Unit-4

Solution of differential Equations

The general first order differential equation

…. (1)

With the initial condition … (2)

In general, the solution of first order differential equation in one of the two forms:

a) A series for y in terms of power of x, from which the value of y can be obtained by direct solution.

b) A set of tabulated values of x and y.

The case (a) is solved by Taylor’s Series or Picard method whereas case (b) is solved by Euler’s, Runge Kutta Methods etc.

Picard’s method-

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.

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

Sol.

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)

Example: Obtain the Picard’s second approximation for the given initial value problem-

Find y (1).

Sol.

The first approximation will be-

Replace y by , we get-

The second approximation is-

The third approximation-

It is very difficult to solve the integration-

This is the disadvantage of the method.

Now we get from the second approximation-

At x = 1-

Euler’s method:

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

Example1: Use Euler’s method to find y (0.4) from the differential equation

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.

Example2: Using Euler’s method solve the differential equation for y at x=1 in five steps

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

Example3: 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

Modified Euler’s Method:

Instead of approximating as in Euler’s method. In the modified Euler’s method, we have the iteration formula

Where is the nth approximation to .The iteration started with the Euler’s formula?

Example1: Use modified Euler’s method to compute y for x=0.05. Given that

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.

Example2: Using modified Euler’s method, obtain a solution of the equation

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

Key takeaways-

1. Picard’s method-

2. Euler’s method:

, n=0,1, 2...

3. Modified Euler’s Method:

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 apowwer 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).

Example1: Solve, using Taylor’s series method and compute .

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

Example2: Using Taylor’s series method, find the solution of

At ?

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

Example3: Solve numerically, start from and carry to using Taylor’s series method.

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

.

Runge-kutta methods-

This method is more accurate than Euler’s method.

Consider the differential equation of first order

Let be the first interval.

A second order Runge Kutta formula

Where

Rewrite as

A fourth order Runge Kutta formula:

Where

Example1: Use Runge Kutta method to find y when x=1.2 in step of h=0.1 given that

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:

Example2: Apply Runge Kutta fourth order method to find an approximate value of y for x=0.2 in step of 0.1, if

Here

Also

By Runge Kutta formula for first interval

A fourth order Runge Kutta formula:

Again

A fourth order Runge Kutta formula:

Example3: Using Runge Kutta method of fourth order, solve

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

Simultaneous equation using Runge Kutta method of 2 orders:

The second order differential equation

Let then the above equation reduces to first order simultaneous differential equation

Then

This can be solved as we discuss above by Runge Kutta Method. Here for and for .

A fourth order Runge Kutta formula:

Where

Example1: Using Runge Kutta method of order four, solve to find

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:

Example2: Using Runge Kutta method, solve

for correct to four decimal places with initial condition .

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

.

Example3: Solve the differential equations

for

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

.

Key takeaways-

Runge-kutta methods-

A second order Runge Kutta formula

Where

A fourth order Runge Kutta formula:

Where

Milne’s predictor corrector formula-

For given dy/dx = f (x, y) and y = and x = , to find the value of y for x = , by using Milne’s method,

We follow the steps given below-

The value being given, here we calculate-

By Taylor’s series or Picard’s method.

Now we calculate-

Then to find

We substitute Newton’s forward interpolation formula-

In the relation-

By putting x = , dx = h dn

Neglecting fourth and higher order differences and expressing in terms of the function values, we get-

This is called a predictor.

Now having found we obtain a first approximation to

Then the better value of is found by simpson’s rule as-

Which is called corrector.

Then an improved value of is computed and again corrector is applied to find a better value of .

We continue this step until remains unchanged.

Once and are obtained to desired degree of accuracy,

is found from the predictor as-

And

is calculated.

Then the better approximation to the value of we get from the corrector as-

We repeat until becomes stationary and we proceed to calculate .

This is called the Milne’s predictor-corrector method.

Adams - Bashforth predictor and corrector formula-

This is called Adams - Bashforth predictor formula.

And

This is called Adams - Bashforth corrector formula.

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

Sol.

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

Example: Solve the initial value problem , y(0) = 1 to find y (0.4) by using Adams-Bashforth method.

Starting solutions required are to be obtained using Runge-Kutta method of order 4 using step value h = 0.1

Sol.

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

Key takeaways-

References