Mathematical Applications
By Matlab
3rd Day
Dr. Wael Khedr
CSI Dept.
Outline of Third Day:
Solve of System of Linear Equations
 Integral (Symbolic & Numeric)
 Differentiation
 Calculus ( Root , Limitation , Polynomial)
 Graphics (Plots: ezmech , ezsurface)

Third Day
2
Solve of System of Linear Equations
• Let’s use the matrix approach to solve the
system of linear equations for example:
• The first step is to write the equations in matrix
form:
• Next, we need to find the inverse of the A matrix:
• To find x and y, multiply the inverse of A by C:
Third Day
4
Graphical Solution
x=linspace(-10, 5, 20);
>> y =(14-2*x)/3;
>> y2= 28+4*x;
>> hold on
>> plot(x,y, 'r')
>> plot(x,y2, 'b')
Third Day
5
MATLAB Solution
>> A = [2 3;-4 1];
>> C = [14;28];
>> X = inv(A)*C
X =
-5
8
Or : >> X = A\C
-5
8
Third Day
6
Example (2)

Consider these two equations:

MATLAB:
>> A = [2 3;4 6];
>> C = [12;28];
>> X = inv(A)*C
Warning: Matrix is singular to working precision.
X =
Inf
Inf
What’s Wrong?

Solve with substitution:
Second equation in terms of y:

Substitute into first equation:

What’s Wrong?

Solve:

No value of x will satisfy this equation
Graphical Solution
Example – 3 Equations

Write these equations in matrix form:
Third Day
Example – 3 Equations

MATLAB solution:
>> A = [12 32 -10; 0 2 3; 7 16 5];
>> C = [-30; 11; 42];
>> X = inv(A)*C
X =
7.0000
-2.0000
5.0000
>>
Third Day
Interpretation of Solution




The graphical solution shows that the two
equations define parallel lines
Since parallel lines never intersect, there is
no point that satisfies both equations
Therefore, there is no solution to these
equations
Note that MATLAB solution will result in the
same error – the inverse of the coefficient
matrix does not exist
Summary


If the inverse of the coefficient matrix exists,
then there is a solution, and that solution is
unique
If the inverse does not exist, then there are
two possibilities:


The equations are incompatible, and so there are
no solutions, or
At least two of the equations are redundant, and
so there are more unknowns than unique
equations. Therefore, there are an infinite number
of solutions
Third Day
Symbolic Calculations in Matlab:
 You must first give MATLAB a list of the variable and function
names that will appear in the symbolic expressions you will be
working with.
>> syms x y z a;
 Determine Integral of function :
z   ydx   x e dx
2 x
>> syms x
>> y = x^2*exp(x)
>> z = int(y)
z = x^2*exp(x)-2*x*exp(x)+2*exp(x)
Third Day
15
Substitute Numerical values
 If you would like to substitute numerical values into a symbolic function,
use the function subs as in the following way.
𝒇 𝒙,𝒚 = 𝒙𝟐 + 𝐬𝐢𝐧 𝒚 ‫إذا أردت التعويض لحساب ناتج الدالة لقيم المتغيرات يمكن حسابها مباشرة‬
>> f = x^2 + sin(y);
>> subs( f, { x , y }, {1 , pi/2 })
Ans = 2
>> int(f,x)
X^3/3 + x sin(y)
>> int(f, y)
y*x^2 – cos(y)
% x ‫ بالنسبة‬f ‫تكامل الدالة‬
% y ‫ بالنسبة‬f ‫تكامل الدالة‬
% x ‫ بالنسبة‬f ‫تكامل محدود الدالة‬
( (a-1)*(a^2 + a + 3*sin(y) +1 ) ) / 3
>> int(f,x 1,a)
Third Day
16
Numerically evaluate Single integral
(1)Trapezoidal : trapz(x,y)
𝝅
Example:
𝐬𝐢𝐧
𝒙
𝟎







dx
%Create a domain vector, X.
>> X = 0: pi /100 : pi;
%Calculate the sine of X and store the result in Y.
>>Y = sin(X);
%Integrate the function values contained
in Y using trapz.
>> Q = trapz(X,Y)
Q = 1.9998
Third Day
17
(2) Using Integral( ) function:
Create the function 𝒇 𝒙 = 𝒆
−𝒙𝟐
+ 𝒍𝒐𝒈 𝒙𝟐 .
%Evaluate the integral from x=0 to x=Inf.
Solution:
>> fun = @(x) exp(-x.^2).*log(x).^2;
>>q = integral(fun , 0 , Inf)
q = 1.9475
Third Day
18
Numerically evaluate Double integral
q = integral2(fun, xmin ,xmax , ymin , ymax)
Example:
𝑦𝑚𝑎𝑥 1
1
dx dy
0
0 𝑥+𝑦(1+𝑥+𝑦)
where: ymax= 1-x
Solution:
>> fun = @(x,y) 1./( sqrt(x + y) .* (1 + x + y).^2 )
>> ymax = @(x) 1 - x
>> q = integral2(fun,0,1,0,ymax)
q = 0.2854
Third Day
19
 Determine Differentiation of function :
>> f = x^2 + sin(y);
>> diff(f,x)
% x ‫ بالنسبة‬f ‫تفاضل الدالة‬
Ans= 2*x
>> diff(f,y)
Ans= cos(y)
% y ‫ بالنسبة‬f ‫تفاضل الدالة‬
 second derivative of f with respect to x
>>diff( f, x, 2);
Ans= 2
20
 Limitation:
𝒍𝒊𝒎 𝒇(𝒙)
𝒙→𝟎
≫ 𝒍𝒊𝒎𝒊𝒕 𝒇, 𝒙, 𝟎
Ans= sin(y)
 We can solve the system of equations x + y = 0 and
x + 2y = −1
using the following command
>> f1 = x + y;
>> f2 = x + 2*y + 1;
>> xy_solution = solve(f1,f2,’x,y’);
>> x_solution = xy_solution.x
1
>> y_solution = xy_solution.y;
-1
Third Day
21
Finding roots of any function:.

>> syms x
>> ff =2*x^2 + 4*x - 8;
>> solve(ff , x)

ans =

5^(1/2)-1
-1-5^(1/2)
>> double(ans)
-3.2361
1.2361


 Consider the following Polynomial

>> ff =[2 5 -8];

>>roots( f )
Third Day
22
Thank You…
23
To demonstrate these plot types create a
symbolic version of “peaks”
We broke this function up into three parts to make it
easier to enter into the computer. Notice that there are
no “dot” operators used in these expressions, since they
are all symbolic.
When we created the same plots using a
standard MATLAB approach it was
necessary to define an array of both x and y
values, mesh them together, and calculate
the values of z based on the two
dimensional arrays. The symbolic plotting
capability contained in the symbolic toolbox
makes creating these graphs much easier.
All of these graphs can be annotated using
the standard MATLAB functions such as
title, xlabel, text, etc.
These contour plots are a
two-dimensional
representation of the threedimensional peaks function
The polar graph requires us
to define a new function
Any of these ezplot
graphs can handle
parameterized
equations