Compre-Sol (2024-2025) - PART-B

1. For the following system of equations, compute y (0:2) and z (0:2) by using RK-4 method (take
h = 0:2):
dy
= yz + x;
dx
dz
= xz + y;
dx
with the initial conditions y (0) = 1; z (0) = .
1

Sol. We have

k1 hf1 (x0 ; y0 ; z0 ) = 0:2f1 (0; 1; 1) = 0:2;

= [1]
m1 = hf2 (x0 ; y0 ; z0 ) = 0:2f2 (0; 1; 1) = 0:2; [1]
h k1 m1
k2 = hf1 (xn + ; yn + ; zn + ) = 0:2f1 (0:1; 0:9; :
0 9) = :
0 142 ; [1]
2 2 2
h k1 m1
m2 = hf2 (xn + ; yn + ; zn + : f
) = 0 2 2 (0 1 0 9 : ; : ; : :
0 9) = 0 162 ; [1]
2 2 2
h k2 m2
k3 = hf1 (xn + ; yn + ; zn + ) = 0 1508 : ; [1]
2 2 2
h k2 m2
m3 = hf2 (xn + ; yn + ; zn + ) = 0 1674: ; [1]
2 2 2

k4 = hf1 (xn + h; yn + k3 ; zn + m3 ) = 0:1014; [1]

m4 = hf2 (xn + h; yn + k3 ; zn + m3 ) = 0:1365: [1]

Thus
1
y1 = y0 + k
[ 1 + 2( 2 + k k3 ) + k4 ] = 0:8522; [1]
2
1
z1 = z0 + m1 + 2(m2 + m3 ) + m4 ] =
[ :
0 8341 : [1]
2
 2. Use Adams-Moulton’s method to approximate the value of y (1:4) as a solution of

y0 = x2 (1 + y ); y (1) = 1;

with h = 0:1. Given y (1:1) = 1:233; y (1:2) = 1:548; y (1:3) = 1:979.

Sol. The starting values for Adams-Moulton’s method are given in the table: The predictor formula

xi yi fi
1 0 2
1.1 1.233 2.702
1.2 1.548 3.669
1.3 1.979 5.035

gives
h
y4 = y3 + [55 3f f
59 2 + 37 1 f f
9 0] [2]
24
:
0 1
:
= 1 979 + :
[55(5 035) :
59(3 669) + 37(2 702) : 9(2)]
24

:
= 2 573 [3]

So f4 = f (x4 ; y4 ) = f (1:4; 2:573) = 7:004. Now the corrector formula gives
h
y4 = y3 + f
[9 4 + 19 3 f f
5 2 + f1 ] [2]
24
:
0 1
:
= 1 979 + :
[9(7 004) + 19(5 035) : : :
5(3 669) + 2 702]
24

:
= 2 575 [3]

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3. Consider the boundary-value problem

x2 y 00 + 2xy 0 6 y = x; y (0) = 0; y (1) = 0:

Find a trial function in the form y (x) = c1 1 (x) + c2 2 (x), where 1 (x) and 2 (x) are appropriate
linearly independent functions satisfying the given boundary conditions. Hence solve the boundary
value problem by using Rayleigh-Ritz method. Find the maximum absolute error in the approximate
solution within the interval [0; 1].
Sol. The given equation can be written as (x2 y 0 )0 + 6y = x. On comparing with (p(x)y 0 )0 +
q (x)y = f (x), here p(x) = x2 ; q (x) = 6 and f (x) = x. Thus the corresponding functional I [y ] can
be derived as
Z 1 Z 1

x2 y 0 (x))2 + 6y (x)] dx +
1
I [y ] = [( xy (x) dx: [1]
2 0 0

Since the boundary conditions are homogeneous, we have to take 1 and 2 satisfying zero BCs.
Thus, take 1 (x) = x(1 x) and 2 (x) = x2 (1 x). Then, we have

y (x) = c1 x(1 x) + c2 x2 (1 x): [2]

On substituting y (x) in I [y ], we get

Z 1

x2 (fc1 (1 x2 )g)2 + 6fc1 x(1 x)g2

1
I [y ] = [ 2x) + c2 (2x 3 x) + c2 x2 (1
2 0

+2 xfc1 x(1 x) + c2 x2 (1 x)g] dx

1 2 1 1 1 1
= c1 + c22 + c1 c2 + c1 + c2 : [2]
6 14 5 12 20

For the minimum of I [y ], we have

@I 1 1 1
= c1 + c2 + = 0 [1]
@c1 3 5 12

@I 1 1 1
= c1 + c2 + = 0 : [1]
@c2 5 7 20

The solution gives c1 =

1
4
; c2 = 0 and so y (x) = 1
4
x(1 x) as the first approximate solution. [2]
Note that this is same as exact solution and hence error is zero. [1]

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4. Using the nodal points x = 0;  generates Hermite’s interpolation formula to find a cubic polynomial
approximating f (x) = cos x. Then, use it to approximate f (x) at x = =2. Also, compute an error
bound for the approximation.
Sol. Given f (x) = cos x, so f 0 (x) = sin x and

x f (x) f 0 (x)
0 1 0

 1 0

Hence

Table 1: 4 M for table

x f (x) f[,] f[„] f[„ ,]
0 1

0 1
2
2 = :
0 2026

2
 :
= 0 2026
4
3 :
= 0 1290

 1
2
2
0

 1

Hence,
 

P3 (x)  f (x) = 1 + x
0 + x 2 2 4
+ x2 (x )
 2 3
x
6
2
x
4
3
= 1 + = 1 :
0 6079 x2 + 0:03225x2 x3 [2]
2 3
 

 
2 3
)P 3
2
= 1
6

2 4 3
+
4

8
= 0 [2]

x2 (x  )2 f iv ( ) x2 (x  )2 cos( )
Error: = =
4! 24

 2 2 4
Error at
2
=
4
 4
 1 =24 =
384
= 0 2537 : : [2]

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5. Consider the following boundary value problem

x2 y 00 + 2xy 0 y = 1+ x; y (0) = 0; y (1) + y 0 (1) = 0:

Using central difference formulae for the derivatives given in the equation and backward difference
formula for the boundary condition, compute y (1=3); y (2=3) and y (1). Take the uniform stepsize
h = 1=3.
Sol.
x2 y 00 + 2xy 0 y = 1 + x:
Applying second order formula
yi+1yi + yi 2 1 yi+1 yi 1
x2i +2 xi yi = 1 + xi for i = 1; 2
 
h2 
2 h
 
) yi 1 x2i hxi + yi 2x2i h2 + yi+1 x2i + hxi = (1 + xi ) h2 for i = 1; 2 [2]

       

For, i=1 y0
1

9
1

3
 1

3
+ y1
2

9
1

9
+ y2
1

9
+
1

9
= 1+
1

3
1

) 3

9
y1 + y2
2

9
=
4

3
 1

   
) 
y
9 1 +6 2 = 4 y
  
[2]

For, i=2 y1
4

9
1

3
 2

3
+ y2 2  4

9
1

9
+ y3
4

9
+
1

3
 2

3
= 1+
2

3
1

) y 2

9
1
9

9
y2 + y3
6

9
=
5

3
 1

) y 6 1 y
27 2 + 18 3 = 5 y [2]
Also; y (1) + y 0 (1) = 0
y3 y2
by applying Backward difference formula )y 3 +
h
= 0

) (1 + h) y3 y2 = 0 ) 4 3 y y
3 2 = 0 [2]
2 32 3 2 3
9 6 0y1 4
6 76 7 6 7
4 6 27 185 4y2 5 = 455

0 3 4 y3 0

)y 1 =
56
57
= :
0 9825 ; y2 =
46
57
= :
0 8070 ; y3 =
23
38
= :
0 6053 . [2]

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6. Use power method to find the numerically largest eigenvalue and associated eigenvector of the matrix
2 3
3 2 2
6 7
A = 42 2 05 ;
2 0 4

starting with X (0) = (1 1 1) ; ; T . Perform five iterations.

Sol. We have
2 32 3 2 3 2 3
3 2 2 1 7 1
6 76 7 6 7 6 7
X (1) = AX (0) = 42 2 05 415 = 445 = 7 40 57145 : [2]
2 0 4 1 6 0:8571
2 32 3 2 3 2 3
3 2 2 1 : 5 857 1
6 76 7 6 7 6 7
X (2) = AX (1) = 42 2 05 40:57145 = 43:1435 = 5:857 40:53665 [2]
2 0 4 0:8571 5:428 0:9268
2 32 3 2 3 2 3
3 2 2 1 : 5 927 1
6 76 7 6 7 6 7
X (3) = AX (2) = 42 2 05 40:53665 = 43:0735 = 5:927 40:51855 [2]
2 0 4 0:9268 5:707 0:9629
2 32 3 2 3 2 3
3 2 2 1 : 5 963 1
6 76 7 6 7 6 7
X (4) = AX (3) = 42 2 05 40:51855 = 43:0375 = 5:963 40:50935 [2]
2 0 4 0:9629 5:852 0:9814
2 32 3 2 3 2 3
3 2 2 1 : 5 981 1
6 76 7 6 7 6 7
X (5) = AX (4) = 42 2 05 40:50935 = 43:0195 = 5:981 40:50485 [2]
2 0 4 0:9814 5:926 0:9908

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