plot a graph of year vs rate

so i have a code :

function ydot= fossil_fuels(t,y)
p = y(1); sigmas = y(2); sigmad = y(3); alphas = y(4); alphad = y(5);
r = [0.0 0.0 1.0 4.0 5.0 8.0 10.0 10.5 10.0 8.0 3.5 2.0 0.0 0.0];
yr = [1000 1850 1950 1980 2000 2050 2080 2100 2120 2150 2225 2300 2500 5000];
f = pchip(yr,r);
d = 8.65;
u1 = 4.95 * 10^2;
u2 = 4.95 * 10^-2;
vs = 0.12;
vd = 1.23;
w = 10^-3;
k1 = 2.19 * 10^-4;
k2 = 6.12 * 10^-5;
k3 = 0.997148;
k4 = 6.79 * 10^-2;
hs = (sigmas - (sigmas^2 - k3*alphas*(2*sigmas - alphas)))^0.5/k3;
cs = (alphas - hs)/2;
ps = k4*hs^2/cs;
ydot = [(ps - p)/d + ppval(f,t)/u1; 1/vs * ((sigmad-sigmas)*w - k1 - u2*(ps-p)/d);...
    1/vd(k1-(sigmad-sigmas)*w);(1/vs) * ((alphad - alphas) * w-k2);...
    1/vd * (k2-(alphad - alphas)*w)];

using the plot command to plot yearly values in the variable u against the rates in the variable v. The output of pchips may be evaluated using ppval ( to evaluate the interpolated function at these time values).

this is what i have done :

u = yr;
v = r;
plot(u,v)

but it gives me an error of

Not enough input arguments.

Error in fossil_fuels (line 2)

p = y(1); sigmas = y(2); sigmad = y(3); alphas = y(4); alphad = y(5);

[Walter Roberson]

How are you invoking fossil_fuels ? Are you just clicking the green Run button?

[NIMA RAHMANI MEHDIABADI]

yes

Answer 1

[Walter Roberson]

You cannot invoke that code by just clicking the green run button. You need to invoke it from other code, or by going to the command line and calling upon it passing in appropriate input values. For example,

p = 10*rand;
alphas = rand;
alphad = rand;
sigmas = randn();
sigmad = randn();

t = 0;
ydot = fossil_fuels(t, [p, sigmas, sigmad, alphas, alphad]);

Normally you would have ode45 invoke the routine. For example,

tspan = [0 10];
x0 = [p, sigmas, sigmad, alphas, alphad];
ode45(@fossil_fuels, tspan, x0)