Numerical Methods, Matlab, Solvers, Programming, Computational methods

1. NM [DR D.P.BHASKAR, SANJIVANI COLLEGE OF ENGINEERING KOPARGAON]
1
Numerical Methods by MATLAB
Mechanical Engineering
Prepared by
Dr.D.P.Bhaskar
Department of Mechanical Engineering
Sanjivani College of Engineering, Kopargaon
Maharastra
(An Autonomous Institute SP Pune University)

2. NM [DR D.P.BHASKAR, SANJIVANI COLLEGE OF ENGINEERING KOPARGAON]
2
Unit 1 Roots of Equation
%Bisection Method
clc
clear all
f=inline('x‐cos(x)'); acc=0.001;
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
for i=1:100
x1=input('Enter new x1');
x2=input('Enter new x2');
if(f(x1)*f(x2)<0) break
end
end
fprintf('n‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐n');
fprintf('I x1 x2 x3 Acc Error)');
fprintf('n‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐');
for i=1:100
x3=(x1+x2)/2;
b=abs(x1‐x2); c=(f(x1)*f(x3));
fprintf('n%d %.4f %.4f %.4f %.4f %.4f ',i,x1,x2,x3,b,c);
if(b<acc) break
else if(c>0) x1=x3; else x2=x3; end
end
end
fprintf('n‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐');
fprintf('n Root=%f',x3);

3. NM [DR D.P.BHASKAR, SANJIVANI COLLEGE OF ENGINEERING KOPARGAON]
3
%Newton Raphson Method
clc
clear all
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
f=inline('x.^3‐cos(x)*cos(x)');
fd=inline('2*cos(x)*sin(x) + 3*x.^2');
acc=0.001; x1=1;
fprintf('n‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐n');
fprintf('I x1 f(x1) fd(x1) x2 Acc');
fprintf('n‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐');
abs(f(x1))
abs(fd(x1))
for i=1:100
x2=x1‐f(x1)/fd(x1);
b=abs(x1‐x2);
fprintf('n%d %.4f %.4f %.4f %.4f %.4f',i,x1, f(x1), fd(x1),x2,b);
if(b <acc) break
else x1=x2;
end
end
fprintf('‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐');
fprintf('n Root=%f',x2);

4. NM [DR D.P.BHASKAR, SANJIVANI COLLEGE OF ENGINEERING KOPARGAON]
4
Unit 2 Simultaneous Equation
%2.0 GEM
clc
clear all
n=3; m=n+1;
a=[1 2 2 7;2 ‐4 1 ‐5;1 1 2 5];
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
for k=1:(n‐1) %Ip=Jp=1 2 3
for i=k+1:n
PVT= a(k,k); TL= a(i,k); p=TL/PVT;
for j=k:m
a(i,j)=a(i,j)‐p*a(k,j);
end
end
end
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
for i=n:‐1:1
sum=0.0;
for j=i+1:n
sum=sum+a(i,j)*x(j);
end
x(i)=(a(i,n+1)‐sum)/a(i,i);
end
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
for i=1:n
fprintf('x(%d)=%0.2fn',i,x(i));
end

5. NM [DR D.P.BHASKAR, SANJIVANI COLLEGE OF ENGINEERING KOPARGAON]
5
% GEM with Partial Pivoting
clc
clear all
n=3; m=n+1;
a=[1 2 2 7;2 ‐4 1 ‐5;1 1 2 5];
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
for k = 1:n‐1
for i = k+1:n
if (abs(a(k,k)) < abs(a(i,k)))
a([k i],:) = a([i k],:);
end
end
end
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
for k=1:(n‐1) %Ip=Jp=1 2 3
for i=k+1:n
PVT= a(k,k); TL= a(i,k); p=TL/PVT;
for j=k:m
a(i,j)=a(i,j)‐p*a(k,j);
end
end
end
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
for i=n:‐1:1
sum=0.0;
for j=i+1:n
sum=sum+a(i,j)*x(j);
end
x(i)=(a(i,n+1)‐sum)/a(i,i);
end
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
for i=1:n
fprintf('x(%d)=%0.2fn',i,x(i));
end
%SOLVER
clc
clear all
a=[1 2 2 ;2 ‐4 1 ;1 1 2];
b=[4;5;6];
x=inv(x)*b

6. NM [DR D.P.BHASKAR, SANJIVANI COLLEGE OF ENGINEERING KOPARGAON]
6
%GAUSS SEIDAL METHOD
clc
clear all
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
f1=inline('(13‐4*y‐3*z)/2');
f2=inline('(16‐3*x‐z)/‐6');
f3=inline('(9‐x‐3*y)/2');
x1=0; y1=0; z1=0; acc=0.001;
fprintf('n‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐');
fprintf('n I X Y Z ');
fprintf('n‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐');
for i=1:100
x2=f1(y1,z1); y2=f2(x2,z1); z2=f3(x2,y2);
fprintf('n %d %f %f %fn',i,x2,y2,z2);
a=abs(x1‐x2); b=abs(y1‐y2); c=abs(z1‐z2);
if (a<acc && b<acc && c<acc) break
else x1=x2; z1=z2; y1=y2;
end
end
fprintf('‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐');

7. NM [DR D.P.BHASKAR, SANJIVANI COLLEGE OF ENGINEERING KOPARGAON]
7
Unit 3 Curve Fitting
%3.0 Curve fitting F Line Yf=C0+C1(X)
clc
clear all
%‐‐‐‐‐‐‐‐‐‐‐‐‐Direct input of Data‐‐‐‐‐‐‐‐‐‐‐
x=[1 2 3 4];
y=[2 6 12 20];
n=length(x);
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐OR‐User UInput of data‐‐‐‐‐
% n=input('Enter n=');
% for i=1:n
% fprintf('Enter x(%d)=',i);
% x(i)=input('');
% fprintf('Enter y(%d)=',i);
% y(i)=input('');
% end
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
s1=sum(x); s2=sum(x.^2); s3=sum(y);
s4=sum(x.*y);
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
d=[n s1;
s1 s2];
d1=[s3 s1;
s4 s2];
d2=[n s3;
s1 s4];
d=det(d); d1=det(d1); d2=det(d2);
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
c0=d1/d; c1=d2/d;
fprintf('BEST FIT ,Yn=%0.3f+%0.3f(X)n',c0,c1)
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
yn=c0+c1*x;
plot(x,y,'*',x,yn)
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐

8. NM [DR D.P.BHASKAR, SANJIVANI COLLEGE OF ENGINEERING KOPARGAON]
8
% Quadratic FITTING Yf=C0+C1(X)+C2(x^2);
clc
clear all
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
x=[1 2 3 4 5 6 7 8 9];
y=[2 6 7 8 10 11 11 10 9];
n=9;
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
x2=x.^2; x3=x.^3; x4=x.^4; xy=x.*y; x2y=(x.^2).*y;
s1=sum(x); s2=sum(x2); s3=sum(x3); s4=sum(x4); s5=sum(y); s6=sum(xy); s7=sum(x2y);
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
d=[n s1 s2;
s1 s2 s3
s2 s3 s4];
d1=[s5 s1 s2;
s6 s2 s3
s7 s3 s4];
d2=[n s5 s2;
s1 s6 s3
s2 s7 s4];
d3=[n s1 s5;
s1 s2 s6
s2 s3 s7];
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
d=det(d); d1=det(d1); d2=det(d2); d3=det(d3); c0=d1/d; c1=d2/d; c2=d3/d;
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
fprintf('BEST FIT ,Yn=%0.3f+%0.3f(X)+%0.3f(x^2) /n',c0,c1,c2)
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
yn=c0+c1*x+c2*x2;
plot(x,y,'*',x,yn,'r')
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐

9. NM [DR D.P.BHASKAR, SANJIVANI COLLEGE OF ENGINEERING KOPARGAON]
9
%Power Equation
clc
clear all
%‐‐‐‐‐‐‐‐‐‐‐‐‐Power Equation‐‐‐‐‐‐‐‐‐‐‐
%PV^Gamma=C
p=[0.5 1.0 1.5 2.0 2.5 3.0];
v=[1.62 1.0 0.75 0.62 0.52 0.46];
n=6;
y=log10(p);
x=log10(v);
xy=x.*y;
x2=x.^2;
s1=sum(x); s2=sum(x2); s3=sum(y); s4=sum(xy);
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
d=n*s2‐s1*s1;
d1=s3*s2‐s4*s1;
d2=n*s4‐s1*s3;
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
c0=d1/d; c1=d2/d;
gamma=‐c1;
c=exp(c0);
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
fprintf('PV^%0.3f=%f/n',gamma,c)
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
%plot(p,v);
for i=1:n
fprintf('%f %f %f %f %f %fn',v(i),p(i),x(i),y(i),x2(i),xy(i));
end

10. NM [DR D.P.BHASKAR, SANJIVANI COLLEGE OF ENGINEERING KOPARGAON]
10
%Exponential
clc
clear all
%‐‐‐‐‐‐‐‐‐‐‐‐‐Exponential‐‐‐‐‐‐‐‐‐‐‐
%y=ae^bx
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
x=[2 4 6 8];
y=[25 38 56 84];
n=length(x);
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
y1=log(y);
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
x2=x.^2; xy1=x.*y1;
s1=sum(x); s2=sum(x2); s3=sum(y1); s4=sum(xy1);
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
d=n*s2‐s1*s1;
d1=s3*s2‐s4*s1;
d2=n*s4‐s1*s3;
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
c0=d1/d; c1=d2/d;
a=exp(c0);
b=c1;
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
fprintf('y=%0.3f e^(%0.3fx) n',a,b)
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
for i=1:n
fprintf('%f %f %f %f %f n',x(i),y(i),y1(i),x2(i),xy1(i));
end
%Solver
clear all
x=[19 25 30 36 40 45 50];
y=[76 77 79 80 82 83 85];
d=1; %d=1 (line) d=2(parabola)
c=polyfit(x,y,d)
% yn = polyval(c, x);
yn=c(2)+c(1)*x;
plot(x,y,'*',x,yn)

11. NM [DR D.P.BHASKAR, SANJIVANI COLLEGE OF ENGINEERING KOPARGAON]
11
Unit 4 ODE & PDE
%EULERS METHOD
clc
clear all
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
f=inline('x*x*y+y*y');
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
x1=input('ENTER x1='); y1=input('ENTER y1=');
xn=input('ENTER xn='); n=input('ENTER n=');
h=(xn‐x1)/n;
fprintf('n‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ ');
fprintf('n X Y ');
fprintf('n‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ ');
fprintf('n %0.3f %0.3f',x1,y1);
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
for i=1:n
x2=x1+h;
y2=y1+h*f(x1,y1);
fprintf('n %0.3f %0.3f',x2,y2);
x1=x2; y1=y2;
end
fprintf('n‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ ');

12. NM [DR D.P.BHASKAR, SANJIVANI COLLEGE OF ENGINEERING KOPARGAON]
12
%RK4
clc
clear all
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
f=inline('x*x*y+y*y');%(Sample function)
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
x1=input('ENTER x1='); y1=input('ENTER y1='); % BCns
xn=input('ENTER xn='); n=input('ENTER n=');
h=(xn‐x1)/n;
fprintf('n‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ ');
fprintf('n X Y ');
fprintf('n‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ ');
fprintf('n %0.3f %0.3f',x1,y1);
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
for i=1:n
s1=h*f(x1,y1); s2=h*f(x1+h/2,y1+s1/2);
s3=h*f(x1+h/2,y1+s2/2); s4=h*f(x1+h,y1+s3);
s=(s1+2*s2+2*s3+s4)/6;
y2=y1+s; x2=x1+h;
fprintf('n %0.3f %0.3f',x2,y2);
x1=x2; y1=y2;
end
fprintf('n‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ ');
%SOLVER
clc
clear all
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
f=inline('x+y');
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
x1=0; xn=0.2; y1=2; h=0.1;
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
[x,y]=ode23(f,[x1:h:xn],y1)
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐

13. NM [DR D.P.BHASKAR, SANJIVANI COLLEGE OF ENGINEERING KOPARGAON]
13
%Simultaneous ODE %RK2
clc
clear all
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
f1=inline('x+(y*z)');%(Sample function)
f2=inline('(x*x)‐(y*y)');%(Sample function)
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
x1=0; y1=1; z1=0; xn=0.2;% BCns Direct Input
n=2; % Number of strips Direct Input
h=(xn‐x1)/n;
fprintf('n‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ ');
fprintf('n X Y ');
fprintf('n‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ ');
fprintf('n %0.3f %0.3f',x1,y1);
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
for i=1:n
s1=h*f1(x1,y1,z1); k1=h*f2(x1,y1,z1);
s2=h*f1(x1+h,y1+s1,z1+k1); k2=h*f2(x1+h,y1+s1,z1+k1);
s=(s1+s2)/2; k=(k1+k2)/2;
y2=y1+s; z2=z1+k; x2=x1+h;
fprintf('n %0.3f %0.3f %0.3f',x2,y2,z2);
x1=x2; y1=y2; z1=z2;
end

14. NM [DR D.P.BHASKAR, SANJIVANI COLLEGE OF ENGINEERING KOPARGAON]
14
%Elliptical Laplace Equation
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
clc
clear all
rh=0;lh=100;top=100;bot=0;itr=4;
n=3;m=3;h=1;k=1;
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
for i=1:n+1
for j=1:m+1
a(i,j)=0;
end
end
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
for i=2:(n)
a(i,1)=bot; a(i,m+1)=top;
end
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
for j=2:(m)
a(1,j)=lh; a(m+1,j)=rh;
end
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
for k=1:itr
for i=2:n
for j=2:m
a(i,j)=(a(i‐1,j)+ a(i+1,j)+a(i,j‐1)+ a(i,j+1))/4;
end
end
end
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
disp(a)

15. NM [DR D.P.BHASKAR, SANJIVANI COLLEGE OF ENGINEERING KOPARGAON]
15
Unit 5 Interpolation
%LAGRANGES Method
clc
clear all
xp=1.1;
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
x=[1 1.2 1.3 1.5];
y=[1 1.0954 1.1402 1.2247];
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐OR‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
n=length(x);
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
sum=0;
for i=1:n
term=y(i);
for j=1:n
if(i~=j)
term=term*(xp‐x(j))/(x(i)‐x(j))
end
end
sum=sum+term
end
fprintf('x=%.3f,yp=%.5f',xp,sum)
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
% n=input('Enter n=');
% for i=1:n
% fprintf('Enter x(%d)=',i);
% x(i)=input('');
% fprintf('Enter y(%d)=',i);
% y(i)=input('');
% end

16. NM [DR D.P.BHASKAR, SANJIVANI COLLEGE OF ENGINEERING KOPARGAON]
16
%Newton Forward Method
clc
clear all
x=[0 1 2 3 4 5]; y=[1 2 9 28 65 126];
% x=input('Enter [x]='); y=input('Enter [y]=');
n=length(x); h=x(2)‐x(1);
xp=input('xp=');
%‐‐‐‐‐‐‐‐‐‐‐‐‐Table Calculations‐‐‐
for j=1:(n‐1)
for i=1:(n‐j)
if(j==1) f(i,j)=y(i+1)‐y(i);
else f(i,j)=f(i+1,j‐1)‐f(i,j‐1);
end
end
end
fprintf('n‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐NF TABLE ‐‐‐‐‐‐‐‐')
for i=1:n
fprintf('n %0.2f t %0.2f',x(i),y(i));
for j=1:(n‐i)
fprintf('tt %0.2f',f(i,j));
end
end
fprintf('‐‐‐‐‐‐‐‐‐‐‐‐‐‐TABLE ENDS‐‐‐‐‐‐‐')
sum=y(1);
for i=1:(n‐1)
prod=f(1,i);
for j=1:i
prod=prod*(xp‐x(j))/(j*h);
end
sum=sum+prod;
end
fprintf(' YP=%f',sum);

17. NM [DR D.P.BHASKAR, SANJIVANI COLLEGE OF ENGINEERING KOPARGAON]
17
Unit 6 Integration
%TRAP
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
clc
clear all
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
f=inline('x.^3');
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
a=input('Enter Lower Limit,a='); b=input('Enter Upper Limit,b=');
ns=1; % depends on method
na=input('Enter Number of application ,na=');
n=na*ns;
h=(b‐a)/n;
sum=0;
x1=a;
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
for i=1:na
sum=sum+f(x1)*1+f(x1+h)*1; % depends on method
x1=(x1+h);
end
Area=sum*h/2; % depends on method
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
fprintf('n Area of integration=%0.2f',Area)
% Solver
clc
clear all
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
f =inline('x.^3');
a=2;b=3;
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
Q = quad(f,a,b)

18. NM [DR D.P.BHASKAR, SANJIVANI COLLEGE OF ENGINEERING KOPARGAON]
18
% Double Trap Integration
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
clc
clear all
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
f=inline('x+y');
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
ns=1;
a=1;b=3;c=0;d=2;na=2;ma=2;
n=na*ns;m=ns*ma;
h=(b‐a)/n; k=(d‐c)/m;
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
y1=c;
for j=1:m+1
x1=a;
sum=0;
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
for i=1:na
sum=sum+f(x1,y1)+f((x1+h),y1);
x1=(x1+h);
end
s(j)= h/2*sum;
y1=y1+k;
end
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
sum=0;
for j=2:ma
sum=sum+s(j);
end
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
sum=sum*2;
sum=sum+s(1)+s(m+1);
sum=sum*k/2
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
fprintf('Volume=%f',sum);
%Solver Double Integration (dblquad)
clc
clear all
f =inline('x+y')
a=1;b=3;c=0;d=2;
%‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
Volume=dblquad(f,a,b,c,d);
fprintf(' Volume of integration=%0.2f',Volume);