Question

I need the matlab codes for following question

(1) (a). Solve the following second-order differential equations by a pair of first-order equations, xyʹʹ − yʹ − 8x³y³ = 0; with initial conditions y = 0.5 and yʹ = −0.5 at x = 1.

(b). Solve the problem in part (a) above using MATLAB built-in functions ode23 and ode45, within the range of 1 to 4, and compare with the exact solution of y = 1/(1 + x²)

[Hint: ode23 à 0.0456, ode45 à 0.0588]

(c). How can we improve the accuracy for the solutions obtained in parts (a) and (b) above?

Answer 1

Let d 92 2 2. 2 dyレク 82,3 dM

Matlab code for solving ode clear all close all Answering question b. %Initial conditions for ode yo-[0.5;-0.51: minimum and maximum x xspan-[1 4]; %Solution of ODEs using ode45 matïab function soll-ode45 e (x, y) odefcnl (x,y), xspan, y0) Equally splitting x into small intervals x1linspace(xspan (1), xspan (end),501) tyy is the corresponding x y v and z yyl -deval soll,x1) figure(1) hold on plot (x1,yyl(1,,'linewidth',2) gminimum and maximum x xspan-[1 4; %Solution of ODEs using ode45 matlab function so12-ode23 ( (x,y) odefcnl (x,y), xspan, y0); %Equally splitting x into small intervals x1 = linspace(xs pan (1),xspan (end), 501); yy is the corresponding x y v and z yy2 - deval so12, x1) figure (1) plot (x1,yy2(1,:), 'linewidth',2) exact solution y ext-e(x) 1./(1+x.*2); plot (xl,y_ext(x1), 'linewidth',2) hold off title 'Plot for y(x) vs.x solution') xlabel ( ' x' ) ylabel('y(x)') legend("ode45', "ode23 "Exact solution') box on grid on fprint f ("AtAt x. % 2.f solution using ode45 is %2.5f. n,x1 (end), yy1 (1,end)) fprint f ( .\n\tAt x.82.f solution using ode23 is %2.5f. n,x1 (end) , yy2 (1,end)) fprint f ( "\n\tAt x=82. f solution using exact analytic is %2.5f. n',x1 (end),y_ext (xl (end)))

Function for evaluating the ODE function dydx oder cn 1 (x,y) eql- y (2); %Evaluate the ODE for our present problem dydx = [ eq 1 ; eq 2 ] ; At x= 4 solution us ing ode4 5 is 0.05845. At x= 4 solution using ode23 is 0.02541. At x= 4 solution using exact analytic is 0.05882. Plot for y vs. x solution 0.5 ode45 045 Exact solution 0.4 0.35 0.3 025 0.2 0.15 0.1 0.05 1.5 2.5 3.5 Published with MATLAB R2018a

%%Matlab code for solving ode
clear all
close all
%Answering question b.
%Initial conditions for ode
y0=[0.5;-0.5];
      
        %minimum and maximum x
        xspan=[1 4];
        %Solution of ODEs using ode45 matlab function
        sol1= ode45(@(x,y) odefcn1(x,y), xspan, y0);
        %Equally splitting x into small intervals
        x1 = linspace(xspan(1),xspan(end),501);
        %yy is the corresponding x y v and z
        yy1 = deval(sol1,x1);
        figure(1)
        hold on
        plot(x1,yy1(1,:),'linewidth',2)
      
        %minimum and maximum x
        xspan=[1 4];
        %Solution of ODEs using ode45 matlab function
        sol2= ode23(@(x,y) odefcn1(x,y), xspan, y0);
        %Equally splitting x into small intervals
        x1 = linspace(xspan(1),xspan(end),501);
        %yy is the corresponding x y v and z
        yy2 = deval(sol2,x1);
        figure(1)
     
        plot(x1,yy2(1,:),'linewidth',2)
     
      
        %exact solution
        y_ext=@(x) 1./(1+x.^2);
        plot(x1,y_ext(x1),'linewidth',2)
        hold off
      
title('Plot for y(x) vs. x solution')
xlabel('x')
ylabel('y(x)')
legend('ode45','ode23','Exact solution')
box on
grid on

fprintf('\tAt x=%2.f solution using ode45 is %2.5f.\n',x1(end),yy1(1,end))
fprintf('\n\tAt x=%2.f solution using ode23 is %2.5f.\n',x1(end),yy2(1,end))
fprintf('\n\tAt x=%2.f solution using exact analytic is %2.5f.\n',x1(end),y_ext(x1(end)))

%Function for evaluating the ODE
function dydx = odefcn1(x,y)

    eq1= y(2);
    eq2= (y(2)./x)+8.*x.^2.*(y(1)).^3;
  

    %Evaluate the ODE for our present problem
    dydx = [eq1;eq2];
  
end

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