Q1) Solve, using Taylor’s series method and compute .

A1)

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

Q2) Using Taylor’s series method, find the solution of

At ?

A2)

Here

At implies that or or

Differentiating, we get

implies that or .

implies that or

implies that or

implies that or

The   Taylor’s series at ,

  (1)

At   in equation (1) we get

At in equation (1) we get

Q3) Use Euler’s method to find y(0.4) from the differential equation

    with h=0.1

A3)

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.

Q4) Given  with the initial condition y=1 at   x=0. Find y for x=0.1 by Euler’s   method (five steps).

A4)

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  

Q5) Use modified Euler’s method to compute y for x=0.05.  Given that

A5)

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 third and fourth approximation are equal.

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

Q6) Using modified Euler’s method, obtain a solution of the equation

A6)

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

Q7) Use Runge Kutta method to find y when x=1.2 in step of h=0.1 given that

A7)

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:

Q8) Using Runge Kutta method of order four, solve   to find

A8)

Given second order differential equation is

Let   then above equation reduces to

Or

(say)

Or .

By Runge Kutta Method we have

A fourth order Runge Kutta formula:

Q9) Use Jacobi’s method to solve the system of equations:

A9)

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

Q10) Use Jacobi’s method to solve the system of the equations

A10)

Rewrite the given equations

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

Q11) Solve the following system of equations

  By Gauss-Seidel method.

A11)

Rewrite the given system of equations as

Let the initial approximation be

Thus the required solution is