Unit-4

Solution of differential equations

Q1) Explain Picard’s method.

A1) Let us suppose the first order equation-

It is required to find out that particular solution of equation (1) which assumes the value when ,

Now integrate (1) between limits, we get-

This is equivalent to equation (1),

For it contains the not-known y under the integral sign,

As a first approximation to the solution, put in f (x, y) and integrate (2),

For second approximation-

 Similarly-

And so on.

Q2) Find the value of y for x = 0.1 by using Picard’s method, given that-

A2) We have-

For first approximation, we put y = 1, then-

Second approximation-

We find it very hard to integrate.

Hence, we use the first approximation and take x = 0.1 in (1)

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

A4) 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 

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

Result correct to three decimal places.

A5) 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) Solve, using Taylor’s series method and compute .

A7) 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

Q8) Using Taylor’s series method, find the solution of

At ?

A8) 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

Q9) Write down the fourth order Runge kutta method.

A9) A fourth order Runge Kutta formula:

Where 

Q10) Use Runge Kutta method to find y when x=1.2 in step of h=0.1 given that

A10) 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:

Q11) Apply Runge Kutta fourth order method to find an approximate value of y for x=0.2   in step of 0.1, if

A11) Given equation 

Here 

Also

By Runge Kutta formula for first interval

A fourth order Runge Kutta formula:

Again

A fourth order Runge Kutta formula:

Q12) Using Runge Kutta method of fourth order, solve

A12) 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

Q13) Using Runge Kutta method of order four, solve   to find

A13) 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:

Q14) Using Runge Kutta method, solve

  for correct to four decimal places with initial condition .

A14) 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:

And

.

Q15) Solve the differential equations

       for

A15) Using four order Runge Kutta method with initial conditions

Given differential equation are

Let

And

Also

By Runge Kutta Method we have

A fourth order Runge Kutta formula:

And  

 .

Q16) What are Adams - Bashforth predictor formula and Adams - Bashforth corrector formula.

A16) Adams - Bashforth predictor and corrector formula-

This is called Adams - Bashforth predictor formula.

And

This is called Adams - Bashforth corrector formula.

Q17) Find the solution of the differential equation   in the range for the boundary conditions y = 0 and x = 0 by using Milne’s method.

A17) By using Picard’s method-

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

Q18) A second order Runge Kutta formula is-

A18) A second order Runge Kutta formula  

Where

Rewrite as 

Q19) Explain Taylor’s series method.

A19) Taylor’s Series Method:

The general first order differential equation

…. (1)

With the initial condition … (2)

Let be the exact solution of equation (1), then the Taylor’s series for around is given by

   (3)

If the values of  are known, then equation (3) gives power series   for y. By total derivatives   we have

,

And other higher derivatives of y.  The method can easily be extended to   simultaneous   and higher –order differential equations. In   general,

Putting   in these above results, we can obtain the values of finally, we substitute these values of  in equation (2) and obtain the approximate value of y; i.e., the solutions of (1).

Q20) Explain Euler’s method.

A20) In this method the solution is in the form of a tabulated values

Integrating both side of the equation (i) we get

Assuming that  in  this gives Euler’s formula

In general formula

, n=0,1, 2….

Error estimate for the Euler’s method