how can i solve this problem with rk4 method?

In this case, m=20, k=20, c=40 and initial x=1, initial v=0, t is 0 to 15. please tell me what's wrong in my code

This is my code when 'c=40' please tell me what's wrong in it

f=@(t,x,v) -0.25*v -x; %dvdt

g=@(t,x,v) v; %dxdt

t0=1;

tl=15;

h=0.1;

n=(tl-t0)/h;

T=t0:h:tl;

X=zeros(1,n+1);

V=zeros(1,n+1);

X(1)=1;

V(1)=0;

for j = 1: n

k1f = f(T(j),X(j),V(j));

k1g = g(T(j),X(j),V(j));

k2f = f(T(j)+h/2,X(j)+k1g*h/2,V(j)+k1f*h/2);

k2g = g(T(j)+h/2,X(j)+k1g*h/2,V(j)+k1f*h/2);

k3f = f(T(j)+h/2,X(j)+k2g*h/2,V(j)+k2f*h/2);

k3g = g(T(j)+h/2,X(j)+k2g*h/2,V(j)+k2f*h/2);

k4f = f(T(j)+h,X(j)+k3g*h,V(j)+k3f*h);

k4g = g(T(j)+h,X(j)+k3g*h,V(j)+k3f*h);

X(j+1)= X(j) + (k1g + 2*k2g + 2*k3g + k4g)/6;

V(j+1)= V(j) + (k1f + 2*k2f + 2*k3f + k4f)/6;

end

plot(T,V)

[jameslk]

i think graph is wrong

what should i fix?

Answer 1

[Alan Stevens]

If m=20, k=20, c=40 then function f should be f = -(c/m)*v - k/m

f=@(t,x,v) -2*v -x; %dvdt