Chapter 4-1 Merged

7/29/2019 Chapter 4-1 Merged

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4.1 Preliminary Theory Homogeneous LinearEquations

Homogeneous linearnth-order DE

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

n nn na x y a x y a x y a x y a x y

+ + + + + =

Non-Homogeneous linearnth-order DE

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

n nn na x y a x y a x y a x y a x y g x

+ + + + + =


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Linear Dependence / Linear Independence

Definition 1

A set of functions f1(x), f2(x), , fn(x) is said to be

linearly dependent on an interval I if there exist

1,

2, ..,

n,

for every x in the interval.

In other words, the set of functions is said to be

linearly independent if

1 1 2 2( ) ( ) ... ( ) 0n nc f x c f x c f x+ + + =

1 2 ... 0nc c c= = = =


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Linear Dependence / Linear Independence

Definition 2

A set of two functions f1(x) and f2(x) is said to be

linearly dependent on an interval I when one

.

In other words, the set of functions is said to belinearly independent if

1

2

( )constant

( )

f x

f x=

1

2

( )constant

( )

f x

f x


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Example 1

Determine whether the functions are linear

independent on the interval (-,):2 3

1 2 3

2 2

1) ( ) , ( ) , ( )

f x x f x x f x x= = =

= = =

1 2

, ,

3) ( ) 2 , ( ) 2f x x f x x= + = +


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Wronskian

Suppose each of the functions f1(x), f2(x), , fn(x)

possesses at leastn

-1 derivatives. The determinant

21

...

...

nfff

where the primes denote derivatives, is called the

Wronskian of the functions.

)1()1(2

)1(1

21

...

::::,...,,

=

nn

nn

n

fff


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Linear Independence Using Wronskian

Theorem

Let y1, y2,.,yn be n solutions of the homogeneouslinear nth-order differential equation on an interval I.

Then the solutions are linearl inde endent on I if

and only if :

for every x in the interval

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


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Fundamental set of solutions

Definition

Any set y1, y2,.,yn of n linearly independent solutions

of the homogeneous linear nth-order differential

of solutions on the interval.


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General Solution

Theorem

Let y1, y2,.,yn be a fundamental set of solutions of the

homogeneous linear nth-order differential equation on

an interva.

en t e genera so ution o t e equation

on the interval is

where ci , i = 1,2,..,n are arbitrary constants.

1 1 2 2( ) ( ) ... ( )n nc y x c y x c y x= + + +


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Example 2

Verify that the given functions form a fundamental set

of solutions of the DE on the indicated interval. Form

the general solution.

3 4'' ' x x , , ,


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Solution for Example 2

Step 1:

Verify that the given functions are solutions for the DE

3 4'' ' 12 0; , , ( , )x xy y y e e =

3 3 31 1 1

3 3 3

Let: 3 9

Substitute into the DE: '' ' 12 0

x x x

x x x

y e y e y e

y y y

= = =

=

42

0 0 (verified)

Let:

e e e

y e

=

=

=4 4

2 2

4 4 4

4 16

Substitute into the DE: '' ' 12 0

16 (4 ) 12( ) 0

0 0 (verified)

x x x

x x x

y e y e

y y y

e e e

= =

=

=

=


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Solution for Example 2

Step 2 :

Verify that the solutions are linearly independent on I

3 4'' ' 12 0; , , ( , )x xy y y e e =

3 43 4

3 4( , )

3 4

x xx x

x x

e eW e e

e e

=

3 4

7 0

and are linearly independent on ( , )

x

x x

e e

e

e e

=

=

Therefore e-3x and e4x form a fundamental set of

solutions. Hence the general solution is:

y = c1y1 + c2y2 = c1e-3x + c2e4x


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Example 3

Verify that the given functions form a fundamental set

of solutions of the DE on the indicated interval. Form

the general solution.

2 '' ' , , ,


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4.2 Reduction of Order

Suppose y1(x) is a solution of the homogeneous linear

second-order DE:

2 1 0( ) ( ) ( ) 0a x y a x y a x y + + =

We can construct a second solution y2(x) by the

method of reduction of order so that y1(x) and y2(x)

are linearly independent on I.


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Formula for Reduction of Order

Consider the homogeneous second order linear DE

If iven the first solution x , then b reduction of

( ) ( ) 0y p x y q x y + + =

Reference: A First Course in Differential Equations with Modelling Applications, (9th Ed) Dennis G. Zill, Brooks/Cole Publishing Company

order method:

( )

2 1 21

( ) ( )( )

p x dxe

y x y x dxy x

=


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Example 1

Given that y1(x) = e2x is a solution of the DE on (-,):

y 4y + 4y = 0

Use a formula of reduction oforder to find the second

solution y2(x). Hence, write the general solution of the

differential equation.


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Example 2

Given that y1(x) = sin 3x is a solution of the DE on (-,):

y + 9y = 0

Use a formula of reduction of order to find the second




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Example 3

Given that y1(x) = x4 is a solution of the DE on (0,):

x2y 7xy + 16y = 0

Use a formula of reduction oforder to find the second




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4.3 Homogeneous Linear DE with ConstantCoefficients

Consider the homogeneous linear second order DE :

where a, b, c are constant coefficients.

0 (1)ay by cy + + =

Try a solution for the DE of the form:

Calculate:

mxe=

2

,mx mx

me y m e = =


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Substitute y, y and y into the DE (1):

0 (1)ay by cy + + =

2

2( ) ( ) 0

( ) 0

mx mx mx

mxa m e b me cee am bm c

+ + =

+ + =

mx ,

This equation is called an auxiliary equation of the DE (1)

where m is the root of quadratic equation (2).

2 0 (2)am bm c+ + =


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In general, if given 2nd

order linear homogeneous DE:

The corresponding auxiliary equation for the DE (1) is

given by:2 0 (2)am bm c+ + =

0 (1)ay by cy + + =

Therefore is a solution of the DE (1) if and only

if m satisfies auxiliary equation (2).

mxe=


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The two roots of equation (2) can be obtained by

factoring method or by quadratic formula:

Roots of auxiliary equation

2 4b b ac

2 0 (2)am bm c+ + =

where the roots may be:

real and distinct roots

real but repeated roots

Complex conjugate roots

2m a=


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The two solutions are:

The general solution is:

Real and distinct roots : (m1 , m2)

1 21 2,

m x m xe y e= =

1 21 1 2 2 1 2

m x m xc y c y c e c e= + = +

Real and repeated roots : (m = m1 = m2)



1 2,mx mxe y xe= =

1 1 2 2 1 2mx mxc y c y c e c xe= + = +


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Complex conjugate roots : (m = ii)

1 2cos , sinx xe x y e x = =

1 1 2 2 1 2cos sinx xc y c y c e x c e x = + = +


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Example 1

Find the general solution of the DE:

) '' 4 ' 5 0a y y y+ + =

) '' 10 ' 25 0b y y y + =

''


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Higher Order linear homogeneous DE

Consider nth order linear homogeneous DE:

The corresponding auxiliary equation is:

( ) ( 1)

1 2 1 0..... 0n n

n na y a y a y a y a y

+ + + + + =

The general solution for the DE:

1 1 2 2 ..... n ny c y c y c y= + + +

1 2 1 0..... 0n na m a m a m a m a

+ + + + + =


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Example 2

Find the general solution of the DE:

) ''' 4 '' 5 ' 0a y y y =

(4)) 2 '' 0b y y y + =

=


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Assignment#1 (5%)

Due: 19 Nov 2012, Monday , 5.00pm at BA-3-71

Find the general solution of the homogeneous DE:

1) 81 0

2) 3 4 12 0

y y

y y y y

+ =

+ =

(5) (4)

(8) (6)

3) 5 36 0

4) 5 2 10 5 0

5) 0

y y y

y y y y y y

y y

=

+ + + =

+ =

www.mnoraini.weebly.com


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4.4 The Method of Undetermined Coefficients Superposition Approach

The general solution of nth order non-homogeneous

linear DE:

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

n nn na y a y a y a y g x

+ + + + =

is given by: ( ) ( ) ( )c px y x y x= +

where yc(x) is a complementary function of DE (1) that

can be obtained by solving the associated homogeneous( ) ( 1)

1 1 0... 0n n

n na y a y a y a y

+ + + + =


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and yp(x) is a particular solution of DE (1)

The particular solution yp(x) can be determined by: The method of undetermined coefficients

Superposition approach Annihilator approach

Variation of parameters


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To find the particular solution, yp for the DE:

can be used if and onl if x is either/combination of

The Method of Undetermined Coefficients

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


+ + + + =

these functions:

Constant, linear, polynomials, sine, cosine functions,

exponential functions.


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For DE:

Choose a trial x that is similar to the function x

The Method of Undetermined Coefficients

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


+ + + + =

and involving unknown coefficients to be determined

by substituting the choice of yp(x) into (1)


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List of trial yp(x) by Superposition Approach

Form of

1.

2.

3.

4.

)(xgPy

75 +x BAx +

23 2 x CBxAx ++2

13 +xx DCxBxAx +++23

constant)(any4

5.

6.

7.

8.

9.

10.

x4sin xBx 4sin4cos +

xBxA 4sin4cos +x4cos

xe

5 xAe5

xxexeBAx + )(

xe x 3cos2 xBexAe xx 3sin3cos 22 +

xx 4sin5 2 xFExDxxCBxAx 4sin)(4cos)( 22 +++++

4


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Important rules to determine the final yp(x) by

superposition approach

Rule 1:

If g(x) consists of m different functions,then1 2( ) ( ) ( ) .... ( )kg x g x g x g x= + + +

Rule 2:

No duplicate terms between yc(x) and yp(x).

If exist, yp(x) must be multiplied by xn where n is the

smallest positive integer that eliminate the duplication.

1 2....

kp p p px y x y x y x=


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Example 1

Solve the given non-homogeneous linear DE by

superposition approach:

y 10y + 25y = 30x + 3


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Solution Example 1

y 10y + 25y = 30x + 3

Step 1: Find yc(x):y 10y + 25y = 0

Roots: m1 = m2 = 5 (repeated)

yc(x) = c1e5x + c2xe5x


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Solution Example 1

y 10y + 25y = 30x + 3

Step 2: Find yp(x) using superposition approach

g(x) = 30x + 3Trial: yp(x) = Ax + B

c p ,

Final: yp(x) = Ax + B

Solve all constants in yp(x) :yp(x) = Ax + B , yp(x) = A , yp(x) = 0


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Solution Example 1 - continue

Substitute into the given DE:

y 10y + 25y = 30x + 3

0 10A + 25(Ax + B) = 30x + 3(-10A + 25B) + 25Ax = 30x + 3

-10A + 25B = 3 . (1)

25A = 30 .. (2)

Solve (1) & (2): A= 6/5, B = 3/5yp(x) = (6/5)x + 3/5


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Solution Example 1 - continue

General solution is:

y(x) = yc(x) + yp(x)

y(x) = c1e5x + c2xe5x + (6/5)x + 3/5


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Example 2

Determine the form of particular solution for the given

non-homogeneous linear DE by superposition

approach. (Do not solve the constants)

2 2 x x

4

2 4 2 xy y x xe

=

= +( )

)


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4.6 Variation of Parameters

Applicable for

Linear differential equations with constant or variable

coefficients.

inear i erentia equations w ere g x is any unctions.

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

n nn na x y a x y a x y a x y g x

+ + + + =


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Solving the non-homogeneous DE by variation of

parameters

( ) ( ) ( )p x y q x y f x + + =

Given a second order non-homogeneous linear differential

equation in the form:

Step 1: Find yc by solving :

1 1 2 2c c y c y= +

( ) ( ) 0y p x y q x y + + =


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1 2y y

Step 2 : Find yp by variation of parameters

2(a): Calculate the following Wronskians:


parameters

1 2

21

2

12

1

0

( )

0

( )

y yy

Wf x y

yW

f x

=

=


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11 1 1

Wu u u dxW

= =

Step 2 (b): find u1 and u2 to obtain the particular solution


parameters

2 2 2

1 1 2 2p

u u u dxW

y u y u y

= =

= +

Step 3:The general solution for the DE is:

c py y y= +


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Example 1

Solve the given non-homogeneous linear DE by

variation of parameters.

'' tany x+ =


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Example 1 - solution

'' tany x+ =

Step 1: Find yc by solving:

0y y + =

1 2sin cosc

m m

c x c x

+ = =

= +


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Example 1 - continue

'' tany x+ =

Step 2: Find yp by variation of parameters

Define:

1 2sin , cos , ( ) tanx y x f x x= = =

Calculate:

1 2 2 2

1 2

sin cossin cos 1

cos sin

y y x xW x x

y y x x

= = = =


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'' tany x+ =


Calculate:

0 0 cosx1

2

cos tan s n

( ) tan sin

x x x

f x y x x

= = = =

21

2 1

0 sin 0 sinsin tan

( ) cos tan cos

y x xW x x

y f x x x x= = = =


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'' tany x+ =


Find u1:

1 sinW x = = =

1

1sin cos

Wu x dx x

= =


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'' tany x+ =


Find u2:

2 2sin / cos sinW x x x2

2 2

2

2

1 cos

sin 1 cossec cos

cos cos

ln sec tan sin

u

W x

x xu dx dx x x dx

x x

u x x x

= = =

= = =

= + +


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1 1 2 2

cos [sin ] [ ln sec tan sin ]cos

py u y u y

x x x x x x

= +

= + + +

=

Therefore:


1 2sin cos ln sec tan

c py y y

y c x c x x x

= +

= + +

The general solution for the DE is:


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4.7 Cauchy-Euler Equation (C-E)

It is a linear DE with variable coefficients.

A C-E equation has the following form:

22 1 0

( )... ( )

n nna x y a x y a xy a y g x + + + + =

The C-E equation has a special characteristic such that

the degree of the monomial coefficient matches

the order k of differentiationk

k

dx

yd

kx


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Solution for Second-order Cauchy-Euler Equation

Consider the homogeneous linear second order DE :

where a, b, c are constant coefficients.

2 0 (1)ax y bxy cy + + =

Try a solution for the Cauchy-Euler (1) of the form:

Calculate:

mx=

1 2, ( 1)m mmx y m m x = =


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Substitute y, y and y into the DE (1):

2 2 1[ ( 1) ] [ ] 0

( 1) 0

m m m

m m m

m

ax m m x bx mx cx

am m x bmx cx

+ + =

+ + =

=

2 0 (1)ax y bxy cy + + =

Since: for real value of x, therefore:

This equation is called an auxiliary equation of the CE (1).

2

( 1) 0 (2)

( ) 0

am m bm c

am b a m c

+ + =

+ + =

0mx


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In general, the auxiliary equation for the Cauchy-Euler DE:

is given by:

2 0 (1)ax y bxy cy + + =

2

( 1) 0 or

0 2

am m bm c

am b a m c

+ + =

+ + =

The roots of the equation can be real and distinct, real but

repeated or complex conjugates.

Therefore y = xm is a solution of (1) if m is the root of

auxiliary equation (2).


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General Solutions of Homogeneous Cauchy-Euler

Linear Second Order Differential Equation

Homogeneous 2Homogeneous 2ndnd

Order CEOrder CE

AuxiliaryAuxiliary

equationequation

02 =++ cyybxyax

2 ( ) 0+ + =am b a m c

Distinct realDistinct realrootsroots

1 2

21

1 2

1 2

1 2GS:

,m m

mm

m m

y x y x

y c x c x

= =

= +


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Order CEOrder CE

AuxiliaryAuxiliary

equationequation

02 =++ cyybxyax

2 ( ) 0+ + =am b a m c

Repeated realRepeated realrootsroots 1 2

1 2

1 2GS:

m m

m m

m m m

y x y x x

y c x c x x

= =

= =

= +

, ln

ln

G l S l i f H C h E l


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Order CEOrder CE

AuxiliaryAuxiliary

equationequation

02 =++ cyybxyax

2 ( ) 0+ + =am b a m c

ComplexComplexconjugates rootsconjugates roots 1 2

1 2

1 2GS:

= + =

= =

= +

,

cos( ln ), sin( ln )

cos( ln ) sin( ln )

m i m i

y x x y x x

y c x x c x x


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Example:

Solve the following Cauchy-Euler equation:

2

2

(a) 5 3 0

(b) 5 4 0

+ + =

+ + =

x y xy y

x y xy y

2

2

(c) 7 41 0

(d) 2

+ =

+ =

x y xy y

x y xy y x

S l i


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Solution:

2(a) 5 3 0 1 5 3 + + = = = =, ,x y xy y a b c

auxilliary equation:

2

2

0+ + =

( )am b a m c

Roots:

2 4 3 0 3 1 0+ + = + + =( )( )m m m m

1 2

3 11 2

3 11 2

3 1

= =

= =

= +

,

,

m m

y x y x

y c x c x

S l ti


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Solution:

2(b) 5 4 0 1 5 4 + + = = = =, ,x y xy y a b c


2

2

0+ + =

( )am b a m c

Roots:

2 4 4 0 2 2 0+ + = + + =( )( )m m m m

1 2

2 21 2

2 21 2

2 2

= =

= =

= +

,

, ln

ln

m m

y x y x x

y c x c x x

S l ti


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Solution:

2(c) 7 41 0 1 7 41 + = = = =, ,x y xy y a b c


2

2

0+ + =

( )am b a m c

2 8 41 0 4 5 + = = m m m i

4 41 2

4 41 2

4 5

5 5

5 5

= =

= =

= +

,

cos( ln ), sin( ln )

cos( ln ) sin( ln )

y x x y x x

y c x x c x x

Sol tion (d) non homo eneo s Ca ch E ler


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Solution (d): non-homogeneous Cauchy-Euler

2 2 + =x y xy y x

Find yc :

2 0 + =x y xy y

2 =

Therefore:

2

21 2

2 1 0

1 0 1

+ =

= = =( )

m m

m m m

1 2= + lncy c x c x x

Solution (d): non homogeneous Cauchy Euler


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22

1 1 22 + = + =x y xy y x y y yx xx

Find yp using variation of parameters:

Write the DE in the form of:

Define:

1 22

= = =, ln , ( )y x y x x f xx

Solution (d): non homogeneous Cauchy Euler


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Calculate all the Wronskians:

1 2

1 2

ln

(1 ln ) ln1 1 ln

0 0 ln

y y x x x

W x x x x xy y x

x x

= = = + = +

12

12

1

2 n

( ) 2 / 1 ln

0 02

( ) 1 2 /

W x

f x y x x

y xW

y f x x

= = =

+

= = =



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Find u1 and u2 :

1

1

2 21

2ln

2ln2 (ln )

W xu

W x

xu dx udu u x

= =

= = = = x

22 1

2 22ln

Wu u dx x

W x x = = = =



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Write the particular solution:

1 1 2 2

2

2

(ln ) 2 ln ( ln )

py u y u y

x x x x x

= +

= +

=

General solution is:

21 2 ln (ln )c py y c x c x x x x= + = + +

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Chapter 4-1 Merged