Question

Problem # 1 P-1 Solve the following initial value problem using a4 order RK scheme: dy dx=tan(x), y(0)= 0.0 - Compare your results by calcudating the error andploting with the equation analytical solution y = In Isec(x)| for a = 0 to b = π/4 and step size 0.01 π: b- Solve the same problem with an accurate library scheme that can improve the answer 03 03 07

use matlab only please

Answer 1

Matlab code using RK4 for 1st order differential equation clear all close all Program for RK4 %function of 1st order diffrent!aï egn f-L(x,y) tan (x)i %step size h=0.01*pi; %all final time steps x(1)-0 ;y(1)-0; %initial conditions x in-x (1); x-max=pi/4; %Runge Kutta 4 iterations n-(x_max-x_in)/h; for i= 1:n+1 %initial x %Final x k2-h 슛 f ( x ( i) + ( 1 /2 ) *h,y(i)t(1/2)-k1); k3=h * f (x(i)+h,y(i)-k2 ) ; x(i+1)=xin+1*h; y(1+1)=double(y(İ)+(1/6) * ( k0+2+k1+2+k2-k3)); %printing the result n',x end),h, double(y (end) )) %function for exact solution fprint f ( "\tThe value of y at x=%2.2f for h= %2.2f is %f ext-e (x) log (abs(sec (x)) tprint f ( "\tThe exact value of y at x-82.2f is %f\n n',x_max, y_ext (x max) ) %plotting the function exact and numerical hold on plot(x,y) plot (x,y ext(x)) ⅝solving using ode45 matlab function tspan _ [0 pi/4]; y0 = 0; [xx,yy] -ode45 (e (x,y) tan (x), tspan, y0); fprint f ("AtThe value of y at x-%2.2f using ode45 is %f\n n,xx (end), yy (end)) plot (xx, yy,"-o xlabel ( ' x ' ) ylabel('y')

title('x vs. y plot legend('RK4 solution', "Exact solution',"ode45 solution, 'location,'best) tprint f ( "Error in RK4 solution with exact solution is %e.\n',norm(y- y ext(x))) fprint f ("Error in ode45 solution with exact solution is %e. n',norm (yy-y_ext(xx))) 88888888888888888888 End of Code 움8888888888888888 The value of y at x-0.79 for h. 0.03 is 0.346574 The exact value of y at x-0.79 is 0.346574 The value of y at x-0.79 using ode45 is 0.346574 Error in RK4 solution with exact solution is 8.711695e-09. Error in ode45 solution with exact solution is 2.562880e-08 x vs. y plot 0.35 -R A solution Exact solution odel5 soluton 0.3 025 0.2 0.15 0.1 0.05 0.1 0.6 0.7 02 0.3 0.4 0.5 0.8 Published with MATLAB R2018a

%%Matlab code using RK4 for 1st order differential equation
clear all
close all

%Program for RK4
%function of 1st order diffrential eqn
f=@(x,y) tan(x);
%step size
h=0.01*pi;
%all final time steps

    x(1)=0;y(1)=0; %initial conditions
    x_in=x(1);     %Initial x
    x_max=pi/4;    %Final x
    %Runge Kutta 4 iterations
    n=(x_max-x_in)/h;
    for i=1:n+1
     
        k0=h*f(x(i),y(i));
        k1=h*f(x(i)+(1/2)*h,y(i)+(1/2)*k0);
        k2=h*f(x(i)+(1/2)*h,y(i)+(1/2)*k1);
        k3=h*f(x(i)+h,y(i)+k2);
        x(i+1)=x_in+i*h;
        y(i+1)=double(y(i)+(1/6)*(k0+2*k1+2*k2+k3));
      
    end
    %printing the result
    fprintf('\tThe value of y at x=%2.2f for h= %2.2f is %f\n',x(end),h,double(y(end)))
  
%function for exact solution  
y_ext=@(x) log(abs(sec(x)));

    fprintf('\tThe exact value of y at x=%2.2f is %f\n\n',x_max,y_ext(x_max))
  
%plotting the function exact and numerical
hold on
plot(x,y)
plot(x,y_ext(x))

%solving using ode45 matlab function
tspan = [0 pi/4];
y0 = 0;
[xx,yy] = ode45(@(x,y) tan(x), tspan, y0);

fprintf('\tThe value of y at x=%2.2f using ode45 is %f\n\n',xx(end),yy(end))

plot(xx,yy,'-o')
xlabel('x')
ylabel('y')
title('x vs. y plot')
legend('RK4 solution','Exact solution','ode45 solution','location','best')

fprintf('Error in RK4 solution with exact solution is %e.\n',norm(y-y_ext(x)))
fprintf('Error in ode45 solution with exact solution is %e.\n',norm(yy-y_ext(xx)))

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