Question

Consider the following initial value problem:

1. Use Euler's explicit scheme to solve the above initial value problem with time step h= 0.5. Express all the computed results with a precision of three decimal places.

2. The analytical or exact solution is compute the absolute error at each t_i value. Express all the computed results with a precision of three decimal places.

3. Write a matlab function that solves the above (IVP) using (RK2.M) for arbitrary time-step h.

0.3t 43 1.2t 70 _0.3t--e

Answer 1

Matlab code for solving 1st order ode using RK4 method and its plot clear all close all 酱88,,888888움움88움움888888888888888888888888888888888888888888888888 %function for ode f-L(t,y) -1.2.y+7.*exp(-0.3.*t); fprintf For the function dy/dt- -1.2.*y+7.*exp(-0.3.*t) n) %All values for solving the equation %Euler Method for h-0.5 h-0.5; % amount of intervals initial t % initial y y-3; t_eval-1.5; nm(t-eval-t ) /h; t_va 11 ( 1 )=t; y_vall (1)-y: at what point we have to evaluate % Number of steps fprintf('y (t) vs.t tabular form Euler Methodn') fprint f (Att 82.2f \t y %f\n',t,y) for i-1:n %Euler STEPS dy-h double(f (t,y)); ttth ; t-val I ( i+1)-t ; y-val I ( i+1)-y ; fprintf('\tt=%2.2f \t y=% f\n',t,y) end fprintfn tThe solution using Euler Method 钍\n',h,t-val 1 ( end),y-val I ( end)) for h-8.2f at t(8.2f) is %function for exact solution f ext-e(t) (70/9).*exp (-0.3. *t)-(43/9).*exp(-1.2.*t) fprintf( Exact solution\n') for j-l:length (t_vall) y_ext (j)-f_ext (t_vall(j)); fprint f ('\tt鴫2.2E \t y= % f \n',t-vall (j),y-ext (j) ) end

fprint f ( "\n\tThe solution at t(3.2f) is %f Exact n,_v vall (end), y ext(end)) %Rk2 solution h-0.5; amount of intervals % initial t t_in-0; Y in 3; % initial y fprintf'y (t) vs. t tabular form RK4 Methodin') for j-l:length(t_rk) fprint f ("\tt-82.2f \t y-%f\n',t_rk (j),y-rk ( j)) end fprintf("\n\tThe RK2 solution for h-%f at t(8.2E) is %f n',h,t_rk (end),y rk(end)) 용음음응용음음응용음음응용음음응용음음응용음음용용음음용용음음용용음음용용음음용욤음음各욤음응용욤음응용욤음응용음 RK2 function for solving function [t _rk,y rk]-RK2 (f, t_in,y in,t_ eval,h) amount of intervals -in initial t in initial y teval at what point we have to evaluate n=(t-eval-t-in) /h; t_rk (1) tin; y_rk(1)-y_in; % Number of steps for i-1:n %RK2 Steps k1-h*double (f (t_in,y_ in)); k2-h double(f ((t_inth/2), (y_in+kl/2)))i t in-t inth; y in-y intk2; trk ( i+1 ) =t-in ; y_rk ( i+1 )-y-in ; end 88888888888888888888888888888888888888888888 88888888웅 For the function dy/dt= -1.2.*yt7.*exp (-0.3. *t) y(t) vs. t tabular form Euler Method t0.00 y 3.000000 t 0.50 y-4.700000 t-1.00 y-4.892478 t-1.50y-4.549855 The solution us ing Euler Method for h=0.50 at t(1.50) is 4.549855 Exact solution

t-0.00 y-3.000000 t-0.50y-4.072295 t=1.00 y=4.322880 t-1.50 y-4.169569 The Exact solution at t (1.50) is 4.169569 Y(t) vs. t tabular form RK4 Method t-o.00 y-3.000000 t0.50 y-3.937102 t-1.00 y-4.174583 t-1.50 y-4.048911 The RK2 solution for h-0.500000 at t(1.50) is 4.048911 Published with MATLABE R2018a

%Matlab code for solving 1st order ode using RK4 method and its plot
clear all
close all

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%function for ode
f=@(t,y) -1.2.*y+7.*exp(-0.3.*t);

fprintf('For the function dy/dt= -1.2.*y+7.*exp(-0.3.*t) \n')
%All values for solving the equation

%Euler Method for h=0.5
h=0.5;           % amount of intervals
t=0;             % initial t
y=3;           % initial y

t_eval=1.5;        % at what point we have to evaluate
n=(t_eval-t)/h; % Number of steps
t_val1(1)=t;
y_val1(1)=y;

fprintf('y(t) vs. t tabular form Euler Method\n')
fprintf('\tt=%2.2f \t y=%f\n',t,y)
for i=1:n
  
    %Euler STEPS
  

   dy=h*double(f(t,y));


   t=t+h;
   y=y+dy;
   t_val1(i+1)=t;
   y_val1(i+1)=y;
   fprintf('\tt=%2.2f \t y=%f\n',t,y)

          
end
fprintf('\n\tThe solution using Euler Method for h=%.2f at t(%.2f) is %f\n',h,t_val1(end),y_val1(end))

%function for exact solution
f_ext=@(t) (70/9).*exp(-0.3.*t)-(43/9).*exp(-1.2.*t);

fprintf('Exact solution\n')
for j=1:length(t_val1)
    y_ext(j)=f_ext(t_val1(j));
    fprintf('\tt=%2.2f \t y=%f\n',t_val1(j),y_ext(j))
end

fprintf('\n\tThe Exact solution at t(%.2f) is %f\n',t_val1(end),y_ext(end))

%Rk2 solution
h=0.5;           % amount of intervals
t_in=0;             % initial t
y_in=3;           % initial y

[t_rk,y_rk]= RK2(f,t_in,y_in,t_eval,h);
fprintf('y(t) vs. t tabular form RK4 Method\n')

for j=1:length(t_rk)
  
    fprintf('\tt=%2.2f \t y=%f\n',t_rk(j),y_rk(j))
end

fprintf('\n\tThe RK2 solution for h=%f at t(%.2f) is %f\n',h,t_rk(end),y_rk(end))

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%RK2 function for solving
function [t_rk,y_rk]= RK2(f,t_in,y_in,t_eval,h)
    %h amount of intervals
    %t_in initial t
    %y_in initial y
    %t_eval at what point we have to evaluate
    n=(t_eval-t_in)/h; % Number of steps
    t_rk(1)=t_in;
    y_rk(1)=y_in;
  
    for i=1:n
    %RK2 Steps
       k1=h*double(f(t_in,y_in));
       k2=h*double(f((t_in+h/2),(y_in+k1/2)));
     
       t_in=t_in+h;
       y_in=y_in+k2;
       t_rk(i+1)=t_in;
       y_rk(i+1)=y_in;
    
    end
  
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%