,

Here This implies that .

Differentiating, we get

     .

     .

    .

The   Taylor’s series at ,

  (1)

At in equation (1) we get

At in equation (1) we get

Here .

We have

Differentiating, we get

  implies that or

  implies that or .

   implies that

  implies that

The   Taylor’s series at ,

 Or

Here

The   Taylor’s series

       .

                                            with h=0.1

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.

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  

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

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:

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

Given second order differential equation is

Let   then above equation reduces to

     Or

(say)

Or .

By RungeKutta Method we have

A fourth  orderRungeKutta formula:

for

Given differential equation are

Let

And

Also

 By RungeKutta Method we have

A fourth  orderRungeKutta formula:

And  

.

Where

To get the first approximation-

We put y = 0 in f(x, y),

Giving-

In order to find the second approximation, we put y  = in f(x,y)

Giving-

And the third approximation-

Now determine the starting  values of the Milne’s method from equation  (1), by choosing h = 0.2

Now using the predictor-

X = 0.8

,      

And the corrector-

,                 ................(2)

Now again using corrector-

Using predictor-

X = 1.0,  

,      

And the corrector-

,      

Again using corrector-

,       which is same as before

Hence

,  y(0) = 1

Here we have-

Here

So that-

Thus

To find y(0.2)-

Here

Thus,

Y(0.2) = 

To find y(0.3)-

Here

Thus,

Y(0.3) = 

Now the starting values of Adam’s method with h = 0.1-

Using predictor-

Using corrector-

Hence