1. No category

Ordinary Differential Equations
What is Differential Equation?
An differential equation or Ordinary differential equation that involves one or more derivatives of an unknown function.
Taylor's series(point­wise method)
The Taylor's series expansion of y(s) about x=x0 is given by
y(x)=y(x0)+(x­x0)y'(x0)+
or
What is solution to DE?
A solution of a differential equation is a specific function that satisfies the equation.
What is initial value problem?
A problem with the given condtion or initial value.
Why initial value?
 x−x 0 2
y''(x0)+...
2! 
y(x0+h)=y(x1)=y1
=y0+
h
h2
h3
y0'+
y0''+ y0''+...
1 !
2 !
3 !
In general, if hn+1 and higher power of h are omitted then the error will be proportional to (n+1)th power of the step size.
That is, khn+1, k is step size.
An arbitrary contant in the solution of DE Problem 01:
indicate that a DE does not, in general, dy
determine a unique solution function. For Using Taylor's series solve =1+xy with dx
this reason, a DE is usually accompanied by y(0)=2. Find y(0.1), y(0.2) and y(0.3)
auxiliary conditions that specify the unknown functin precisely.
Solution:
In this chapter, we concentrate on one type of differential equation and one type of auxiliary condition: the initial value problem for a first order differential equation.
The standard form adopted is dy
=f x , y  with initial conditon y  x 0 =x 0
dx
the Taylor's series(1st app.,) is y1=y0+
h
h2
h3
y0'+
y0''+ y0''+...
1 !
2 !
3 !
here intial condition arex0=0, y0=2 and h=0.1 given y'=1+xy
y''=y+xy
y'''=y'+y'+xy'' = 2y'+xy''
The above system will yield the solutions in for the first approximation, substitute (x0, y0)in the above derivatives.
one of the two forms:
1. A series for y interms of powers of x, That is,
from which the value of y can be y0'=1+x0y0=1+0*2=1
obtained by direct substitution.
y0''=y0+x0y0'=2+0*1=2
2. A set of tabulated values of x and y.
y0'''=2y0'+x0y0''=2*1+0*1=2
The methods of Taylor and Picard belong to substitute these values in Taylor's series of class (1), whereas Eulter, Runge­Kutta 1st approximation
belongs to class(2).
PES College, PESIT Campus, Department of BCA
1
Ordinary Differential Equations
y1=y0+
h
h2
h3
y0'+
y0''+ y0''+...
1 !
2 !
3 !
0.1
0.12
0.13
=2+
+
∗2 +
∗2
1 !
2 !
3 !
0.1
0.12
1.4486+
2.53272+ 1 !
2 !
y3=2.2430+
0.13
3.4037
3 !
 y(0.3)=2.4011
=2.1103
 y(0.1)=2.1103
Problem 02:
nd
The Taylor's algorithm for the next(2 ) approximation is
y2=y1+
h
h2
h3
y1'+
y1''+ y1''+...
1 !
2 !
3 !
here x1=x0+h=0+0.1=0.1
Solution:
Taylor's algorithm is,
y1=2.1103
y1'=1+x1y1=1+(0.1)*2.1103=1.21103
y1=y0+
y1''=y1+x1y1'=2.1103+0.1*(1.21103)=2.2314
y1'''=2y1'+x1y1''=2*(1.21103)+0.1*2.2314=2.6452
substitute these values in Taylor's series of 2nd approximation
2
0.1
0.1
y2=2.1103+
1.21103+
2.2314+ 1 !
2 !
0.13
2.6452
3 !
 y(0.2)=2.2430
h
h2
h3
y0'+
y0''+ y0'''+...
1 !
2 !
3 !
here x0=0, y0=0, h=0.1
y'=1­2xy
y''=­2[ y­xy' ]
y'''=­2[ y'+y'+xy'' ]=­2[ xy''+2y' ]
y0'=1­2x0y0
= 1
y0''=­2[ y0­x0y0' ] = 0 y0'''=­2[ x0y0''+2y0' ] =­4
rd
The Taylor's algorithm for the next(3 ) approximation is
y3=y2+
Using Taylor's method, find y(0.1) correct to dy
2xy=1, y 0 =0
3 decimal places from dx
2
3
h
h
h
y2'+
y2''+ y2''+...
1 !
2 !
3 !
substitute above values in taylor's 1st approximation formula
y1=1+0.1+
2
3
0.1
0.1
0+ x­4 = 0.0993
2 !
3 !
here x2=x1+h=0.1+0.1=0.2
Exercise:
y1=2.2430
Using Taylor's find y at x=1.1 and 1.2 by dy 2
2
=x  y given y(1)=2.3
solving dx
y2'=1+x2y2=1.4486
y2''=y2+x2y2'=2.53272
y2'''=2y2'+x2y2''=3.4037
substitute these values in Taylor's series of 3rd approximation
PES College, PESIT Campus, Department of BCA
Ans: y(1.1)=3.1209
y(1.2)=4.6823
2
Ordinary Differential Equations
Picard's Method

2
4
3
= 1 x x − x − x
2 3∗4  3
In this method, to solve dy
=f x , y  with initial conditon y  x 0 =x 0
dx
= 1x
x
∫ f  x , y  dx
integrate y=y0+
x

0
x 2 x 4 x3
− −
2 12 3
x
rd
(3)
3 approximation y =y0+
x0
x0
x
the 1st approximation y(1)=y0+
∫ f  x 0, y 0 dx
x0
x

∫ f  x 0, y
 1
0
y =1+
x
3rd approximation y(3)=y0+
x
dx
x0
∫ f  x 0, y02 dx
x0
and so on.
∫  y02−x 2  dx
(3)
x
2nd approximation y(2)=y0+
∫ f  x 0, y02 dx
= 1 + ∫
0
[
0
1x
y(3) = 1x
 ]
x2 x 4 x3
− − −x 2 dx
2 12 3
x2 x3 x4 x5
− − −
2 6 12 60
put x=0.1 in the above equation,
Problem 03:
y(0.1)=1+(0.1)+
0.1 2 0.1 3 0.1 4 0.15
−
−
−
2
6
12
60
dy
= y−x 2 , y(0)=1 by Picard's = 1.1048249
dx
method upto the third approximation. Hence put x=0.2 in the above equation,
find the value of y(0.1) and y(0.2).
0.2 2 0.2 3 0.24 0.25
y(0.2)=1+(0.2)+
−
−
−
2
6
12
60
Soultion:
= 1.218528
Given that x0=0, y0=1 and f(x,y)=y­x2.
Solve x
st
(1)
1 approximation y =y0+
∫ f  x 0, y 0 dx
x0
∫ 1−x 2 dx
0
x
 
x3
x−
3
= 1 + = 1x−
0
x3
3
x
2nd approximation y(2)=y0+
∫ f  x 0, y01 dx
x0
x
(2)
 y =1+
∫  y 1−x 2 dx
0
x
= 1+
∫
0
[
1x−
dy
=x y 2 , y(0)=1 by dx
Picard's method upto the 2nd approximation. Hence find the value of y(0.1). [ans: 1.1156255]
1. Solve x
 y(1)=1+
Exercise:
 ]
x3
−x 2 dx
3
PES College, PESIT Campus, Department of BCA
dy
=x y , y(0)=1 by Picard's dx
method. Hence find the value of y(0.1)&y(0.2) 2. Solve dy
=2x− y , y(1)=3 by dx
Picard's method. Hence find the value of y(1.1). 3. Solve Note: if the level of approximatiion is not given, assume it according to the marks.
3
Ordinary Differential Equations
Euler's Method
The Euler's Successive approximation to solve y'=f(x,y) with the initial conditions y(x0)=y0 are as follows:­
dy
=1xy with initial condition dx
x0=0 and y0=2.
Given DE The Euler's formula for 1st approximation is y1=y0+ h f(x0, y0)
y1=y0+ h f(x0, y0)=2+ 0.1 (1+xy)
y2=y1+ h f(x1, y1), ...
=2+0.1*(1+0) = 2.1
In general, that is, x1=0.1 and y1=2.1.
yn+1=yn+ h f(xn, yn), n=0,1,2, ...
y2=y1+ h f(x1, y1)=2.1+0.1*(1+0.1*2.1)
Problem 04:
=2.221
dy
=1− y , y(0)=0 at x­0.1 and x=0.2 that is, when x1=0.2 and y1=2.221
Solve dx
using Euler's method.
y3=y2+ h f(x2, y2)=2.221+0.1*(1+0.2*2.221)
Solution:
dy
=f x , y  is The solution of dx
=2.36542
that is, y(0.3)=2.36542
yn+1=yn+ h f(xn, yn)
dy
=1− y with initial conditon Modified Euler's Method
dx
is x0=0, y0=0 and h=0.1
For a reasonable accuracy, we need to take a small value for h and hencethe process y1=y0+ h f(x0, y0)=0+0.1(1­0)=0.1
will become very slow. Because of this y2=y1+ h f(x1, y1)=0.1+0.1(1­0.1) = 0.19
restriction on h, the method is not suitable for practical use and a modification of it, hence y(0.1)=0.1
known as the modified Euler's method, y(0.2)=0.19
which gives more accurate reslts.
Given DE is Problem 05:
dy
=1xy
dx
with y(0)=2. Find y(0.1), y(0.2) and y(0.3). Using Euler's method solve Solution:
The solution of dy
=f x , y  is dx
yn+1=yn+ h f(xn, yn) PES College, PESIT Campus, Department of BCA
In Modified Euler's method, instead of approximating f(x,y) by f(x0,y0), we 1
approximating it by [f  x 0, y 0f  x 1, y 1] , 2
which is the mean of the slopes of the tangents at the points corresponding to x=x0 and x=x1. Thus we obtain
h
y1(1)= y 0 [f  x0, y 0 f  x 1, y 1]
2
In general,
h
 n
y1(n+1)= y 0 [f  x0, y 0 f  x 1, y 1 ]
2
4
Ordinary Differential Equations
Problem 06:
Using modified Euler's method solve dy
=1xy with y(0)=2. Find y(0.1), y(0.2) dx
and y(0.3). Now find y(0.2):
h
y2(1)= y 1 [f  x 1, y 1 f x 2, y 2 ] ... (2)
2
here x1=0.1, y1=2.1105
x2=x1+0.1=0.2, y2 find using Euler's method.
Solution:
That is,
The first approximation is
y2=y1+ h f(x1, y1)
h
y1(1)= y 0 [f x 0, y 0 f  x 1, y 1 ] ...(1)
2
=2.1105+0.1 * (1+x1y1)
here f(x,y)=1+xy
=2.2316
 f(x0,y0)=1+x0y0=1+0*2=1
=2.1105+0.1 * (1+0.1*2.1105)
f(x1, y1)=(1+x1y1)=(1+0.1*2.1105)=1.21105
next f(x1,y1)
f(x2, y2)=(1+x2y2)=1+0.2*2.2316=1.4463
x1=x0+h=0+0.1=0.1
substitute these value in (2), we get
y1=?. We need to find y1 using Euler's method.
y2(1)= 2.1105
The Euler's formula for 1st approximation is =2.2434
y1=y0+ h f(x0, y0)=2+ 0.1 (1+xy)
continuiing this process, 0.1
[1.211051.4463 ]
2
=2+0.1*(1+0) = 2.1
y2(2)=2.2435
that is, x1=0.1 and y1=2.1.
y2(3)=2.2434
Again, f(x1,y1)=1+x1y1=1+0.1*2.1=1.21
y2(4)=2.2434
Now apply in equation (1),
y1(1)= 2
0.1
[11.21] =2.1105
2
h
 1
y1(2)= y 0 [f  x0, y 0 f  x 1, y 1 ]
2
f  x1, y 11 =1+x1y1(1)=1+0.1*2.1105
=1.21105
y1(2)= 2
0.1
[11.21105] =2.1105
2
therefore, the value of y2=2.2434
Similarly, find y(0.3). [Ans: y3=2.4018]
Exercise:
Given dy y 1
 = y 1=1.
dx x x 2,
Evaluate y(1.3) by modified Euler's method.
Hint: here x0=1, take h=0.1 [in case of 7marks question) and find when
since differentence between two successive x1=1.1 what is y1? [ans: y1=0.9960]
approximation is less than desired accuracy, x2=1.2 what is y2? [ans: y2=0.9857]
the final value of y1=2.1105 when x1=0.1
x3=1.3 what is y3? [ans: y1=0.9715]
PES College, PESIT Campus, Department of BCA
5
Ordinary Differential Equations
Runge­Kutta Method
The Taylor's series method of solving DE numerically is handicapped by the problem of find higher order derivatives.
Problem 07:
Compute y(0.1) and y(0.2) by RK method dy
2
=xy y , y(0)=1.
dx
Euler's method is less efficient in practical problems since it requires 'h' to be small for Solution:
The formulas for the forth order RK method obtaining resonable accuracy.
are
The Runge­Kutta methods do not require the calculations of higher order derivatives and k1=h f(x0, y0)
they are designed only the function values at k
h
k2=h f x 0 , y 0 1
some selected points on the subinterval.
2
2
These methods agree with Taylor's series k
h
solution upto the terms of hr where r is the k3=h f x 0 , y 0 2
2
2
order of RK method.


Runge Kutta(RK) II order


k4=h f  x 0h , y 0k 3 
1
(k1+2k2+2k3+k4)
6
k1=h f(x0, y0)
∆y=
k2=h f(x0+h, y0+k1)
given f(x, y)= xy+y2; x0=0; y0=1 and h=0.1
y1=y0+
1
 k k 
2 1 2
k1=(0.1) (0+1)=0.1

k2=(0.1) f 0
Runge Kutta IV order (referred as RK method) k1=h f(x0, y0)


k
h
k2=h f x 0 , y 0 1
2
2
k
h
k3=h f x 0 , y 0 2
2
2


k4=h f  x 0h , y 0k 3 
∆y=
1
(k1+2k2+2k3+k4)
6
then the 1st approximatiion is given by
y1=y0+∆y
Similarly, 2nd approximatiion is given by
y2=y1+∆y, where x1=x0+h & y0 by y1
In general, yn+1=yn+∆y, PES College, PESIT Campus, Department of BCA
0.1
0.1
,1
2
2

=f(0.05,0.05)
= (0.1) (0.05*0.05+(0.05)2)=0.1155

k3=(0.1) f 0
0.1
0.1155
,1
2
2

=f(0.05, 0.05775)=0.1172
k4=(0.1) f  x 0h , y 0k 3 
= (0.1) f(0+0.1, 1+0.1172)
= 0.1360
∆y=
1
(0.1+2(0.1155)+2(0.1172)+0.1360)
6
= 0.1169
Therefore, y(0.1)=1.1169
For 2nd approximatiion, we have x1=0.1; y1=1.1169 and h=0.1
k1=0.1359
k2=0.1582
6
Ordinary Differential Equations
k3=0.1610
k4=0.1889
∆y=0.1605
Therefore, y(0.2)=1.2774
Exercies:
1. Use RK method to find y(0.1) given dy
1
=
, y(0)=1 that dx  x y 
[Ans: 1.0914]
2. Find y(0.1), y(0.2) and y(0.3) given dy
=1xy , y(0)=2
that dx
[ans: 2.1086, 2.2416, 2.3997]
dy y 1
 = y 1=1. Evaluate dx x x 2,
y(1.1) [ans: 0.9957]
3. Given PES College, PESIT Campus, Department of BCA
7