Unit-6

Interpolation, Numerical integration, Solution of ordinary differential equations

Question-1: Using Newton’s divided difference formula, find the values of from the following table:

We construct the divided difference table is given by:

By Newton’s Divided difference formula

.

Now, putting in above we get

.

Question-2: Find a polynomial satisfied by , by the use of Newton’s interpolation formula with a divided difference.

Here

We will construct the divided difference  table:

By Newton’s divided difference formula

.

This is the required polynomial.

Question-3: Given  find   , by using  Newton forward interpolation method.

Let  , then

The table of forward finite difference is given below:

By Newton forward difference method

Here initial value = 45, difference of  interval  h = 5 and  the value to be calculated at x=52.

By Formula

Question-4: Find from the following table:

 Consider the backward difference method

  Here

By Newton backward difference formula

Question-5: The following are the marks obtained by492  candidate  in a certain examination

Find  out the number of candidates:

a)     Who secured more than 48  but not more than  50 marks?

b)    Who secured less than 48  but not less than 45 marks?

Consider the  forward difference table given below:

               Here 

 By Newton Forward Difference formula

f

a)     No. of  candidate secured more than 48  but  not more than  50 marks

b)    No. of  candidate secured less than 48 but  not  less  than 45 marks

Question-6: Use the inverse   Lagrange’s method to find the root of the equation , give data

Here , we have the  data

Also.

Lagrange’s inverse interpolation formula is given by

Thus the approximate root of the given equation is .

Question-7: Given that

Find at .

Here the first derivative is to be calculated at the beginning of the table, therefore  forward difference  formula will be used

Forward difference  table is given below:

By Newton’s forward differentiation formula for differentiation

Here 

Question-8: From the following  table of values of x and y find   for 

  Here the value of the derivative is to be calculated at the beginning of the table.

Forward difference table is given by

From Newton’s forward difference formula for differentiation we get

Here

     =0.48763

Question-9: Given that

Find ?

Backward difference table:

Newton’s Backward formula for differentiation

Here

Question-10: Compute the value of   ?

Using the trapezoidal rule with h=0.5, 0.25, and 0.125.

Here

For h=0.5, we construct the data table:

By Trapezoidal rule

For h=0.25, we construct the data table:

By Trapezoidal rule

For h = 0.125, we construct the data table:

By Trapezoidal rule

[(1+0.5)+2(0.98461+0.94117+0.87671+0.8+0.71910+0.64+0.56637)]

Question-11: Estimate the value of the integral

by Simpson’s rule with 4 strips and 8 strips respectively.

For n=4, we have

Construct the data table:

By Simpson’s Rule

For n = 8,  we have

By Simpson’s Rule

Question-12: Using Simpson’s 1/3 rule with h = 1, evaluate

For h = 1, we construct the data table:

By Simpson’s Rule

            = 177.3853

Question-13: Evaluate

By Simpson’s 3/8 rule.

Let us divide the range of the interval [4, 5.2] into six equal parts.

For h=0.2, we construct the data table:

By Simpson’s 3/8 rule

            = 1.8278475

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

with h=0.1

Given  equation

          Here

We break the interval into 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.

Question-15: Given with the initial condition  y=1 at   x=0. Find y for x=0.1 by  Euler’s method(five steps).

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  

Question-16: Use modified  Euler’s method to compute y  for  x=0.05.  Given that

The result correct to three decimal places.

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 the third and fourth approximation is equal.

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

Question-17: Use the Runge-Kutta method to  find y when x=1.2  in the step of h=0.1 given that

Given equation

Here

Also

By  Runge-Kutta formula for first  interval

Again

A fourth  orderRunge-Kutta formula:

To  find y at 

A fourth  orderRunge-Kutta formula:

Question-18: Using the Runge-Kutta method of fourth-order,   solve

Given equation 

Here 

Also

By  Runge-Kutta formula for first  interval

)

 A fourth  orderRunge-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  orderRunge-Kutta formula:

Hence at  x  = 0.4 then  y=1.37527

Question-19: UsingRungeKutta method of order four, solve   to find

Given second order differential equation is

Let   then above equation reduces to

     Or

(say)

Or .

By RungeKutta Method we have

A fourth  orderRungeKutta formula:

Question-20: Solve the differential equations

for

Using the four order RungeKutta method with initial conditions

Given differential equation are

Let

And

Also

 By RungeKutta Method we have

A fourth  orderRungeKutta formula:

And  

.