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    Lab Tasks for engineering

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    All Lab Task

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    Adnan
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    Lab Tasks for engineering

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    Lab Tasks for engineering

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    © All Rights Reserved
    Download as DOCX, PDF, TXT or read online from Scribd
    Download as docx, pdf, or txt
    3 views13 pages

    All Lab Task

    Uploaded by

    Adnan
    Lab Tasks for engineering

    Copyright:

    © All Rights Reserved
    Download as DOCX, PDF, TXT or read online from Scribd
    Download as docx, pdf, or txt
    of 13

    UNIVERSITY OF ENGINEERING AND TECHNOLOGY,

    PESHAWAR PAKISTAN
    Main Campus

    CS-311 Web Programming


    Practical Task 01
    Submitted By
    Name: SAJJAD KHAN
    Registration No. 21PWBCS0832
    Semester: [BS CS 6th
    Section: A

    Submitted To
    Sir Anwar Shan

    DEPARTMENT OF COMPUTER SCIENCE & IT


    UNIVERSITY OF ENGINEERING AND TECHNOLOGY, PESHAWAR, PAKISTAN
    LAB TASK 04
    Q1:
    clc;
    clear all;

    A = [5 2 5 -3 1 2; 3 32 5 -12 2 1; 2 3 25 -2 5 2; 1 12 5 -42 2 5; 4 6 5 4 30
    8; 5 6 7 3 8 42];
    b = [50; 60; 40; 35; 41; 24];
    X1 = inv(A) * b;

    maxn = 20;
    n = length(b);
    tol = 1e-2;
    p=[0 0 0 0]';

    % p = zeros(n, 1);

    % Direct method using MATLAB's mldivide operator


    X_direct = A \ b;
    disp('Direct Method Solution:');
    disp(X_direct);

    R = sum(abs(A), 2);
    D = abs(diag(A));
    W = R - D;
    check = D >= W;
    disp('Check for Diagonal Dominance:');
    disp(check);
    if all(check)
    disp('Matrix is diagonally dominant.');

    % Jacobi iteration
    for k = 1:maxn
    for j = 1:n
    x(j) = (b(j) - A(j, [1:j - 1, j + 1:n]) * p([1:j - 1, j + 1:n])) /
    A(j, j);
    end
    err = abs(norm(x' - p));
    p = x';
    X(:, k) = p;
    if err < tol
    break;
    end
    end
    X
    else
    disp('Matrix is not diagonally dominant.');
    end

    OUTPUT:
    Q2:
    % Gauss- Sediel iterarion Ax=b
    clc, clear all
    A = [5 2 5 -3 1 2; 3 32 5 -12 2 1; 2 3 25 -2 5 2; 1 12 5 -42 2 5; 4 6 5 4 30
    8; 5 6 7 3 8 42];
    b = [50; 60; 40; 35; 41; 24];

    % X1=inv(A)*b
    maxn=20;
    n=length(b);
    tol=10^(-2);
    p=[0 0 0 0]';

    % Direct method using MATLAB's mldivide operator


    X_direct = A \ b;
    disp('Direct Method Solution:');
    disp(X_direct);

    R=sum(abs(A),2);
    D=abs(diag(A));
    W=R-D;
    check=D>=W
    DD=all(check);
    % pause
    if DD==1
    for k=1:maxn
    for j=1:n
    if j==1
    x(1)=(b(1)-A(1,2:n)*p(2:n))/A(1,1);
    elseif j==n
    x(n)=(b(n)-A(n,1:n-1)*x(1:n-1)')/A(n,n);
    else
    x(j)=(b(j)-A(j,1:j-1)*x(1:j-1)'-A(j,j+1:n)*p(j+1:n))/A(j,j);
    end

    end
    err=abs(norm(x'-p));
    p=x';
    X(:,k)=p

    pause
    if err<tol,break
    end % a11x1+a12x2 p= intial values, x=updated values
    % pause
    end
    X
    else
    disp('matrix is not diaganolly dominant')
    end

    OUTPUT:
    Lab Task 05
    Q1: Trapezoidal Rule

    clc, clear all, close all


    f = @(x) x.^3 + x.^2 - 2.*x; % Define the function
    % check if n= 10
    n = 10; % Number of trapeziums
    a = 0; % Lower limit
    b = 3; % Upper limit
    h = (b - a) / n; % Step size

    % Trapezoidal Rule
    sum = 0;
    for i = 1:n-1
    sum = sum + f(a + i*h);
    end
    TR = (h/2) * (f(a) + 2*sum + f(b));
    fprintf('Result using trapezoidal rule with n=10 is: %0.8f\n', TR)

    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    % check if n= 20
    n = 20; % Number of trapeziums
    h = (b - a) / n; % Step size

    % Trapezoidal Rule
    sum = 0;
    for i = 1:n-1
    sum = sum + f(a + i*h);
    end
    TR = (h/2) * (f(a) + 2*sum + f(b));
    fprintf('Result using trapezoidal rule with n=20 is: %0.8f\n', TR)

    OUTPUT:

    Q2: Simpson’s 1/3 Rule


    clc, clear all, close all

    f = @(x) sin(x) + 2 * cos(x); % Define the function properly

    % For n = 10
    n = 10; % Number of intervals, must be even for Simpson's rule
    a = 0; % Lower limit
    % b = 1.571428571428571; % Upper limit
    b = pi/2; % Upper limit
    h = (b - a) / n; % Step size

    % Simpson's 1/3 Rule


    oddsum = 0;
    for j = 1:2:n-1
    oddsum = oddsum + f(a + j * h);
    end
    evensum = 0;
    for k = 2:2:n-2
    evensum = evensum + f(a + k * h);
    end
    sim13 = (h / 3) * (f(a) + 4 * oddsum + 2 * evensum + f(b));
    fprintf('Result using Simpson 1/3rd rule with n=10 is: %0.8f\n', sim13)

    % Repeat the process for n = 20


    n = 20;
    h = (b - a) / n; % Recalculate step size for new n

    % Simpson's 1/3 Rule


    oddsum = 0;
    for j = 1:2:n-1
    oddsum = oddsum + f(a + j * h);
    end
    evensum = 0;
    for k = 2:2:n-2
    evensum = evensum + f(a + k * h);
    end
    sim13 = (h / 3) * (f(a) + 4 * oddsum + 2 * evensum + f(b));
    fprintf('Result using Simpson 1/3rd rule with n=20 is: %0.8f\n', sim13)

    OUTPUT:

    Q3: Simpson’s 3/8 Rule


    clc, clear all, close all
    f = @(x) sin(x).^2 + cos(x); % Define the function properly

    % For n = 10
    n = 10; % Number of intervals
    a = 0; % Lower limit
    b = pi; % Upper limit
    h = (b - a) / n; % Step size

    % Simpson's 3/8 Rule


    sum3 = 0;
    for l = 3:3:n-1
    sum3 = sum3 + f(a + l * h);
    end
    sumall = 0;
    for m = 1:n-1
    sumall = sumall + f(a + m * h);
    end
    sumnot3 = sumall - sum3;
    sim38 = (3 * h / 8) * (f(a) + 3 * sumnot3 + 2 * sum3 + f(b));
    fprintf('Result using Simpson 3/8th rule with n = 10 is: %0.8f\n', sim38)

    % Repeat the process for n = 20


    n = 20;
    h = (b - a) / n; % Recalculate step size for new n

    % Simpson's 3/8 Rule


    sum3 = 0;
    for l = 3:3:n-1
    sum3 = sum3 + f(a + l * h);
    end
    sumall = 0;
    for m = 1:n-1
    sumall = sumall + f(a + m * h);
    end
    sumnot3 = sumall - sum3;
    sim38 = (3 * h / 8) * (f(a) + 3 * sumnot3 + 2 * sum3 + f(b));
    fprintf('Result using Simpson 3/8th rule with n = 20 is: %0.8f\n', sim38)

    OUTPUT:

    Lab task 06:


    Q1:
    %%%%%%%%%%%%%%%%%%%%%%%%%% E1 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    syms y(x)
    eqn1 = diff(y, x) + y*tan(x) == y^3 * sec(x)^4;
    sol1 = dsolve(eqn1);
    sol1final = simplify(sol1)

    % Displaying Problem 1
    disp('Problem 1:')
    disp('Given ODE:')
    disp(eqn1)
    disp('Solution:')
    disp(sol1final)

    %%%%%%%%%%%%%%%%%%%%%%%%%% E2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    syms x y(x)
    equ2=y^4 * diff(y, x) + diff(y, x) +x^2 + 1 == 0;
    sol2=dsolve(equ2);
    sol2final = simplify(sol2)

    % Displaying Problem 2
    disp('Problem 2:')
    disp('Given ODE:')
    disp(equ2)
    disp('Solution:')
    disp(sol2final)

    %%%%%%%%%%%%%%%%%%%%%%%%%% E3 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    syms y(x)
    equ3=diff(y, x, 2) + 3*diff(y, x) + y == exp(x) -3;
    sol3=dsolve(equ3);
    sol3final = simplify(sol3)

    % Displaying Problem 3
    disp('Problem 3:')
    disp('Given ODE:')
    disp(equ3)
    disp('Solution:')
    disp(sol3final)

    %%%%%%%%%%%%%%%%%%%%%%%%%% E4 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    syms y(x)
    equ4=diff(y, x, 4) - y == 2* exp(2*x) + 3*exp(x);
    sol4=dsolve(equ4);
    sol4final = simplify(sol4)

    % Displaying Problem 4
    disp('Problem 4:')
    disp('Given ODE:')
    disp(equ4)
    disp('Solution:')
    disp(sol4final)

    %%%%%%%%%%%%%%%%%%%%%%%%%% E5 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    syms x y(x)
    Dy = diff(y, x);
    eqn5 = diff(y, x, 2) * (x + 3) * Dy - y == 0;
    cond5 = [y(0) == 1, Dy(0) == 2];
    gsol5 = simplify(dsolve(eqn5))
    psol5 = simplify(dsolve(eqn5, cond5))

    % Displaying Problem 5
    disp('Problem 5:')
    disp('Given ODE:')
    disp(eqn5)
    disp('General Solution:')
    disp(gsol5)
    disp('Particular Solution with Initial Conditions:')
    disp(psol5)

    %%%%%%%%%%%%%%%%%%%%%%%%%% E6 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    syms y(x)
    Dy = diff(y, x);
    D2y = diff(y, x, 2);
    D3y = diff(y, x, 3);
    D4y = diff(y, x, 4);
    eqn6 = D4y == -sin(x) + cos(x);
    cond6 = [y(0) == 0, Dy(0) == -1, D2y(0) == -1, D3y(0) == 7];
    gsol6 = simplify(dsolve(eqn6))
    psol6 = simplify(dsolve(eqn6, cond6))

    % Displaying Problem 6
    disp('Problem 6:')
    disp('Given ODE:')
    disp(eqn6)
    disp('General Solution:')
    disp(gsol6)
    disp('Particular Solution with Initial Conditions:')
    disp(psol6)

    %%%%%%%%%%%%%%%%%%%%%%%%%% E7 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    syms x y(x)
    Dy = diff(y, x);
    D2y = diff(y, x, 2);
    eqn7 = D2y == 2 - 6*x;
    cond7 = [y(0) == 1, Dy(0) == 4];
    gsol7 = simplify(dsolve(eqn7))
    psol7 = simplify(dsolve(eqn7, cond7))

    % Displaying Problem 7
    disp('Problem 7:')
    disp('Given ODE:')
    disp(eqn7)
    disp('General Solution:')
    disp(gsol7)
    disp('Particular Solution with Initial Conditions:')
    disp(psol7)

    OUTPUT:

    sol1final =

    1/((C3 - 2*tan(x))/cos(x)^2)^(1/2)
    -1/((C3 - 2*tan(x))/cos(x)^2)^(1/2)

    Problem 1:

    Given ODE:

    D(y)(x) + tan(x)*y(x) == y(x)^3/cos(x)^4

    symbolic function inputs: x

    Solution:

    1/((C3 - 2*tan(x))/cos(x)^2)^(1/2)

    -1/((C3 - 2*tan(x))/cos(x)^2)^(1/2)

    Warning: Explicit solution could not be found; implicit solution returned.

    > In dsolve at 201

    In problem12 at 17

    sol2final =

    RootOf(z^5 + 5*z + (5*x*(x^2 + 3))/3 - 5*C6, z)

    Problem 2:

    Given ODE:

    D(y)(x)*y(x)^4 + D(y)(x) + x^2 + 1 == 0

    symbolic function inputs: x

    Solution:

    RootOf(z^5 + 5*z + (5*x*(x^2 + 3))/3 - 5*C6, z)

    sol3final =

    exp(x)/10 - exp(x*(5^(1/2)/2 - 3/2))*(exp(-(x*(5^(1/2) - 3))/2)*((9*5^(1/2))/10 + 3/2) - exp(-(x*(5^(1/2) -


    5))/2)*(5^(1/2)/10 + 1/10)) + (9*5^(1/2))/10 - (5^(1/2)*exp(x))/10 + C9*exp(x*(5^(1/2)/2 - 3/2)) + C10*exp(-
    x*(5^(1/2)/2 + 3/2)) - 3/2

    Problem 3:

    Given ODE:

    3*D(y)(x) + D(D(y))(x) + y(x) == exp(x) – 3

    symbolic function inputs: x

    Solution:

    exp(x)/10 - exp(x*(5^(1/2)/2 - 3/2))*(exp(-(x*(5^(1/2) - 3))/2)*((9*5^(1/2))/10 + 3/2) - exp(-(x*(5^(1/2) -


    5))/2)*(5^(1/2)/10 + 1/10)) + (9*5^(1/2))/10 - (5^(1/2)*exp(x))/10 + C9*exp(x*(5^(1/2)/2 - 3/2)) + C10*exp(-
    x*(5^(1/2)/2 + 3/2)) - 3/2
    sol4final =

    (2*exp(2*x))/15 - (9*exp(x))/8 + C12*cos(x) + C15*exp(x) + C13*sin(x) + (3*x*exp(x))/4 + C14*exp(-x)

    Problem 4:

    Given ODE:

    D(D(D(D(y))))(x) - y(x) == 2*exp(2*x) + 3*exp(x)

    symbolic function inputs: x

    Solution:

    (2*exp(2*x))/15 - (9*exp(x))/8 + C12*cos(x) + C15*exp(x) + C13*sin(x) + (3*x*exp(x))/4 + C14*exp(-x)

    Warning: Explicit solution could not be found.

    > In dsolve at 194

    In problem12 at 58

    gsol5 =

    [ empty sym ]

    Warning: Explicit solution could not be found.

    > In dsolve at 194

    In problem12 at 59

    psol5 =

    [ empty sym ]

    Problem 5:

    Given ODE:

    D(y)(x)*D(D(y))(x)*(x + 3) - y(x) == 0

    symbolic function inputs: x

    General Solution:

    [ empty sym ]

    Particular Solution with Initial Conditions:

    [ empty sym ]

    gsol6 =

    C26 + cos(x) - sin(x) + C25*x + (C23*x^3)/6 + (C24*x^2)/2

    psol6 =

    cos(x) - sin(x) + x^3 - 1

    Problem 6:
    Given ODE:

    D(D(D(D(y))))(x) == cos(x) - sin(x)

    symbolic function inputs: x

    General Solution:

    C26 + cos(x) - sin(x) + C25*x + (C23*x^3)/6 + (C24*x^2)/2

    Particular Solution with Initial Conditions:

    cos(x) - sin(x) + x^3 - 1

    gsol7 =

    C34 + x*(- x^2 + x + C33)

    psol7 =

    x*(- x^2 + x + 4) + 1

    Problem 7:

    Given ODE:

    D(D(y))(x) == 2 - 6*x

    symbolic function inputs: x

    General Solution:

    C34 + x*(- x^2 + x + C33)

    Particular Solution with Initial Conditions:

    x*(- x^2 + x + 4) + 1

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