MATLAB Solving One Equation on Matlab: Finding Parameter p

[FrancescoMi]

Hi, I have to solve one equation on Matlab. The equation is: (p-c)*psi'-psi=0 I have to find p. Where psi is:

and psi' is the derivative of psi. I've written this code:

rho = 0.04; %In the image is lambda
c = 25;
eta = 0.02;
mu = 40;
sigma  = 0.57;
alpha = -rho/eta;

syms p t
F = t^(-alpha-1)*exp(-t^2/2+((p-mu)/sigma*sqrt(2*eta))*t); %Funzione integranda non derivata
F_d = t^(-alpha-1)*(t*sqrt(2*eta))/sigma*exp(-t^2/2+((p-mu)/sigma*sqrt(2*eta))*t); %Funzione integranda derivata
D = (exp(((p-mu)/sigma*sqrt(2*eta))^2/4)/gamma(-alpha))*int(F,t,0,inf); %Funzione cilindrica non derivata
D_d = (exp((((p-mu)/sigma*sqrt(2*eta))^2)/4)/gamma(-alpha))*int(F_d,t,0,inf)+((eta*(p-mu))/(4*(sigma)^2))*(exp((((p-mu)/sigma*sqrt(2*eta))^2)/4)/gamma(-alpha))*int(F,t,0,inf); %Funzione cilindrica derivata
psi = exp((eta*(p-mu)^2)/(2*(sigma)^2))*D;
psi_d = (eta*(p-mu))/(sigma^2)*exp((eta*(p-mu)^2)/(2*(sigma)^2))*D + exp((eta*(p-mu)^2)/(2*(sigma)^2))*D_d; %Derivata di psi
Theta = (p-c)*psi_d-psi;

FreeBoundary = solve(Theta,p)

I've calculated the derivatives by myself, and I hope they are ok. What do you think about the code? It works only if rho>eta. Moreover for different values of sigma it give me some strange results.
Do you think the code is right and the derivatives are correct?

 

[kreil]

I think the problem with this post is the question posed:

"Do you think the code is right and the derivatives are correct?"

It is unreasonable to request that someone manually check all of your work and then debug your code on this scale. This could easily take a few hours, which is why nobody has responded.

Let's separate the math from the code: if you have a problem with the derivative or equations, post that in the Calculus forums. Once you nail down the expression you want to solve, I suggest you repost here with some updated code. The code you posted didn't run for me, so there is at least a little more work to be done:

Warning: Cannot solve symbolically. Returning a numeric
approximation instead. 
> In solve at 306 

FreeBoundary =

1.0

Any updates or other information you can provide?