how to ues the exponentials and derivatives in bvp4c

I have a coupled non-linear differential equations

u''-b1*(t')*(u')+(1+b1*t)*[G1*F1*t+G2*F1*p-F3*P]=0;

t''-b2*(t')^2+B*F6*(u')^2+(b2-b1)*t*B*F6*(u')^2-b2*b1*B*F6*t^2*(u')^2=0;

p''- A*p=0

and the boundary conditions are

u=0,t=1+m,p=1+n at y=-1 and u=0,t=1,p=1 at y=1.

The code is:

clc;

p=0.01;

Betaf= 207;

Betas = 17;

Beta = 0.5;

kof = 0.613;

kos = 400;

m = -1;

b2 = 0.5;

b1 = 0.5;

G1 = 5;

G2 = 5;

A = 0.5;

Rhof = 997.1;

Rhos = 8933;

P1 = 0.5;

n=0.5;

B=0.01;

A1 = (1-p).^2.5;

A2 = 1/(1 + 1/Beta);

A3 = (1-p)+p.*((Rhos.*Betas)./(Rhof.*Betaf));

F1 = A2.*A3;

F3 = A1.*A2;

F4 = (kos + 2*kof - 2*p.*(kof - kos))/(kos + 2*kof + p.*(kof - kos));

F5 = (1 + 1/Beta)./A1;

F6 = F5./F4;

dydx=@(x,y)[y(4);

y(5);

y(6);

(b1.*y(4).*y(5)-(1+b1.*y(2)).*(G1.*F1.*y(2)+G2.*F1.*y(3)-F3.*P1));

b2.*y(5).^2-B.*F6.*y(4).^2-(b2-b1).*y(2).*B.*F6.*y(4).^2+b2.*b1.*B.*F6.*y(2).^2*y(4).^2;

A.*y(3)];

BC=@(ya,yb)[ya(1);yb(1);ya(2)-(1+m);yb(2)-1.0;ya(3)-(1+n);yb(3)-1.0];

yinit=[0.01;0.01;0.01;0.01;0.01;0.01];

solinit=bvpinit(linspace(-1,1,50),yinit);

S=bvp4c(dydx,BC,solinit)

I need to find T1= e^(-b1*(1+m))*(du/dy), here (du/dy) should be found at y = -1

T2= e^(-b1)*(du/dy), here (du/dy) should be found at y = 1

N1= -(1+b2*(1+m))*(dt/dy), here (dt/dy) should be found at y = -1

N1= -(1+b2)*(dt/dy), here (dt/dy) should be found at y = 1

how to write T1,T2,N1,N2 in the program

Please help me to find these values

Thank you

Answer 1

[Torsten]

yp1 = deval(S,1)
ym1 = deval(S,-1);
T1= exp(-b1*(1+m))*ym1(4);
T2= exp(-b1)*yp1(4);
N1= -(1+b2*(1+m))*ym1(5);
N2= -(1+b2)*yp1(5)

[Syed Mohiuddin]

Thanks a lot