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
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