Question

3In modelling the velocity y of a chain slipping off a horizontal platform, the differential equation y, 10/y-y/x is derived. Suppose the initial condition is y( 1-1 (a) Euler's method for solving yf(x), y(xoyo, is given by yn+n+hf(an,yn), where h is a fixed stepsize, xn xo + nh, and yn y(xn). Apply one step of Euler's method to the initial value problem given above (b) Apply one step of the improved Euler method given by [f(xn, yn) + f(xn + h, yn + hf(xn , yn)] Yn+,-Yn + with h 0.1 to the initial value problem given above (c) For the initial value problem given above, the two methods given in part (a) and (b) as well as the Runge-Kutta method of order 4 were run on a computer withh 0.05 and h 0.025 to obtain approximations to the solution at x 2" The absolute errors of these approximations are shown in the following table. Note that the notation in the table is such that an entry like 0.95166(-3) is shorthand for 0.95166 × 10-3 Absolute error of approximations to y(2) Method h0.05 h0.025 Ratio Euler's method 0.40 I 78(-l ) | 0. I 9541 (-1 ) 2.056 Improved Euler method 0.95166(-3) 0.26047-3 3.654 Runge-Kutta method 0.16374(-4) .94541-6 17.319 The last column of "Ratio" contains for each method the ratio Absolute error with h0.05 Absolute error with h 0.025 For each of the three methods, it is known that the absolute error using a stepsize of h is approximately Ch for some positive constant C and integer p. Use the given ratios to estimate the value of p for each method.

hand written solution only (not computerised) if not possible then please refund the question becs i have already recieved a computerised solution from you but i dont understand.

Answer 1

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ANSWER:

CODE:MATLAB

%%Matlab code for Euler, Improved Euler and RK4 method
clear all
close all
%Function for which solution have to do
f=@(x,y) 10./y-y./x;
hh=[0.05 0.025];

for ii=1:2
    %Euler method
    h=hh(ii);         % amount of intervals
    x=1;             % initial x
    y=1;             % initial y
    x_eval=2;        % at what point we have to evaluate
    n=(x_eval-x)/h; % Number of steps
    x2(1)=x;
    y2(1)=y;
    for i=1:n
        %Eular Steps
        m=double(f(x,y));
        x=x+h;
        y=y+h*m;
        x2(i+1)=x;
        y2(i+1)=y;
    end
    yy(ii,1)=y2(end);
    fprintf('\n\tThe solution using Euler Method for h=%.3f at x(%.1f) is %f\n',h,x2(end),y2(end))

    %Improved Euler method
    x=1;             % initial x
    y=1;             % initial y
    x_eval=2;        % at what point we have to evaluate
    n=(x_eval-x)/h; % step size
    x3(1)=x;
    y3(1)=y;
    for i=1:n
    %improved Euler steps
       m1=double(f(x,y));
       m2=double(f((x+h),(y+h*m1)));
       y=y+double(h*((m1+m2)/2));
       x=x+h;
       y3(i+1)=y;
       x3(i+1)=x;
    end
    yy(ii,2)=y3(end);
  
    fprintf('\tThe solution using improved Euler Method for h=%.3f at x(%.1f) is %f\n',h,x3(end),y3(end))
  
    %RK4 method
    x=1;             % initial x
    y=1;             % initial y
    x_eval=2;        % at what point we have to evaluate
    n=(x_eval-x)/h; % Number of steps
    x4(1)=x;
    y4(1)=y;
    for i=1:n
    %RK4 Steps
       k1=h*double(f(x,y));
       k2=h*double(f((x+h/2),(y+k1/2)));
       k3=h*double(f((x+h/2),(y+k2/2)));
       k4=h*double(f((x+h),(y+k3)));
       dx=(1/6)*(k1+2*k2+2*k3+k4);
       x=x+h;
       y=y+dx;
       x4(i+1)=x;
       y4(i+1)=y;

    end
    yy(ii,3)=y4(end);
    fprintf('\tThe solution using Runge Kutta 4 for h=%.3f at x(%.1f) is %f\n',h,x4(end),y4(end))

end
%exact solution
syms y(x)
eqn = diff(y,x) == 10/y-y/x;
cond = y(1) == 1;
y_ext(x)=dsolve(eqn,cond);

fprintf('\n\tExact solution for y(2) at x=2 is %f.\n',y_ext(2))

%Table for absolute error
fprintf('\n\n\t Absolute error of Approximations to y(2)\n')
fprintf('\tMethod        \th=0.005\th=0.025\tRatio\n\n')
v1=abs(y_ext(2)-yy(1,1)); v2=abs(y_ext(2)-yy(2,1));
fprintf('\tEuler         \t%e\t%e\t%f\n',v1,v2,v1/v2)
v1=abs(y_ext(2)-yy(1,2)); v2=abs(y_ext(2)-yy(2,2));
fprintf('\tImproved Euler\t%e\t%e\t%f\n',v1,v2,v1/v2)
v1=abs(y_ext(2)-yy(1,3)); v2=abs(y_ext(2)-yy(2,3));
fprintf('\tRunge kutta   \t%e\t%e\t%f\n',v1,v2,v1/v2)

%%Plotting solution using Euler method
figure(1)
hold on
plot(x2,y2,'Linewidth',2)
plot(x3,y3,'Linewidth',2)
plot(x4,y4,'Linewidth',2)
xlabel('x')
ylabel('y(x)')
title('Solution plot y vs. x')
legend('Euler Method','Modified Euler','RK4 Method','Location','northwest')
grid on


%%%%%%%%%%%%%%%%% End of Code %%%%%%%%%%%%%%%

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