Runge-Kutta 2
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Nagy Csaba Norbert
el 21 de Abr. de 2021
Comentada: Nagy Csaba Norbert
el 22 de Abr. de 2021
Hi, i have a homwork for school including the Runge-Kutta 2.

I have to do it in symbolic, i would be very greatfull if someone can help me in this taks.
intervalmin = 0;
intervalmax = 1;
h1 = 0.1;
g = 0.02;
X0 = [0, 0];
[t1, x1] = fuggveny(intervalmin, X0(1), X0(2), h1, intervalmax);
[t2, x2] = fuggveny(intervalmin, X0(1), X0(2), g, intervalmax);
syms x(t);
dz = diff(x,t);
ode = diff(x,t,2) + 5.*diff(x,t,1) + 4.*x(t) == 3 - 2.*t - t.^2;
cond1 = x(0) == 1;
cond2 = dz(0) == 1;
RK2(t) = dsolve(ode,cond1,cond2);
plot(t1, x1, '-y');
hold on;
plot(t2, x2, '--r');
hold on;
fplot(RK2,'-*b');
hold on
legend('h=0.1','h=0.02','ode')
ax = gca;
ax.XAxisLocation = 'origin';
ax.YAxisLocation = 'origin';
grid on;
function dX = f(t, x1, x2)
X1 = x2;
X2 = -5.*x2 + 4.*x1 + 3 - 2.*t - t.^2;
dX = [X1, X2];
end
function [t, x] = fuggveny(intervalmin, X0_1, X0_2, h, intervalmax)
t = (intervalmin:h:intervalmax);
X1 = zeros(size(t));
X2 = zeros(size(t));
X1(1) = X0_1;
X2(1) = X0_2;
for i = 1:1:length(t) - 1
k1 = f(t(i), X1(i), X2(i));
k2 = f(t(i) + h/2, X1(i) + h/2 * k1(1), X2(i) + h/2 * k1(2));
k3 = f(t(i) + h, X1(i) - h*k1(1) + 2*h*k2(1), X2(i) - h*k1(2) + 2*h*k2(2));
X1(i + 1) = X1(i) + h/6 * (k1(1) + 4*k2(1) + k3(1));
X2(i + 1) = X2(i) + h/6 * (k1(2) + 4*k2(2) + k3(2));
end
x = X1;
end
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Respuesta aceptada
James Tursa
el 21 de Abr. de 2021
Editada: James Tursa
el 21 de Abr. de 2021
The + 4.*x1 should be - 4.*x1 in your derivative function. Also, using both X1 and x1 and X2 and x2, even though MATLAB is case sensitive and these are different variables, is confusing to read. Suggest using different names such as x1dot and x2dot, or even just form the derivative vector directly. E.g.,
function dX = f(t, x1, x2)
dX = [x2; -5.*x2 - 4.*x1 + 3 - 2.*t - t.^2];
end
Also, just to clarify, you are asked to solve this both symbolically and numerically?
Not sure why the assignment says Runge-Kutta II when it looks like you are using a 3rd order method.
3 comentarios
James Tursa
el 22 de Abr. de 2021
Are you getting errors? The output is not what you expected? Or ...?
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