Question

2. Differential Equation (5 points) Using (i) Euler's method and (ii) modified Euler's method, both with step size h-0.01, to construct an approximate solution from t-0 to t-2 for xt 2 , 42 with initial condition x(0)-1. Compare both results by plotting them in the same figure. 3. Simulink (5 points) Solve the above differential equation using simplink. Present the model and result.

Please show Matlab code and Simulink screenshots

Answer 1

dxdt-@ (t , X) 3qrt (x*t/ (Χ. ^2+t . ^2) ) ; h -0.01 t = 0:h:2; xeuler zero3 (1, length (t) ) ; x mod euler zero3 (1, length (t)); % Initializing modified Eulerョolution % time vector (t0, t1, tn) % Initializing Euler solution vector having 3ame size as time vector - % Initial Conditions x euler (1) - 1; x_mod euler (1)-1: or 2:length (t) x-euler (k) =x_culer (k-1) + h*dxdt(t(k-1), X-euler (k-1)); x mod euler (k) = x mod euler (k-1) + (h/2) * (dxdt (t (k-1) , x mod end euler (k-1) ) + dxdt (t (k) , x euler (k) ) ) ; - % Comparing Results in table (intermediate values) fprintf("in %83 %163 %203\n", "t", "Euler", "Modified Euler") tprintf ( " %8 . 2f %16 . 6f %20 . 6f\n", [t (1 : 20 : end) ' , x euler ( 1 : 20 : end) ' , x mod euler ( 1 : 20 : end) ' . ' ) Ploting Both on Same Graph = plot (t , x euler, '-, ' , t , x mod euler, '--' ) ; %p ( 1 ) . Linewidth = 1 . 6 ; legend (["Euler Method", "Modified Euler Method"], 'Location', "Best' xlabel ('t'ylabel("x(t)"): p (2) . Linewidth 1.6; title ("Euler Method vs Modiied Euler Method")

Euler 1.000000 1.056033 1.155296 1.273355 1.401232 1.534560 1.671064 1.809486 1.949097 2.089454 2.230281 5 Modified Euler 1.000000 1.058150 1.158007 1.276342 1.404352 1.537748 1.674287 1.812728 1.952348 2.092711 2.233540 0.00 0.20 0.60 0.80 1.00 1.20 1.60 1.80 2.00

Euler Method vs Modifed Euler Method 2.4 r 2.2 1.8 1.6 1.4 Euler Method Modified Euler Method 1.2 0 0.2 0.40.6 0.8 1 12 1.416 1.8 2

Text :-

dxdt = @(t,x) sqrt(x*t/(x.^2+t.^2));
h = 0.01;
t = 0:h:2;      % time vector (t0,t1,...tn)
x_euler = zeros(1,length(t));      % Initializing Euler solution vector having same size as time vector
x_mod_euler = zeros(1,length(t));  % Initializing modified Euler solution

x_euler(1) = 1;             % Initial Conditions
x_mod_euler(1)=1;

for k = 2:length(t)
x_euler(k) = x_euler(k-1) + h*dxdt(t(k-1),x_euler(k-1));
x_mod_euler(k) = x_mod_euler(k-1) + (h/2)*(dxdt(t(k-1),x_mod_euler(k-1)) + dxdt(t(k),x_euler(k)));
end

% Comparing Results in table (Intermediate Values)
fprintf("\n %8s %16s %20s\n","t","Euler","Modified Euler")
fprintf(" %8.2f %16.6f %20.6f\n",[t(1:20:end)',x_euler(1:20:end)',x_mod_euler(1:20:end)'].')
% Ploting Both on Same Graph
p = plot(t,x_euler,'-.',t,x_mod_euler,'--');
p(1).LineWidth = 1.6;       p(2).LineWidth = 1.6;
legend(["Euler Method","Modified Euler Method"],'Location','Best')
xlabel('t');    ylabel("x(t)");      title("Euler Method vs Modifed Euler Method")

Simulink :-

1 Scope Integrator Clock u(2) sqrt(u(1) u(2)/(u(1)2+ u(2)A2)) Fcn Bus Creator

Scope O/P :-

2.2 1.2 0.2 0.4 0.6 0.8 1.2