How to solve a non linear 3rd order differential equation

Hi, I need to solve a 3rd order, non linear differential equation. I would preferably like a symbolic solution for the equation, as this equation is the input for another script. The equation is as follows:

where m is a constant (for my purposes, m =1)

f is a function of variable eta, and the initial conditions are as follows:

;

;

and

I tried using the code mentioned below, but it is unable to give me a symbolic solution. Can anybody please help me solve this problem?

m = 1;
syms f(eta)
Df = diff(f,eta);
D2f = diff(f,eta,2);
ode = diff(f,eta,3)+0.5*(m+1)*f*D2f+m*(1-Df^2) == 0;
cond1 = f(0) == 0;
cond2 = Df(0) == 0;
cond3 = D2f(10^100) == 1;
conds = [cond1 cond2 cond3];
uSol(eta) = dsolve(ode,conds)

[darova]

What about numerical approach? maybe ode45?

[madhan ravi]

It’s a BVP, probably you need bvpc4.

[Sumedh Soman]

Yes, that can be done, but I would prefer to have a symbolic solution. The function which I get is the starting point of another code, whereas, getting a numerical solution with ODE45 might not work in that next code, which is why I insist upon a symbolic solution. Besides, I don't know about the interval over which the solution needs to be integrated, as it is not required in the problem I'm trying to solve via this code. The function is an initiatializing value for that next calculation.

[Sumedh Soman]

I am not familiar with bvp4c actually; Nevertheless, I tried to use it, but matlab gives me an error, saying that, "Unable to solve the collocation equations -- a singular Jacobian encountered". I do not know how the solver actually works,can anyone help? The documentation page failed to help me understand how to use this solver.

[Sumedh Soman]

this is the solver code I used:

xmesh = linspace(0,100,5);
solinit = bvpinit(xmesh, @guess);
sol = bvp4c(@bvpfcn, @bcfcn, solinit);
plot(sol.x, sol.y, '-o');
function dydx = bvpfcn(x,y) % equation to solve
dydx = zeros(3,1);
dydx = [y(2) y(3) (-((1+1)/2)*(y(3)*y(1))-1*(1-(y(2))^2))];
end

function res = bcfcn(ya,yb) 
res = [ya(1) ya(2) yb(3)];
end

function g = guess(x) 
g = [x 1 0];
end

I saw that y = x was a solution of this equation, which is why I have used it as guess(x)

[darova]

here are some corrections