I want to solve some coupled differential equations using ode45, but here I have to pass some constants in that function as well. How will the code be written.
MATLAB: I want to solve some coupled differential equations using ode45, but here I have to pass some constants in that function as well. How will the code be written.
ode45retag
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T= 323.15; A1= 4.01814;B1= 1203.835;C1= -53.226; A2= 4.08245;B2= 1346.382;C2= -53.508; Psat1= 10^(A1-(B1/(T+C1)));Psat2= 10^(A2-(B2/(T+C2)));x= [0:0.05:1]y= ((Psat1).*(x))./((x).*(Psat1)+(1.-x).*(Psat2))plot(x,y,'r+')Hi,
if you search a value for k that complies with the boundary conditions, you could use fzero to solve the problem numerically:
k = fzero(@calculate_k, 0.001);disp(['k = ', num2str(k)])[t,y] = ode45(@(t,y) k*y*exp(450/y), [0 300], 40);plot(t,y,'r')hold onscatter(15,95,'ob')text(23,96,'y(t=15) = 95','Color','b')hold offfunction k_value = calculate_k(x)tspan = [0 15];y0 = 40;[~,y] = ode45(@(t,y) x*y*exp(450/y), tspan, y0);k_value = y(end) - 95;endThis will give you as result:
k = 0.00010435or if you need it more accurate:
k = 1.043495007807761e-04and the plot which belongs to this result looks like the boundary conditions are met (To be precise, it keeps the boundary conditions. This is because fzero is a powerful tool and also because you can think of a very good initial value for x0 with just a few estimates...):
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Best regards
Stephan
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