% Coded by NFR 4.25.19
% This code solves a boundary value (linear equation) ODE
%
span = [0 20];
y0 = [5 2]; % The second value is a guess for the inital value of Z
[tp, yp] = ode45(@dydt,span,y0);
CheckVal1 = yp(end,1)
y0 = [5 -1]; % The second value is a second guess for the inital value of Z
[tp, yp] = ode45(@dydt,span,y0);
CheckVal2 = yp(end,1)
% Interpolation - see equation 24.11
Slope = (-1-2)/(CheckVal2-CheckVal1);
Zinterp = Slope*(8-CheckVal1)+2;
% Use the interpolated value of z to get your solution
y0 = [5 Zinterp]; % The second value is a second guess for the inital value of Z
[tp, yp] = ode45(@dydt,span,y0);
CheckVal3 = yp(end,1)

function dy = dydt(t,y)
% t = x, y(1) = y, y(2) = z
dy = [y(2);
    (2*y(2)+y(1)-t)/7];
end