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
