MATLAB: Solving 4th order ode

ode

Hello everyone! I would appreciate your help in this section.
I have this 4th order,non-linear ,homogeneous ode that I would like to solve:
(2B+1+y)y''-k(1+y)y''''=0 with: y(0)=y(L)=0;y'(0)=y'(L)=1; where B,k and L are constants.
This is my matlab code so far:
L=1;B=0.25;k=0.1;
dydt = zeros(4,1);
dydt(1)=y(2);
dydt(2)=y(3);
dydt(3)=y(4);
dydt(4)=((2.*B+1+y(1)).*y(3))./(k.*(1+y(1)));
y1(1)=0; y1(end)=0; y2(1)=1; y2(end)=1;yint=[y1(1);y1(end);y2(1);y2(end)];
tspan = [0 1];
[t,y] = ode45(@(t,y) dydt, tspan, yint);
plot(t,y,'-o')
I think that something is not right. I have almost no expericance with solving ode's with MATLAB, especially with this high order so basically I gathered pieces of codes from things I saw online with the hope that it will assemble to a reasonable solution. I'm not sure I'm in a good path at all. I would be grateful for some guidance. Roi

Best Answer

Use bvp4c()
You already have written your equation as system of first-order equations
function dydt = ode4(t,y)
    L=1;B=0.25;k=0.1;
    dydt = zeros(4,1);
    dydt(1)=y(2);
    dydt(2)=y(3);
    dydt(3)=y(4);
    dydt(4)=(2.*B+1+y(1)).*y(3))./(k.*(1+y(1));
end
Here is function for your boundary conditions
function res = bc4(y0, y1)
    res = [y0(1)-0        % y(0) = 0
           y0(2)-1        % y'(0) = 1
           y1(1)-0        % y(end) = 0
           y1(2)-1];      % y'(end) = 1
end
See bvp4c
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