

cn

f n

( x) = 0

If not linearly dependent, it is linearly independent.

DEFINITION 1

Linear Dependence and Linear Independence


Homogeneous Equations with Constant Coefficients




Diff_Eq_4_2011 20

Suppose each of the functions f 1 ( x), f

2 ( x), …, f

n ( x)

possesses at least n – 1 derivatives. The determinant

is called the Wronskian of the functions.

DEFINITION 2Wronskian

)1()1()1(

21

21

1

21

'''),...,(

−−−

=

nn

nn

n

n

n

f f f

f f f

f f f

f f W

L

MMM

L

L

Higher Order Linear EquationsHomogeneous Equations with Constant Coefficients




Diff_Eq_4_2011 21

Let y1

( x), y2

( x), …, yn

( x) on an interval I . This set of

functions is linearly independent if and only if

W ( y1

, y2

, …, yn

) ≠ 0 for every x in the interval.

THEOREM 3Criterion for Linear Independence

Any set y1

( x), y2

( x), …, yn

( x) of n linearly independent

solutions is said to be a fundamental set of functions.

DEFINITION 3

Fundamental Set





Diff_Eq_4_2011 22

• Ini t ial-value Problem An nth-order initial problem is

Solve:

• Subject to n initial conditions :

1

1 1 01[ ] ( ) ( ) ( ) ( ) ( )

n n

n nn n

d y d dy L y a x a x a x a x y g x

dx dx dx

−

− −= + + + + =L

0 0

'

0 0

( 1) 1

0 0

( )( ) , ,

( )

n n

y x y y x y

y x y

− −

=′ =

=

L





Diff_Eq_4_2011 23

• The following DE

is said to be homogeneous;

• with g(x) not identically zero, isnonhomogeneous .

( )1

1 1 01( ) ( ) ( ) ( ) 0

n n

n nn n

d y d y dy L y a x a x a x a x y

dx dx dx

−

− −= + + + + =L

( )1

1 1 01( ) ( ) ( ) ( ) ( )

n n

n nn nd y d y dy L y a x a x a x a x y g xdx dx dx

−

− −= + + + + =L


L( y ) = g( x )

L( y ) = 0




Diff_Eq_4_2011 24

1

1 1 01[ ] ( ) ( ) ( ) ( ) ( )

n n

n nn nd y d dy L y a x a x a x a x y g xdx dx dx

−

− −= + + + + =L

10)1(

1000 )(,,)(,)( −− ==′= n

n y x y y x y y x y L


Let an

( x), an-1

( x), …, a0

( x), and g ( x) be continuous on I,

an

( x) ≠

0

for all x on I . If x = x0

is any point in this

interval, then a solution y( x) exists on the intervaland is unique.

THEOREM 1Existence and Uniqueness




Diff_Eq_4_2011 25

There exists a fundamental set of solutions for DE on an

interval I.

THEOREMExistence of a Fundamental Set


(A) y = cy1

is also a solution if y1

is a solution.

(B) A homogeneous linear DE always possesses thetrivial solution y = 0.

COROLLARY Corollary




Diff_Eq_4_2011 26

Let y1

, y2

, …, yk

be a solutions of the homogeneous

nth-order differential equation on an interval I .

Then the linear combination y = c1

y1

( x) + c2

y2

( x) + …+ ck

yk

( x)

where the ci , i = 1, 2, …, k are arbitrary constants, isalso a solution on the interval.

THEOREM

Superposition Principles – Homogeneous Equations





Diff_Eq_4_2011 27

Let y1

( x),

y2

( x), …,

yn

( x) be a fundamental set of

solutions of homogeneous DE on an interval I . Thenthe general solution is y

= c1

y1

( x) +

c2

y2

( x) + …

+

cn

yn

( x)

where ci

are arbitrary constants.

THEOREM

General Solution – Homogeneous Equations


E l



Diff_Eq_4_2011 28

• The functions y1 = e3x, y2 = e-3x aresolutions of

y″ – 9y = 0 on (-∞

,∞

)Now

for every x.So y = c1e3x + c2e

-3x is the general

solution.

Example

0633),( 33

3333

≠−=−= −

−−

x x

x x x x

ee

ee

eeW

E l



Diff_Eq_4_2011 29

• The functions y1 = ex, y2 = e2x , y3 = e3x

are solutions of y″′ – 6y″ + 11y′ – 6y = 0

on (-∞

,∞

).Since

for every real value of x.

So y = c1ex + c2 e2x + c3e

3x is the general

solution on (-∞, ∞).

Example

02

94

32),,( 6

32

32

32

32 ≠== x

x x x

x x x

x x x

x x x e

eee

eee

eee

eeeW




Diff_Eq_4_2011 30

• If y1,…, yn are solutions to ODE, then so

is linear combination

• Every solution can be expressed in thisform, with coefficients determined by

initial conditions, iff we can solve:

1 1 2 2( ) ( ) ( ) ( )n n y x c y x c y x c y x= + + +L

1 1 0 0 0

1 1 0 0 0

( 1) ( 1) ( 1)

1 1 0 0 0

( ) ( )( ) ( )

( ) ( )

n n

n n

n n n

n n

c y x c y x yc y x c y x y

c y x c y x y− − −

+ + =′ ′ ′+ + =

+ + =

LL

M

L






Diff_Eq_4_2011 31

( )( )

1 0 2 0 0

1 0 2 0 0

1 2 0

( 1) ( 1) ( 1)

1 0 2 0 0

( ) ( ) ( )

( ) ( ) ( )

, , ,

( ) ( ) ( )

n

n

n

n n n

n

y x y x y x

y x y x y x

W y y y x

x y x y x− − −

′ ′ ′

=

L

L

K M M O M

L



• The system of equations on the previous slide has aunique solution iff its determinant, or Wronskian, is

nonzero at x0:

COROLLARYCorollar




Diff_Eq_4_2011 32

•

Solution is:

• The Wronskian of y1 and y2 is

• Since W ≠ 0 for all t, linear combinationsof y1 and y2 can be used to construct

solutions of the IVP for any initial value

t .

1 2

1 2

,

( )

t t

t t

y e y e

y t c e c e

−

−= =

= +

22 0

2121

21

21 −=−=−−=′−′=

′′

= −− eeeee y y y y

y y

y yW t t t t

( ) ( )

0,

0 3, 0 1

y y

y y

′′ − =

′= =EXAMPLE

SOLUTION

g e O de ea quat o s

SOLUTIONS TO HOMOGENEOUS



Diff_Eq_4_2011 33

SOLUTIONS TO HOMOGENEOUS

LINEAR EQUATIONS WITHCONSTANT COEFFICIENTS

All solutions of the differential equation

are, or are constructed from, exponentialfunctions of the form

y = ert .

NOTE: an

, an−1

, . . . , a1

, a0

are constants.

1

1 1 010

n n

n nn n

d y d y dya a a a y

dt dt dt

−

− −+ + + + =L




Diff_Eq_4_2011 34

• Consider the nth order linear homogeneous differentialequation with constant, real coefficients:

• By the fundamental theorem of algebra, a polynomial of degree n has n roots r1, r2, …, rn, and hence

[ ] 01

)1(

1

)(

0 =+′+++= −− ya ya ya ya y L nn

nn L

0)( polynomialsticcharacteri

1

1

10 =++++= −

−

4 4 4 4 4 34 4 4 4 4 21 Lr Z

nn

nnrt rt

ar ar ar aee L

)())(()( 210 nr r r r r r ar Z −−−= L

g q





Diff_Eq_4_2011 35

• If roots of Z(r) are real and unequal, then

there are n distinct solutions:

• Then general solution of differential

equation is

• The Wronskian can be used to determine

linear independence of solutions.

t r t r t r neee ,,, 21 K

t r

n

t r t r nececect y +++= K21

21

)(

g q

ADistinct Roots




Diff_Eq_4_2011 36

• Assuming exponential solution leads to

characteristic equation:

• Thus the general solution is

( )( )( )( ) 04321

02414132)( 234

=+−+−⇔

=+−−+⇒=

r r r r

r r r r et y rt

2 3 4

1 2 3 4( ) t t t t y t c e c e c e c e− −= + + +

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

02414132)4(

−=′′′=′′−=′=

=+′−′′−′′′+

y y y y

y y y y y

g q

EXAMPLE

SOLUTION

•Consider the initial value problem




Diff_Eq_4_2011 37

• The initial conditions

yield

• Solving,

• Hence

164278

01694

1432

1

4321

4321

4321

4321

−=−+−

=+++

−=−+−

=+++

cccc

cccc

cccc

cccc

1)0(,0)0(,1)0(,1)0( −=′′′=′′−=′= y y y y

t t t t

ecececect y4

3

3

3

2

21)(−−

+++=

7

1,70

11,5

4,2

14321 −=−=== cccc

t t t t eeeet y 432

71

7011

54

21)( −− −−+=

g q

SOLUTION

(Cont’d)




Diff_Eq_4_2011 38

Complex Roots• If Z(r) has complex roots, then they must

occur in conjugate pairs, λ ±

iμ .

• Solutions corresponding to complex roots have

the form

• We use the real-valued solutions

( )

( ) t iet ee

t iet eet t t i

t t t i

μ μ

μ μ λ λ μ λ

λ λ μ λ

sincos

sincos

−=+=

−

+

t et e

t t

μ μ

λ λ

sin,cos

B




Diff_Eq_4_2011 39

• Consider the equation

• Then

• Now


( )( ) 01101)(3

=++−⇔=−⇒= r r r r et yrt

0=−′′′ y y

2

1,2

1 1 4 1 3 1 31 0

2 2 2 2

ir r r i

− ± − − ±+ + = ⇔ = = = − ±

2/3sin2/3cos)( 2/3

2/21 t ect ecect y t t t −− ++=

EXAMPLE

SOLUTION




Diff_Eq_4_2011 40

• Consider the initial value problem

• The roots are 1, -1, i, -i. Thus the generalsolution is

• Using the initial conditions, we obtain

• The graph of solution is given on right.

( )( ) 01101)( 224 =+−⇔=−⇒= r r r et y rt

2)0(,2/5)0(,4)0(,2/7)0(,0)4( −=′′′=′′−=′==− y y y y y y

( ) ( )t ct cecect y t t sincos)(4321

+++= −

( ) ( )t t eet y t t sincos2

130)( −++= −

EXAMPLE

SOLUTION




Diff_Eq_4_2011 41

Repeated Roots• Suppose a real root rk of characteristic polynomial Z(r) is

a repeated root with multiplicty s.

• Then linearly independent solutions corresponding to this

repeated root have the form

t r st r t r t r k k k k

et et tee

12

,,,,

−

K

C




Diff_Eq_4_2011 42

Repeated Roots• If a complex root λ + iμ is repeated s times, then so is its

conjugate λ - iμ .

• There are 2s corresponding linearly independent solns,

derived from real and imaginary parts of

or

1 1

cos , sin ,cos , sin , ,

cos , sin ,k k

t t

t t

r t r t s s t

e t e t te t te t

t e t t e e t

λ λ

λ λ

λ

μ μ μ μ

μ μ

− −

K

( ) ( ) ( ) ( )t iu st iut iut iu

et et tee+−+++ λ λ λ λ 12

,,,, K

C




Diff_Eq_4_2011 43

Example 4: Repeated Roots

• Consider the equation

• Then

• The roots are 2i, 2i, -2i, -2i.


( )( ) 0440168)( 224 =++⇔=++⇒= r r r r et y rt

0168)4( =+′′+ y y y

( ) ( )t t ct t ct ct ct y 2sin2cos2sin2cos)( 4321 +++=

EXAMPLE

SOLUTION

Second Order Differential Equations



Diff_Eq_4_2011 44

In general, it is not easy to discover particular

solutions to a second-order linear equation.

But it is always possible to do so if the

coefficient functions P , Q and R are constant

functions, that is, if the differential equation hasthe form:

0)()()(2

2

=++ y x R

dx

dy xQ

dx

yd x P

0ay by cy′′ ′+ + =

SOLUTIONS TO THE



Diff_Eq_4_2011 45

SOLUTIONS TO THE

HOMOGENEOUS LINEAR 2ND

-ORDER DE

The solutions to the homogeneous linear second-

order differential equation depend on the solutionsto the characteristic equation. The solutions to the

characteristic equation fall into three cases:• Distinct (unequal) real roots

• Repeated (equal) real roots

•

Conjugate complex roots

THE CHARACTERISTIC



Diff_Eq_4_2011 46

THE CHARACTERISTIC

EQUATION

Substituting y = emx

results in the auxiliary

equation or

characteristic equation:

am2

+ bm + c = 0.

From eq am2 + bm + c = 0 the two



Diff_Eq_4_2011 47

From eq. am2 + bm + c = 0 the two

roots are

(1)b2 – 4ac > 0: two distinct realnumbers.

(2)b

2

– 4ac = 0: two equal realnumbers.

(3)b2 – 4ac < 0: two conjugate complex

numbers.

21 ( 4 ) / 2r b b ac a= − + −

2

2 ( 4 ) / 2r b b ac a= − − −



Diff_Eq_4_2011 48

If the characteristic equation has two distinct

(unequal) real roots m1

and m2

, the two solutionsto the DE are .

This results in the general solution:

DISTINCT REAL ROOTS

xm xm e ye y 21

21 and ==

xm xm ecec y 21

21 +=



Diff_Eq_4_2011 49

When the roots of the characteristic equation are

equal (that is, m1

= m2

), then the two solutions are

The general solution is:

REPEATED REAL ROOTS

.and 11

21

xm xm xe ye y ==

.11

21

xm xm

xecec y +=



Diff_Eq_4_2011 50

CONJUGATE COMPLEX ROOTSIf the roots of the characteristic equation are

conjugate complex roots

m1

= α

+ i β

and m1

= α

−

i β ,

then the two solutions are

y1 = e

α x

cos β x and y2 = e

α x

sin β x,and the general solution is

( ) ( ) ( )α α β β = +1 2e sin e cosx x

y x C x C x




Diff_Eq_4_2011 51

Example 0y y′′ − = CE2 1 0r − = 1 or 1.r r⇒ = = −

1 2e ex xy C C −= +Solution

Example 2 0y y y′′ ′− + = CE2 2 1 0r r− + = 1 (double root).r⇒ =

1 2e ex xy C C x= +Solution

Example 2 5 0y y y′′ ′+ + = CE2 2 5 0r r+ + =

1 2 1 1 2r i⇒ = − ± − = − ±

( ) ( )1 2e sin 2 e cos 2x xy C x C x− −= +

Solution

I i i l l d b d l bl




Diff_Eq_4_2011 52

An initial-value problem for the second-order Equation

consists of finding a solution

y

of the differential equation

that also satisfies initial conditions of the form

where and are given constants. If P, Q, R, and G arecontinuous on an interval and there, then a theorem found in more advanced books guarantees the

existence and uniqueness of a solution to this initial-value problem.

1000 )()( y x y y x y =′=

0 y 1 y0)( ≠ x P

Initial-value and boundary-value problems




Diff_Eq_4_2011 53

A boundary-value problem for Equation 1 consists of finding a solution of the differential equation that also

satisfies boundary conditions of the form

In contrast with the situation for initial-value problems, a

boundary-value problem does not always have asolution.

1100 )()( y x y y x y ==



Diff_Eq_4_2011 54

• Solve the following DEs:

(a)

(b)

(c)

Example 1

03'5"2 =−− y y y x x ecec y 3

2

2

1 += −

025'10" =+− y y y

x x xecec y 52

51 +=

07'4" =++ y y y

2 1 2( cos 3 sin 3 ) x y e c x c x−

= +



Diff_Eq_4_2011 55

Example 2Solve

Solut ion:

2)0(',1)0(,017'4"4 =−==++ y y y y y

)2sin432cos(2/ x xe y x+−= −

diff equation 4 2011 fall high order theory

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