Question

Can someone quickly help me with this? It is a second ODE question that needs to be answered by using MATLAB or python please.

The ODE in Problem #2 and #3 for an oscillating mass is used to approximate the motion of a pendulum when the angle of displacement, theta, is small. That is: 0% = -2206 – m200 Where Og represents the approximate value of 0. The actual equation of motion for the pendulum is given by the trickier second order ODE 0" = -220' - m2 sin(0) Following the same ideas as were explored in Problem #2 (expressing each ODE as a system of two first order ODEs), find alt) and O(t) using Euler's method with step size 0.01 on the time interval [0, 5]. Use: = 0 (0) = Oa (0) 50 (in degrees. Convert this to radians.) e'(0) = 0,0) = 0 2 = 2.8 0 = 10.3 What is the maximum absolute difference between the two numerical solutions?

Answer 1

solution 0." =-200-w²Oa let oa da, and daa dar then, dar daz I مجله w²n tay =-2 Anar nai 2 o =-20 0-w²sino et o = and o'= n2 then ai=x2 3 مدله n2=-2002 - ² sinn, s

Figure 1 File Edit View Insert Tools Desktop Window Help - NE junk2.mxeulersys.mx + clc clear close lam=2.8; w=10.3; to=0; t_end=5; 6 - yO=[50*pi/180;0]; h=0.01; fcnl=@(t, xa) [xa (2);-2*lam*xa (2)-w*2*xa (1)]; 9 [ti, yl] = eulersys (to, yo, t end, h, fcnl); 10 - fcn2=C(t, x) [x (2); -2*lam*x (2)-W*2*sin(x(1))]; 11 - [t2, y2] = eulersys (to, yo, t_end, h, fcn2); 12 plot(ti, yl(:,1), t2, y2(:,1)) 13 - legend ('Problem 2', 'Problem 3') 14 - xlabel('Time'), vlabel('\theta(t)') Problem 2 Problem 0.5 1)0 0 3 -0.5 0 0.5 1 1.5 2 3 3.5 4 4.5 5 2.5 Time TILL Click and drag to move iunk2.m or its tab script

MATLAB script

clc
clear
close
lam=2.8; w=10.3;
t0=0; t_end=5;
y0=[50*pi/180;0];
h=0.01;
fcn1=@(t,xa)[xa(2);-2*lam*xa(2)-w^2*xa(1)];
[t1,y1] = eulersys(t0,y0,t_end,h,fcn1);
fcn2=@(t,x)[x(2); -2*lam*x(2)-w^2*sin(x(1))];
[t2,y2] = eulersys(t0,y0,t_end,h,fcn2);
plot(t1,y1(:,1),t2,y2(:,1))
legend('Problem 2','Problem 3')
xlabel('Time'),ylabel('\theta(t)')

eulersys function

function [t,y] = eulersys(t0,y0,t_end,h,fcn)
m = length(y0);
n = fix((t_end-t0)/h)+1;
t = linspace(t0,t0+(n-1)*h,n)';
y = zeros(n,m);
y(1,:) = y0;
for i = 2:n
% y(i-1,:)
% h*feval(fcn,t(i-1),y(i-1,:))'
y(i,:) = y(i-1,:)+h*feval(fcn,t(i-1),y(i-1,:))';
end