Assignment 1

Maximum Marks 40

Due Date 3rd

May, 2004

Assignment Weight age 2%

Solve the following Differential Equations.

Question 1

Solve

2 5 3

2 4 6

dy x y

dx x y

Solution

,

2 5 3 0

2 4 6 0

1.

Let x X a and y Y b wherea and bobey therelations

a b

a b

which solveto givea b

2 5

2 4

.

Making these substitutions we find

dY X Y

dX X Y

whichis homogeneousODE

Put Y = vX

2

2

2 5

2 4

2 5

2 4

2 5 2 4

2 4

2 7 4

2 4

2 4 1

4 1 2

dv vv X

dX v

dv vX v

dX v

v v v

v

v v

v

vdv dX

v v X

By partial Fraction on left hand side we have,

4 2

3 4 1 3 2

4 2

3 4 1 3 2

dXdv

v v X

dXdv

v v X

2 3

2 3 3

1 2ln 4 1 ln 2 ln

3 3

ln 4 1 2 ln

4 1 2

v v Xc

v v Xc

v v X C C c

But Y= vX, so

2

3

2 3

3

2

2

2

4 1 2

4 2

4 2

,

, .

4 2

4 4 2 2

Y YX C

X X

CY X Y X X

X

Y X Y X C

But x X a and y Y b

X x a Y y b So above become

x a y b y b x a C

x y a b y x a b C

Question 2

Solve

2 21( )

2xdy ydx x y dx

Solution

2

2

2 2

2

2

1

1(1 )

2

1(1 )

2

1tan , .

2

Dividing both sidesby x

xdy ydx ydx

x x

y xdy ydxPut t and dt

x x

dt t dx

yt x c where t

x

Question 3

Solve 2dy y

x ydx x

Solution

2

2

2 22

2

2

, ,

.

1ln ln

dy xy yWehave

dx x

dy dvPut y vx v x

dx dx

dv x vx v xwe get v x v v

dx x

dvor x v

dx

x cv

Question 4

Solve

2 2dyx y x x y

dx

Solution

2 2

2 2 22

, , .

1

The givenequationis

y x x ydy

dx x

dy dvPut y vx v x we get

dx dx

dv vx x x v xv x v x v

dx x

21

dvor dx

v

1

:

sinh

sinh( )

Integrating

v x c

y x x c

Solution Assignment 2

Maximum Marks 40

Due Date 20th May, 2004

Assignment Weight age 2%

Question 1

Find the member of the orthogonal trajectories for 2

13 2 3xy c x that passes through (0,10).

Here 2

13 2 3xy c x --------------(1)

2

1

3 2

3

xyc

x

------------- (2)

Differentiating (1) with respect to x, we get

2

1

22

2

3 6 3

3 22 3 6 3

3

1

3

dyy xy c

dx

dy xyy xy

dx x

dy

dx x y

So the Differential Equation of Orthogonal Trajectory is given by

2

2

2

3

3

3

3

ln 3

dyx y

dx

dyx dx

y

dyx dx

y

y x C

Now member of this passes through (0,10) is obtained by finding corresponding value of C, so putting

(0,10) in (3) we get

ln10 0

ln10

C

C

So the required curve is given by

3

10 xy e

Question 2

When interest is compounded continuously, the amount of money S increases at a rate proportional to

the amount present at any time: dS/dt = rS, where r is the annual rate of interest.

(a) Find the amount of money accrued at the end of 5 years when $ 5000 is deposited in a savings

account drawing 3

5 %4

annual interest compounded continuously.

(b) In how many years will the initial sum deposited be doubled?

Here we have the equation

dSrS

dt

Solving we get

r tS Ce --------------(1)

Since initially $ 5000 deposited, so we have 0 5000S . Here given r the annual interest equals to

35 %

4 i.e. 5.75 / 100 0.0575r . Moreover by first condition, C = 5000.

Thus (1) becomes

0.0575

5000t

S e

(a) Now after 5 years means, we have to calculate S when t = 5.

0.0575 55000 6665.45S e

(b) we have to calculate t when sum = 2 (initial money) = 10000

0.0575

0.0575

10000 5000

2

1ln 2 12.05

0.0575

t

t

e

e

t

So approximately after 12 years the initial sum deposited will be doubled.

Question 3

A thermometer is taken from an inside room to the outside where the air temperature is 05 F.

After 1 minute the thermometer reads 055 F, and after 5 minutes the reading is 030 F. What is the

initial temperature of the room?

By Newton’s Law of Cooling, we have

0

dTk T T

dt

Where k is constant of proportionality and 0T is temperature of surroundings.

Solving this equation we get

0

ktT Ce T .

Here, temperature of the surroundings is given by0 5T F , moreover when time is 1 then T is 55. So

1 55T F , similarly we have other boundary condition as 5 30T F . So using these conditions,

we have

59.4642 0.1733C and k .

So the solution equation is

0.173359.4642 5

tT e

Now initial temperature is obtained by putting t = 0.

So we have

64.4642T F .

Question 4

Find a second solution of the following equation.

2 4

1

2

1

20 0;

1 2 2 1 2 0; 1

a x y y y x

b x x y x y y y x

(a)

Here 2 20 0x y y

2

200y y

x

Comparing with

0y P x y Q x y ,

It gives

0P x

Since

2 1 2

1

0

4

8

9 54

9 9

P x dx

dx

ey y dx

y

ex dx

x

x xx

(b)

Here 2 220 0 1 2 2 1 2 0x y y x x y x y y

2 2

2 1 20

1 2 1 2

xy y y

x x x x

Comparing with

0y P x y Q x y ,

It gives

2

2 1

1 2

xP x

x x

Since

2

2

2

2 1 2

1

2 1

1 2

2

2 1

1 2

2

ln 2 1

2

2

2

2

11

11

11

2 11

1

21 1

1

21

1

1 2

P x dx

xdx

x x

xdx

x x

x x

ey y dx

y

ex dx

x

ex dx

x

ex dx

x

x xx dx

x

x dxx

x xx

x x

- Assignment 3

Maximum Marks 50

Due Date 29th June, 2004

Assignment Weight age 2%

Question 1

The roots of an auxiliary equation are 1 2 3

1, 3 , 3

2m m i m i . What is the corresponding

differential equation?

Solution

Given that

1 2 3

13 3

2

10 ( 3) 0 ( 3) 0

2

m m i m i

m m i m i

The corresponding auxiliary equation will be,

2

3 2

1( )( ( 3) )( ( 3) ) 0

2

1( )( 6 10) 0

2

2 11 14 10 0

m m i m i

m m m

m m m

Finally the corresponding differential equation will be,

3 2(2 11 14 10) 0D D D y

3 2

3 22 11 14 10 0

d y d y dy

dx dx dx

Question 2

(c) Use a trigonometric identity as an aid in finding a particular solution of the given differential equation.

sin cos2y y x x

Solution

For complementary solution, consider the homogeneous part. ' ' 0y y

Auxiliary equation is

2 1 0m

m i

So, the complementary solution is given by

1 2os incy C C x C S x

For particular solution

Here we have

( ) sin cos2g x x x

Using trigonometric identities we can write.

1( ) (2 2 )

2

1( ) [ ( 2 ) ( 2 )]

2

1( ) [ (3 ) ( )]

2

1( ) [ 3 ]

2

g x SinxCos x

g x Sin x x Sin x x

g x Sin x Sin x

g x Sin x Sinx

1 1

( ) 32 2

g x Sin x Sinx

Thus we can divide particular integral into two parts i.e.

1 2p p py y y

Let 1

3 3py ASin x BCos x

And 2

py C Sinx DCosx

Clearly, assumed function 2

py is already in complementary solution

So,

2

py Cx Sinx D xCosx

3 3py ASin x BCos x Cx Sinx D xCosx

After simplification and comparing the co-efficient (students are required to make complete

calculations in their assignments), we get

1 1

316 4

py Sin x xCosx

So the general solution is

1 2

1 1os in 3

16 4y C C x C S x Sin x xCosx

Question 3

Solve the given differential equations subject to the indicated initial conditions.

21 5 6 10 , 0 1, 0 1

2 8cos 2 4sin , / 2 1, / 2 0

xy y y e y y

y y x x y y

Solution

(1) 25 6 10 xy y y e ----------------------------(1)

For complementary solution

Consider 5 6 0y y y

Auxiliary equation is 2 5 6 0m m

Roots are m=1 and m=-6

So, 6

1 2

x x

cy C e C e

Now for particular solution

Suppose 2x

py Ae

Substituting in (1) and after simplification

We get 5

4A

So, 25

4

x

py e

Thus, the general solution is

y= 6

1 2

x xC e C e + 25

4

xe ----------------(2)

and

' 6

1 26x x

cy C e C e + 25

2

xe ----------------(3)

By applying given initial condition to find out values of constants

1=y(0)= 1C + 2C +

5

4

1=y’(0)= 1C -6 2C +5

2

After calculations

1C =3

7

and 2C =

5

28

Hence, the solution is

6 23 5 5

7 28 4

x x xy e e e

Solution

(2) 8cos2 4siny y x x ------------------(1)

For complementary solution

Consider y y 0

Auxiliary equation is 2 1 0m

Roots are

m=i and m=-i

So,

1 2cy C Cosx C Sinx

For particular solution consider g(x)= 8cos2 4sinx x

1 2p p py y y

12 2pLet y ASin x BCos x

2pand y C Sinx DCosx

Clearly, assumed function 2py is already in complementary solution

So,

2pLet y Cx Sinx D xCosx

2 2py ASin x BCos x Cx Sinx D xCosx

Putting in (1) and after simplification, we get

3 2 3 2 2 2 8 2 4ACos x BSin x CSinx DCosx Cos x Sinx

After comparing the co-efficient

We get,

A= 8

3

, B=0, C=2, D=0

Thus, particular solution is

82 2

3py Cos x xCosx

General solution is

y= 1 2C Cosx C Sinx + 8

2 23

Cos x xCosx

Now,

'

1 2

162 2 2

3y C Sinx C Cosx Sin x Cosx xSinx

By applying given initial condition to find out values of constants

2

81 ( )

2 3y C

'

10 ( )2

y C

So,

1C And 2

11

3C

Hence, the solution is

11 82 2

3 3y Cosx Sinx Cos x xCosx

Question 4

Given that 2

1y x and 3

2y x , form a fundamental set of solutions of 2 4 6 0x y xy y on 0, .

Find the general solution of

2 14 6x y xy y

x

Solution

2 3

2 3

2( , )

2 3

x xW x x

x x

2 3 4 4 4( , ) 3 2 0W x x x x x

i.e., 1y and 2y are linearly independent on 0, .

From the given equation

'

2 3

14 6

y yy

x x x

3

1 2

3

0

113

x

Wx

x

2

2

3

01

12

x

Wxx

x

Now, we determine the derivatives of the unknown variables 1u and 2u through the relations

W

Wu

W

Wu 2

21

1 ,

'

1 4

1u

x and '

2 5

1u

x

By integrating both sides, we get

1 3

1

3u

x and 2 4

1

4u

x

Thus,

2 3

3 4

1 1

3 4py x x

x x

1 1

3 4py

x x

1

12py

x

Hence, the general solution is

2 3

1 2

1

12y C x C x

x

Assignment 4

Maximum Marks 50

Due Date 14th July, 2004

Assignment Weight age 2%

Question 1

A 1-kg mass is attached to a spring whose constants is 16 N/m and the entire system is then submerged

in a liquid that imparts a damping force numerically equal to 10 times the instantaneous velocity.

Determine the equations of motion if

a) the weight is released from rest 1 m below the equilibrium position; and

b) the weight is released 1 m below the equilibrium position with an upward velocity of 12 m/s.

Solution:

Here 1

. 16 /

10

10

mass m Kg

Spring cons k N m

dx dxdamping force

dt dt

But differential equation of Motion with damping is given by

2

2

2

2

2

2

16 10

10 16 0

d x dxm kx

dt dt

d x dxx

dt dt

d x dxx

dt dt

If we suppose solution of the form mtx e , then the auxiliary equation is given by

2

2

10 16 0

8 2 16 0

2 8 0

2, 8

m m

m m m

m m

m

The general solution is this given by

2 8

1 2

t tx C e C e

(a) Since when time is zero, then mass is one meter below so

0 1x

Also it starts from rest i.e. velocity is zero at time = 0, so

0 0x .

These two conditions give us

1 2

2 8

2

4 / 3, 1/ 3

4 1

3 3

t t

C C

x e C e

(b) Again here 0 1x , and in this case it starts with an upward velocity of 12 /m s so

0 12x

These conditions give us

1 2

2 8

2

2 / 3, 5 / 3

2 5

3 3

t t

C C

x e C e

Question 2

A mass m is attached to the end of a spring whose constant is k. After the mass reaches equilibrium, its

support begins to oscillate vertically about a horizontal line L according to a formula h(t). The value of

h represents the distance in feet measured from L. See Figure 5.26 ( page 240 Differential Equations

by Dennis G. Zill). Determine the differential equation of motion if the entire system moves through a

medium offering a damping force numerically equal to /dx dt .

Solution:

The forces acting on the system are:

a) Weight of the body mg

b) The restoring force = hxsk

c) The damping effect )/( dtdx

Hence hx denotes the distance of the mass m from the equilibrium position. Thus the total force

acting on the mass m is given by

dt

dxhxskmgForce

By the Newton’s 2nd

law of motion, we have

2

2

dt

xdmmaForce

Therefore

dt

dxkhkxksmg

dt

xdm

2

2

But 0 ksmg

So that 02

2

h

m

kx

m

k

dt

dx

mdt

xd

or hxdt

dx

dt

xd 22

2

2

2

Where, 2 2 and k

m m

.

Question 3

Solve the given differential equations subject to the indicated initial conditions.

2

2

1 0, 1 1, 1 1

2 3 4 0, 1 5, 1 3

x y xy y y y

x y xy y y y

Solution:

(1) Here 2 0x y xy y , to convert this equations into second order differential equation

with constant coefficients, we put

2 22

2 2

ln

1

1

1

,

tx e x t

dt

dx x

dy dy dt dy

dx dt dx x dt

dy dyx

dx dt

Similarly

d y d y dyx

dx dt dt

So the equation becomes

2

2

2

2

0

0

d y dy dyy

dt dt dt

d yy

dt

Its solutions is given by

1 2cos siny C t C t

But by (1) lnt x , so

1 2cosln sin lny C x C x

Now, the initial conditions yields,

1

2

1 1 1

1 1 1

y C

y C

Thus

cosln sin lny x x

(2) Here 2 3 4 0x y xy y , to convert this equations into second order differential equation

with constant coefficients, we put

2 22

2 2

ln

1

1

1

,

tx e x t

dt

dx x

dy dy dt dy

dx dt dx x dt

dy dyx

dx dt

Similarly

d y d y dyx

dx dt dt

So the given equation becomes

2

2

2

2

3 4 0

4 4 0

d y dy dyy

dt dt dt

d y dyy

dt dt

Its solutions is given by

2 2

1 2

t ty C e C te

But by (1) lnt x , so

2ln 2ln

1 2

2 2

1 2

ln

ln

x xy C e C xe

C x C x x

Now, the initial conditions yields,

1

1 2

2

1 5 5

1 3 2 3

3 10 7

y C

y C C

C

Thus

2 25 7t ty e te

Question 4

Find two linearly independent power series solutions about the ordinary point x = 0 for the following

differential equation.

2 1 0x y xy y

Solution Since the singular points are 1x , 0x is the ordinary point, a power series will converge at least

for 1x . The assumption

0n

n

n xcy leads to

2 2 1

2 1 0

( 1) ( 1) n n nn n n

n n n

x n n c x x nc x c x

2

2 2 1 0

( 1) ( 1)n n n nn n n n

n n n n

n n c x n n c x nc x c x

0 2 0

2 3 1 0 1

2 4 2 2

( 1) 2 6 ( 1)n n n n

n n n n

n n n n

n n c x c x c x n n c x c x nc x c x c x c x

k=n k=n-2 k=n k=n

2 0 3 22

2 6 [ ( 1) ( 2)( 1) ] 0kk k k k

k

c c c x k k c k k c kc c x

or 2 0 3 22

2 6 [( 1)( 1) ( 2)( 1) ] 0.kk k

k

c c c x k k c k k c x

Thus 02 02 cc

03 c

2( 1)( 1) ( 2)( 1) 0k kk k c k k c

This implies

022

1cc

03 c

2

( 1),

( 2)k k

kc c

k

2,3,k

Iteration of the last formula gives

02024!22

1

42

1

4

1cccc

05

235 cc

03046!32

31

642

3

6

3cccc

07

457 cc

8 6 0 04

5 3 5 1 3 5

8 2 4 6 8 2 4!c c c c

09

679 cc

050810!52

7531

108642

753

10

7cccc

and so on.

Therefore

2 3 4 5 6 7 85 70 1 2 3 4 6 8y c c x c x c x c x c x c x c x c x

2 4 6 8 101 0 52 3 4

1 1 1 3 1 3 5 1 3 5 7[1 ]

2 2 5!2 2! 2 3! 2 4!y c x c x x x x x

The solutions are

21 0

1

1 3 5 (2 3)( ) [1 ],

2 !

nn nn

ny x c x

1x

2 1( ) .y x c x

Assignment 5

Maximum Marks 30

Due Date 30th July, 2004

Assignment Weight age 2%

Question 1

Show that the indicial roots differ by an integer. Use the method of Frobenius to obtain two linearly

independent series solutions about the regular singular point0 0x . Form the general solution on

0, .

0xy xy y

solution

put

0

' 1

0

'' 2

0

( )

( )( 1)

n r

n

n

n r

n

n

n r

n

n

y C x

y C n r x

y C n r n r x

then the equation becomes

2 1

0 0 0

( )( 1) -x ( ) + n r n r n r

n n n

n n n

xy xy y x C n r n r x C n r x C x

2

0 0 0

[ ( )( 1) - ( ) + ]=0 r n n n

n n n

n n n

x C n r n r x C n r x C x

1

0C ( 1) 0r r x and

1( 1)( 1) - ( 1) =0 k kk r k r C k r C

1

=

( 1)

k

k

CC

k r

1 20 1r r

and

but 2 1 1r r is an integer, then there exist two linearly independent solutions of the form

11 0

0

, 0n r

nn

y c x c

(3 )a

22 1 0

0

( )ln , 0n r

nn

y Cy x x b x b

(3 )b

Where C is a constant that could be zero.

then

1 00

, 0nn

n

y c x c

and

12 1 0

0

( )ln , 0nn

n

y Cy x x b x b

when 1 0r

1

=

( 1)

k

k

CC

k

for 0,1,2,3,4,...k

1 0

01

2

02

3

3 0

4

04

5

5 0

6

6 0

7

C

C2 2

C3 3.2

C

4 4.3.2

C

5 5.4.3.2

C

6 6.5.4.3.2

C

7 7.6.5.4.3.2

C

CC

CC

CC

CC

CC

CC

----------------------------------

----------------------------------

----------------------------------

0

!n

CC

n

So,

01 0

0

, 0!

n

n

Cy x c

n

when 2 1r then

1

=

1

k

k

CC

k

for 0,1,2,3,4,...k

0

1

01

2

02

3

3 0

4

04

5

5 0

6

C2

C3 3.2

C

4 4.3.2

C

5 5.4.3.2

C

6 6.5.4.3.2

C

7 7.6.5.4.3.2

C

CC

CC

CC

CC

CC

------------------------------

------------------------------

------------------------------

0

( 1)!n

CC

n

So,

0

112 1

0

( )ln , 0( 1)!

n

n

Cy Cy x x x C

n

Question 2

Use the change of variable 1/ 2y x v x to find the general solution of the equation

.

Solution

1/ 2y x v x

' 1/ 2 1/ 2 '1

2y x v x x v x

1

'' 5 / 2 3 / 2 ' ''23

( )4

y x v x x v x x v x

by substituting values in the given differential equation, we get

0= 2x [ 1

5 / 2 3 / 2 ' ''23

( )4

x v x x v x x v x

]+2x[ 1/ 2 1/ 2 '1

2x v x x v x ]+ 2 2x [ 1/ 2x v x

]

3 31 1 1

' '' 22 2 2 2 23

( ) ( ) ( ) ( ) ( ) 04

x v x x v x x v x x v x x v x

By multiplying the above equation by 1

2x

, we get 2 '' ' 2 2 1( ) ( ) ( ) ( ) 0

4x v x xv x x v x

By comparing the equation with the general Bessel’s equation , which is 2 '' ' 2 2( ) ( ) ( ) ( ) 0x y x xy x x v y x

we get 2 2 1 1

4 2x x

So, the solution of our equation is

1 1 2 12 2

( ) ( )v C J x C J x

By putting in 1/ 2y x v x , we get

1 12 2

1 1 2 12 2

( ) ( )y C x J x C x J x

which is the required solution

Question 3

Solve the given differential equations subject to the indicated initial conditions.

2

2

2

2

0 (1)

4 0 2

0 1, 0 0,

0 1, 0 5

d x dx dy

dt dtdt

d y dy dx

dt dtdt

x x

y y

Solution: First we write the differential equations of the system in the differential operator form:

2

2

( ) 0

( ) 4 0

D D x Dy

D D y Dx

Then we eliminate one of the dependent variables, say x . Multiplying first equation with 4 and the

second equation with the operator D+1 and then adding, we obtain

2

1 +4 0[ ] D D yD

or 2[( 1) 4 ] 0D D D x

The auxiliary equation of the differential equation found in the previous step is

2[( 1) 4] 0m m

Therefore, roots of the auxiliary equation are

1 2 30, 1 2 , 1 2 m m i m i

So that the complementary function for the retained variable y is

4 5 6( cos2 sin 2 .)ty c e c t c t ---------(3)

Next we eliminate the variable y from the given system. For this purpose we multiply second equation

with 1 while operate on the first equation with the operator D +1 and then subtracting, we obtain

2[( 1) 4 ] 0D D D x

So, 1

2 3 4

0,

( cos 2 2sin 2 .)

1, 2, 5

t

c

y e t t

c c c

since we have given only four initial conditions but there are six constants to be determined, so some

of them must multiple of others, to find out we put x and y in (1) and after simplifying, we get

Coefficients of cos2te t are

5 6 3 22 2 4 0c c c c ---------(5)

Coefficients of sin 2te t are

6 5 2 32 2 4 0c c c c ---------(6)

multiplying (5) by 2 and then subtracting (6)

we get 6 22c c

multiplying (5) by 2 and then adding (6)

we get 5 32c c

by putting these values in (4) we get

4 3 2( 2 cos2 2 sin 2 .)tx c e c t c t -------(7)

after substituting the given initial conditions in (3) and (7) we get the values of constants

1 2 3 40, 1, 2, 5c c c c

then solution becomes

5 (4cos2 2sin 2 .)tx e t t

( cos2 2sin 2 .)ty e t t

Assignment 6

Maximum Marks 30

Due Date 6th August, 2004

Assignment Weight age 2%

Question 1

Use Gauss-Jordan elimination to demonstrate that the given system of equations has no solution.

3 1

4 0

2 2 6

4 7 7 9

x y z p

y z p

x y z p

x y z

Solution

(a) The augmented matrix of the system is

1 1 1 3 1

0 1 1 4 0

1 2 2 1 6

4 7 7 0 9

Multiplying first row with 1 and 4 and then adding to3rd and 4th

row i.e. by 3 1R R and 4 14R R ,

we obtain

1 1 1 3 1

0 1 1 4 0

0 1 1 4 5

0 3 3 12 5

Multiplying second row with 1 ,-1 and 4 and then adding to 1st, 3rd and 4

th row i.e. by

1 2R R , 3 2R R and

4 23R R , we obtain

1 0 0 7 1

0 1 1 4 0

0 0 0 0 5

0 0 0 0 5

7 1

4 0

0 5

0 5

x p

y z p

This shows it has no solution.

Question 2

Find the eigenvalues and eigenvectors of the given matrix.

5 1 0

0 5 9

5 1 0

.

Solution

Eigenvalues

The characteristic equation of the matrix A is

5 1 0

det 0 5 9 0

5 1

A I

Expanding the determinant by the cofactors of the second row, we obtain

3 16 0

2 16 0

Hence the eigenvalues of the matrix are

1 2 30 4 4λ , λ , λ .

Eigenvectors

For 01 we have

5 1 0 0

0 | 0 0 5 9 0

5 1 0 0

A

By 1

5

R

11 0 05

0 5 9 0

5 1 0 0

By 3 15R R

11 0 05

0 5 9 0

0 0 0 0

Thus we have the following equations in 1 2, k k and 3k . The number 3k can be chosen arbitrarily

3 1

25

9k k , 2 15k k

Choosing1 9k , we get 2 45k and 3 25k . Hence, the eigenvector corresponding 01 is

1

9

45

25

K

For 2 4 , we have

1 1 0 0

4 0 0 9 9 0

5 1 4 0

A I

By 3 15R R

1 1 0 0

0 9 9 0

0 4 4 0

By 2

1

9R

1 1 0 0

0 1 1 0

0 4 4 0

By 3 24R R

1 1 0 0

0 1 1 0

0 0 0 0

Hence we obtain the following two equations involving 1 2, k k and 3k .

1 2k k , 1 3k k

Choosing 1 1k , we have 2 31, 1k k . Hence we have an eigenvector corresponding to the

eigenvalue 2 4

2

1

1

1

K

Finally, for 3 4 , we have

09 1 1

4 | 0 0 1 9 0

5 1 4 0

A I

By 1

1

9R

11 0 09

0 1 9 0

5 1 4 0

By 3 1 25 ,( 1)R R R

11 0 09

0 1 9 0

4 05 49

By 3 2

4

9R R

11 0 09

0 1 9 0

0 0 0 0

So that we obtain the equations

1 3 2 1, 9k k k k

The choice 1 1k leads to 2 39, 1k k . Hence, we have the following eigenvector

3

1

9

1

K

Question 3

Find the general solution of the given system.

1 4 2

4 1 2

0 0 6

X X

Solution

Here

1 4 2

4 1 2

0 0 6

0

1 4 2

4 1 2 0

0 0 6

6 3 5 0

6,3, 5.

A I

On simplification

Now for 6

Thus the characteristic equation has real and distinct roots and so are the eigenvalues of the coefficient

matrix A . To find the eigenvectors corresponding to these computed eigenvalues, we need to solve the

following system of linear algebraic equations for 1 2,k k and 3k when 6, 3, -5 , successively.

1

2

3

1 4 2 0

det( ) 0 4 1 2 0

0 0 6 0

k

A I K k

k

For solving this system we use Gauss-Jordon elimination technique, which consists of reducing the

augmented matrix to the reduced echelon form by applying the elementary row operations. The

augmented matrix of the system of linear algebraic equations is

1 4 2 0

4 1 2 0

0 0 6 0

For 6 , the augmented matrix becomes:

7 4 2 0

4 7 2 0

0 0 0 0

Appling the row operation 1 2 2 1 2 1 21

2 , 4 , , 1033

R R R R R R R in succession reduces the

augmented matrix in the reduced echelon form.

21 011 0

20 1 011

00 0 0

So that we have the following equivalent system

0

0

0

00011

210

11

201

3

2

1

k

k

k

or 1 3 2 32 2

,11 11

k k k k

Therefore, the constant 3k can be chosen arbitrarily. If we choose 3 11k , then 1 22, 2k k , So

that the corresponding eigenvector is

1

2

2

11

K

For 32 , the augmented matrix becomes

4 4 2 0

(( - 3 ) | 0) 4 4 2 0

0 0 3 0

A I

We apply elementary row operations to transform the matrix to the following reduced echelon form:

1 1 0 0

0 0 0 0

0 0 1 0

Thus 1 2 3, 0k k k

Again 2k can be chosen arbitrarily, therefore choosing 2 1k we get 1 1k Hence, the second

eigenvector is

2

1

1

0

K

Finally, when 53

the augmented matrix becomes

4 4 2 0

((A + 5 I) | 0) = 4 4 2 0

0 0 11 0

The application of the elementary row operation transforms the augmented matrix to the reduced

echelon form

1 1 0 0

0 0 0 0

0 0 1 0

Thus 1 2 3, 0k k k

If we choose 2 1k , then 1 1k , thus the eigenvector corresponding to 53 is

3

1

1

0

K

Thus we obtain three linearly independent solution vectors

6 3 5

1 2 3

2 1 1

= 2 , 1 , 1

11 0 0

t t tX e X e X e

Hence, the general solution of the given homogeneous system is

6 3 5

1 2 3

2 1 1

2 1 1

11 0 0

t t tX c e c e c e