Question: function ex_with_2eqs t0 = 0; tf = 20; y0 = [10;60]; a = .8; b = .01; c = .6; d = .1; [t,y] =
![function ex_with_2eqs t0 = 0; tf = 20; y0 = [10;60];](https://dsd5zvtm8ll6.cloudfront.net/si.experts.images/questions/2024/09/66fa34576a201_43166fa345719ecb.jpg)
function ex_with_2eqs t0 = 0; tf = 20; y0 = [10;60]; a = .8; b = .01; c = .6; d = .1; [t,y] = ode45(@f,[t0,tf],y0,[],a,b,c,d); u1 = y(:,1); u2 = y(:,2); % y in output has 2 columns corresponding to u1 and u2 figure(1); subplot(2,1,1); plot(t,u1,'b-+'); ylabel('u1'); subplot(2,1,2); plot(t,u2,'ro-'); ylabel('u2'); figure(2) plot(u1,u2); axis square; xlabel('u_1'); ylabel('u_2'); % plot the phase plot end %---------------------------------------------------------------------- function dydt = f(t,y,a,b,c,d) u1 = y(1); u2 = y(2); dydt = [ a*u1-b*u1*u2 ; -c*u2+d*u1*u2 ]; end function LAB04ex1 % NOTE: you CAN run this type of function m file as a script because it requires no inputs. Click the Green arrow above the "Run" tab t0 = 0; tf = 40; % initial and final time y0 = [-1; 0];%[y0, v0] [t,Y] = ode45(@f,[t0,tf],y0);% f=system of diff eqs; [t0,tf]
y = Y(:,1); % the output Y has 2 columns corresponding to y and v v = Y(:,2); % the output Y has 2 columns corresponding to y and v
figure(1); plot(t,y,'b-+'); % plots y(t) hold on plot(t,v,'ro-'); % plots v(t) legend('y (t)','v (t)') grid on; figure(2) plot(y,v); % plot solution in the space of y and v (instead of plotting y and v over time) axis square; % makes the figure window square xlabel('y'); ylabel('v'); ylim([-1.5,1.5]); xlim([-1,1]); grid on end %----------------------------------- function dydt = f(t,Y) y = Y(1); v = Y(2);% for EX4 you will need a third element w=Y(3) dYdt = [v; cos(t)-4*v-3*y];%=[dy/dt; dv/dt], the two derivatives as a function of y and v end
function ex_with_param t0=0; tf=3; y0=1; a=1; [t,y]=ode45(@f,[t0,tf],y0,[],a); disp(['y(',num2str(t(end)),') = ',num2str(y(end))]) disp(['length of y =', num2str(length(y))]) %------------------------------------------------- function dydt=f(t,y,a) dydt=-a*(y-exp(-t))-exp(-t); 3. Solve numerically the IVP +7y_ +5y =sint, h y(0)=-0.5,-(0)=0.5 wit dt dt in the interval 0Sts 50. Include the M-file in your report. Is the behavior of the solution significantly different from that of the solution of (L4.7)? Is MATLAB giving any warning message? Comment
Step by Step Solution
There are 3 Steps involved in it
Get step-by-step solutions from verified subject matter experts
