Question

Is it possible to do this without matlab?

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 y-f(x,y), y(XO-yo, is given byYn+1-yn+hf(xn,Yn) where h is a fixed stepsize, xnxo nh, and yn ~y(x). 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 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 with h - 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 x 10-3. Absolute error of approximations to y(2) Method h0.05 h- 0.025 Ratio Euler's method 0.40178(-1)0.19541(-1) 2.056 Improved Euler method 0.95166(-3) 0.26047(-3)3.654 Runge-Kutta method 0.16374(-4)0.94541(-6) 17.319 The last column of "Ratio" contains for each method the ratio Absolute error with h 0.05 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.

Answer 1

Matlab code for Euler, Improved Euler and RK4 method clear all close all %Function for which solution have to do f-e(x,y) 10./у-у.7x; hh-0.05 0.025] for ii-1:2 Euler method hhh ( ii); amount of intervaïs x-1; % initial x initial y % at what point we have to evaluate x_eval-2: n«(x, eval-x) /h; x2(1)=x; % Number of steps for i=1:n 8Bular Steps m-double(f (x,y)); x-x+h y-y+h*m; x2(i+1)-xi end УУ (ii,1)-y2 ( end); fprint f( "\n\tThe solution is using Euler Method for h=8.3f at x(8.1f) %f\n',h,x2(end ) , y2 ( end)) Improved Euler method % initial x % initial y % at what point we have to evaluate % step size y-l; x eval-2 n-(x-eval-x) /h ; x3 (1)x for i=1:n improved Euler steps m1=double(f(x,y)); m2-double(f ((x+h), (y+h*ml))); y-y+double(h* ((ml+m2) /2)); x-x+h; end yy (ii,2)-y3 (end); fprint f( .\tThe solution using mproved Euler Method x(8.If) is %f\n",n,x3(end ) , y3 ( end)) for h=8.3f at RK4 method initial x initial y y= 1 ;

x eval-2; n=(x-eval-X)/h; % at what point we have to evaluate % Number of steps for i-1:n RK4 Steps k1-h double(f (x,y)) k2-h*double (f ((x+h/2), (ytkl/2)); k 3-h double(f ((x+h/2), (y+k2/2))); k4-h*double (f ( (x+h), (y+k3))); dx=(1/6)" (k1+2*k2+2*k3tk4); x-x+h; Y-Y+dx; end yy ( ії, 3)-y4 ( end); fprintf( "\tThe solution using Runge Kutta is 4 for h-%.3f at x(8.1f) %f\n.,h, x4 (end) , y4 (end)) exact solution syms y(x) eqndiff (y,x)10/y-y/x; cond- y (1)1; y-ext ( x ) =d solve ( eqn, cond ) ; forint f ( \n\tExact solution for y(2) x-2 is %f.\n.y_ext (2)) at %Table for absolute error fprintf(nn\t Absolute error of Approximations to y (2)In') fprintf(tMethod vl-abs (y_ ext (2)-yy (1,1)); v2-abs (y_ext (2)-yy (2,1)) fprintf( ' \tEuler \th-0.005 \ th-o. 02 5 \ tRatio\n\n') tteltse\tsf\n',vl,v2,vl/v2) vl-abs (y_ ext (2)-yy (1,2)); v2-abs (y_ext (2)-yy (2,2)) fprintf tImproved Euler\tteltteltef\n' ,vl, v2,v1/v2) vl-abs (y_ ext (2) -yy (1,3)) v2-abs (y_ext (2)-yy (2,3)) fprintf("\tR unge kutta \tteltteit%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

The solution using Euler Method for h-0.050 at x(2.0) is 3.492230 The solution using improved Euler Method for h=0.050 at x ( 2.0) is 3.451101 The solution using Runge Kutta 4 for h=0.050 at x(2.0) is 3.452069 The solution using Euler Method for h=0.025 at x(2.0) įs 3.471594 The solution using improved Euler Method for h-0.025 at x (2.0) is 3.451792 The solution using Runge Kutta 4 for h·0.025 at x(2.0) IS 3.452053 Exact solution for y(2) at x-2 is 3.452053 Absolute error of Approximations to y(2) Method h 0.005 h 0.025 Ratio Euler 4.017772e-02 1.954147e-02 2. 056024 Improved Euler 9.516568e-04 2.604706e-04 3.653606 Runge kutta 1.637421e-05 9.454053e-07 17.319776 Solution ploty vs. x 3.5 Euler Mefhod Modified Euler RK4 Method 2.5 1.5 1.2 1A 1.3 1.,5 1.6 1.8 1.9 Published with MATLAB R2018a

%%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 %%%%%%%%%%%%%%%