How to solve third order equation using ode45

Hi, I was wondering if you could help me, I'm trying to solve the following third order equation, using ode45 for the range from [0,5]. Apart from that, I'd need to plot the different solutions of y', y'' and y.

My function is this one:

y'''-y'' * y +1 = 0

y(0) =1

y'(0)=0

y''(0)=0.1

I don't know how to apply ode45 for this equation, I'd gladly accept any help.

Thank you very much.

[Torsten]

Setting y(1) = y, y(2) = y' and y(3) = y'', your differential eqation can be written as a system of equations:

dy(1)/dt = y(2)
dy(2)/dt = y(3)
dy(3)/dt = y(3)*y(1) - 1

with initial conditions

y(1)(0) = 1
y(2)(0) = 0
y(3)(0) = 0.1

Now look at the page of ODE45 on how to set up this system:

https://de.mathworks.com/help/matlab/ref/ode45.html

Answer 1

[Star Strider]

I let the Symbolic Math Toolbox do everything —

syms y(t) T Y 
Dy = diff(y);
D2y = diff(Dy);
D3y = diff(D2y);
Eq = D3y - D2y * y + 1 

Eq(t) = 

[VF,Sbs] = odeToVectorField(Eq)

VF = 

Sbs = 

yfcn = matlabFunction(VF, 'Vars',{T,Y})

yfcn = function_handle with value:
    @(T,Y)[Y(2);Y(3);Y(1).*Y(3)-1.0]

ics = [0 0 0];
[t,y] = ode45(yfcn, [0 5], ics);

figure
plot(t,y)
grid
legend(string(Sbs), 'Location','best')

.

Answer 2

[Sam Chak]

@Álvaro Recalde, I'll show an example from

https://www.mathworks.com/help/matlab/ref/ode45.html#bu3uhuk

function dydt = odefcn(t, y)

dydt = zeros(3,1);

dydt(1) = y(2); % y' = ...

dydt(2) = y(3); % y'' = ...

dydt(3) = - 3*y(3) - 3*y(2) - y(1); % y''' = ...

end

and run ode45 to solve it

tspan = [0 10];
y0 = [1 0.5 0];
[t, y] = ode45(@odefcn, tspan, y0);