Unit 2

Numerical Methods

Q(1) Apply Gauss Elimination method to solve the equations:

Ans-

Given                                                    Check Sum (sum of coefficient and constant)

                        -1                            …. (I)

                      -16                          …. (ii)

                          5 …. (iii)

(I)We eliminate x from (ii) and (iii)

Apply eq(ii)-eq(i) and eq(iii)-3eq(i) we get

                        -1                            ….(i)

                      -15                          ….(iv)

                          8            ….(v)

(II) We eliminate  y from eq(v)

Apply

                        -1                            ….(i)

                      -15                          ….(iv)

                      73            ….(vi)

(III) Back Substitution we get

                 From (vi) we get

                  From (iv) we get

                   From (i) we get 

Hence the solution of the given equation is

Q(2) Solve the equation by Gauss Elimination Method:

Ans-

Given

Rewrite the given equation as

                 … (i)

                             ….(ii)

                             ….(iii)

                                 …(iv)

(I)                We eliminate x from (ii),(iii) and (iv) we get

Apply eq(ii) + 6eq(i), eq(iii) -3eq(i), eq(iv)-5eq(i) we  get

                 …(i)

                             ….(v)

                       ….(vi)

                             …(vii)

(II)             We eliminate y   from (vi) and (vii) we get

Apply 3.8 eq(vi)-3.1eq(v) and 3.8eq(vii)+5.5eq(v) we get

                 …(i)

                             ….(v)

                         …(viii)

                             …(ix)

(III)           We eliminate z from eq (ix) we get

Apply 9.3eq (ix) + 8.3eq (viii), we get

                 … (i)

                             ….(v)

                         …(viii)

        350.74u=350.74

       Or u = 1

(IV)          Back Substitution

From eq(viii)

Form eq(v), we get

From eq(i) ,

Hence the solution of the given equation is x=5, y=4, z=-7 and u=1.

Que(3) Apply Gauss Elimination Method to solve the following system of equation:

Ans-

Given                    … (i)

              … (ii)

              … (iii)

(I)                We eliminate x from (ii) and (iii)

Apply we get

                … (i)

              … (iv)

             … (v)

(II)             We eliminate y from  (v)

Apply we get

                … (i)

              … (vi)

                      … (vii)

(III)           Back substitution 

From (vii)  

From (vi)

From (i)

Hence the solution of the equation is

Q(4) Solve by cholesky method

Ans-

Solution. Step I. Consider AX= B

Step 2: Consider

By solving

,     

Put all the values

Step 3: LY = B, Let

By multiplying and equating

g=3

and,

and,

Therefore,

Step 4:

Now we are just cross check our Ans. by putting given equations of question

Given equation:

Our answer are correct

Q(5) Solve by cholesky method

Ans-

Solution.

Step I. consider AX =B

Step 2. Consider

BY solving,

Put all the values we get

Step 3: LY = B, Let

By multiplying and equating

Step 4:

BY MULTIPLYING WE GET

Now we are just cross check our ans by putting values in given equations of question,

Given equation

Our answer is correct.

Q(6) Solve by cholesky method

Ans-

Step I Consider AX = B

Step 2: Consider

Put all the values we get

Step 3:

By multiplying and equating

Step 4:

By multiplying we get

Now we are just cross check our answer by putting values of in equation

Given equation:

And we have

Our correct answer is

Q(7) Solve by cholesky method

Ans-

Solution-

 Step 1. Consider AX = B

Step 2: Consider,

Let,

Put all the values we get

Step 3: LY = B, Let

By multiplying and equating

Step 4:

By Multiplying we get

Now we just cross check our answer by putting values in given equation

Given equation:

Final answer,

Q(8) Use Jacobi’s method to solve the system of equations:

Ans-

Since

So, we express the unknown with large coefficient in terms of other coefficients.

              (1)

Let the initial approximation be

2.35606

0.91666

1.932936

0.831912

3.016873

1.969654

3.010217

1.986010

1.988631

0.915055

1.986532

0.911609

1.985792

0.911547

1.98576

0.911698

Since the approximation in ninth and tenth iteration is same up to three decimal places, hence the solution of the given equations is

Q(9) Solve by Jacobi’s Method, the equations

Ans-

Given equation can be rewrite in the form

     … (i)

    ..(ii)

    ..(iii)

Let the initial approximation be

Putting these values on the right of the equation (i), (ii) and (iii) and so we get

Putting these values on the right of the equation (i), (ii) and (iii) and so we get

Putting these values on the right of the equation (i), (ii) and (iii) and so we get

0.90025

Putting these values on the right of the equation (i), (ii) and (iii) and so we get

Putting these values on the right of the equation (i), (ii) and (iii) and so we get

Hence   solution approximately is

Que(10) Use Jacobi’s method to solve the system of the equations

Ans-

Rewrite the given equations

        (1)

Let the initial approximation be

1.2

1.3

0.9

1.03

0.9946

0.9934

1.0015

Hence the solution of the above equation correct to two decimal places is

Q(11)Use Gauss –Seidel Iteration method to solve the system of equations

Ans-

Since

So, we express the unknown of larger coefficient in terms of the unknowns with smaller coefficients.

Rewrite the above system of equations

      (1)

Let the initial approximation be

3.14814

2.43217

2.42571

2.4260

Hence the solution correct to three decimal places is

Que(12) Solve the following system of equations

  By Gauss-Seidel method.

Ans-

Rewrite the given system of equations as

      (1)

Le t the initial approximation be

Thus the required solution is

Que(13) Solve the following equations by Gauss-Seidel Method

Ans-

Rewrite the above system of equations

          (1)

Let the initial approximation be

Hence the required solution is

Q(14) Use Euler’s method to  find y(0.4) from the differential equation

    with h=0.1

Ans-

      Given  equation  

          Here

      We  break  the  interval in four steps.

             So that

          By Euler’s  formula

, n=0,1,2,3   ……(i)

         For n=0 in equation (i) we  get

          For n=1 in equation (i) we  get

.01

           For n=2 in equation (i) we  get

          For n=3 in equation (i) we  get

      Hence y(0.4)  =1.061106.

Q(15)Using  Euler’s method  solve the differential equation  for  y at x=1 in five steps

 Ans-

Given equation  

           Here 

    No. of steps  n=5 and  so that

    So that

           Also

          By Euler’s  formula

, n=0,1,2,3,4   ……(i)

         For n=0 in equation (i) we  get

      For n=1  in equation (i) we  get

      For n=2  in equation (i) we  get

         For n=3  in equation (i) we  get

       For n=4  in equation (i) we  get

          Hence 

Q(16) Given  with  the initial condition  y=1 at   x=0.Find  y for x=0.1 by  Euler’s   method(five  steps).

Ans-

Given equation is  

Here   

No. of steps  n=5 and so that

So that

           Also

          By Euler’s  formula

, n=0,1,2,3,4   ……(i)

         For n=0 in equation (i) we  get

        For n=1 in equation (i) we  get

       For n=2 in equation (i) we  get

       For n=3 in equation (i) we  get

       For n=4 in equation (i) we  get

  Hence  

Q(17) Use modified  Euler’s method  to compute y  for  x=0.05.  Given that

Result correct  to three decimal places.

Ans-

Given  equation  

Here

Take  h  =  = 0.05

By modified Euler’s  formula the  initial iteration is

)

The iteration  formula by  modified Euler’s method  is

-----(i)

     For n=0  in equation  (i)  we get

Where   and  as  above

For n=1  in equation  (i)  we get

 For n=3  in equation  (i)  we get

Since  third and  fourth approximation are equal .

Hence y=1.0526  at x =  0.05  correct  to  three decimal places.

Q(18) Using modified   Euler’s method ,  obtain a solution of  the equation

Ans-

Given  equation 

Here

By modified Euler’s  formula the  initial iteration is

  The iteration  formula by  modified Euler’s method  is

-----(i)

     For n=0  in equation  (i)  we get

Where   and  as  above

  For n=1  in equation  (i)  we get

  For n=2 in equation  (i)  we get

    For n=3  in equation  (i)  we get

Since third and fourth approximation are equal.

Hence y=0.0952 at  x=0.1

To calculate  the value  of  at x=0.2

By modified Euler’s  formula the  initial iteration is

The iteration  formula by  modified Euler’s method  is

-----(ii)

          For n=0  in equation  (ii)  we get

1814

       For n=1  in equation  (ii)  we get

1814

   Since first and second approximation   are  equal .

  Hence  y  = 0.1814  at  x=0.2

To  calculate the value  of  at x=0.3

By modified Euler’s  formula the  initial iteration is

The iteration  formula by  modified Euler’s method  is

-----(iii)

          For n=0  in equation  (iii)  we get

      For n=1  in equation  (iii)  we get

         For n=2 in equation  (iii)  we get

     For n=3 in equation  (iii)  we get

     Since third and fourth approximation are same.

      Hence y  = 0.25936 at x = 0.3

Q(19) Use Runge-Kutta method to  find y when x=1.2  in step of h=0.1 given that

Ans-

Given equation

Here

Also

By  Runge-Kutta formula for first  interval

Again

A fourth  order Runge-Kutta formula:

To  find y at 

A fourth  order Runge-Kutta formula:

Q(20) Apply Runge-Kutta fourth order method to find an approximate value of  y for  x=0.2   in step of 0.1,  if

Ans-

Given  equation 

Here 

Also

By  Runge-Kutta formula for first  interval

A fourth  order Runge-Kutta formula:

Again

A fourth  order Runge-Kutta formula:

Q(21)Using Runge-Kutta method of fourth order,   solve

Ans-

Given equation 

Here 

Also

By  Runge-Kutta formula for first  interval

)

 A fourth  order Runge-Kutta formula:

Hence at x  = 0.2 then y  = 1.196

To find the  value of y at x=0.4. In this case

 A fourth  order Runge-Kutta formula:

Hence at  x  = 0.4 then  y=1.37527