Shooting method ode45 Fzero

I want to use the fzero function to approximate the value of t=1.65. However since the value of that point is about 501 and not zero, i am confused as how to code it. I am open to any solution as long as it uses the fzero function.

function shooting_method
x=1;
x1=fzero(@solver,x)
end
function F=solver(x)
% Parameters
L=3.6; %Length of stick
T0=315; % Temperature at x=0
TL=445; % temperature at x=L
options=odeset(RelTol=10^-6,AbsTol=10^-6);
[t,u]=ode45(@equation,[0 L],[T0 x],options);
s=length(t);
F=u(s,1)-TL;
figure(1)
plot(t,u(:,1))
end
function dy=equation(t,y)
L=3.6; %Length of stick
dy=zeros(2,1);
dy(1)=y(2);
dy(2)=((280*exp(-((t-L/2)^2))+y(2)/3)/(-(2+t/3)));
end

Answer 1

[Torsten]

Your code tries to determine dT/dt at t=0 such that the solution of the equation

d^2T/dt^2 = ((280*exp(-((t-L/2)^2))+(dT/dt)/3)/(-(2+t/3)))
T(0) = 315

passes through

T(L) = 445

and gets

dT/dt @(t=0) = 156

as solution.

L=3.6; %Length of stick
T0=315; % Temperature at x=0
TL=445; % temperature at x=L
x=1;
flag = 0;
x1=fzero(@(x)solver(x,flag,L,T0,TL),x)

x1 = 156.5218

flag = 1;
solver(x1,flag,L,T0,TL)

ans = 5.6843e-14

function F=solver(x,flag,L,T0,TL)
  options=odeset(RelTol=10^-6,AbsTol=10^-6);
  [t,u]=ode45(@(t,y)equation(t,y,L),[0 L],[T0 x],options);
  s=length(t);
  F=u(s,1)-TL;
  if flag==1
    plot(t,u(:,1))
  end
end
function dy=equation(t,y,L)
  dy=zeros(2,1);
  dy(1)=y(2);
  dy(2)=((280*exp(-((t-L/2)^2))+y(2)/3)/(-(2+t/3)));
end