Help using ode45 to solve 3rd order ODE

I'm having trouble figuring out how to use the ode45 funciton in MATLAB to solve the following 3rd order ODE:

F''' + F*F'' = 0

With initial conditions F(0) = 0, F'(0) = 1, F''(inf) = 0, and F''(0) = -0.6276.

I am new to the ode45 function, and any help would be greatly appreciated

Answer 1

[Ameer Hamza]

Convert your 3rd order ODE into a system of 3-first order ODEs

IC = [0; 1; -0.6276];
tspan = [0 10];
[t, y] = ode45(@odefun, tspan, IC);

plot(t, y);
legend({'$F$', '$\dot{F}$', '$\ddot{F}$'}, 'FontSize', 16, 'Interpreter', 'latex', 'Location', 'best')

function dFdt = odefun(t, F)
    dFdt = zeros(3, 1);
    dFdt(1) = F(2);
    dFdt(2) = F(3);
    dFdt(3) = -F(1)*F(3);
end

[David Goodmanson]

Hi AJD,

Ameer's answer, while correct, does not say anything about the fact that you have a third order differential equation and four boundary conditions. That's one too many conditions, and overspecifies the problem. So you had better hope that if you use the three initial conditions at t=0, the answer will automatically end up with f''(inf) = 0. Fortunately, it does.

[Ameer Hamza]

Correct. This was just a coincidence. This equation can only have 3 boundary conditions.