i make a code of shooting method for 3rd order differential equation but i dont get the accurate result

clear all;
clc;

%%%%%%%%%%%% first guess %%%%%%%%%%%%%%%
f = @(x, y)([y(3); y(2); x * y(1) + (x^3 - 2*x^2 - 5*x - 3) * exp(x)]);
gues1 = 1;
gues3 = 5;
x0_first = [0; gues1; gues3];
x = linspace(0, 1, 100);
[x, x_first] = ode45(f, x, x0_first);
u_10_first = x_first(end, 1);

%%%%%%%%%% 2nd guess %%%%%%%%%%
gues2 = 4;
gues4 = 9;
x0_second = [0; gues2; gues4];
[x, x_second] = ode45(f, x, x0_second);
u_10_second = x_second(end, 1);

%%%%%%%%% Linear interpolation %%%%%%%%%%
sol1 = u_10_first;
sol2 = u_10_second;
targ = 1;
targ2 = -2.718;

% Linear interpolation for the first target
gues5 = gues2 + ((gues1 - gues2) / (sol1 - sol2)) * (targ - sol2);
gues6 = gues4 + ((gues3 - gues4) / (sol1 - sol2)) * (targ2 - sol2);

yi = gues5;
zi = gues6;
x0_final_first_target = [0; yi; zi];
[x, x_final_first_target] = ode45(f, x, x0_final_first_target);
u_10_final_first_target = [x(end,end),x_final_first_target(end, 1)]

u_10_final_first_target = 1x2

1.0000 -2.7180

% Linear interpolation for the second target
gues7 = gues2 + ((gues1 - gues2) / (sol1 - sol2)) * (targ2 - sol2);
gues8 = gues4 + ((gues3 - gues4) / (sol1 - sol2)) * (targ - sol2);

yi = gues7;
zi = gues8;
x0_final_second_target = [0; yi; zi];
[x, x_final_second_target] = ode45(f, x, x0_final_second_target);
u_10_final_second_target = [x(end,end),x_final_second_target(end, 1)]

u_10_final_second_target = 1x2

1.0000 1.0000

% Plotting
figure;
plot(x, x_first(:, 1), 'r', x, x_second(:, 1), 'g', x, x_final_first_target(:, 1), 'k', x, x_final_second_target(:, 1), 'b');
legend('first guess', 'second guess', 'final value (target 1)', 'final value (target 2)');
grid on;

[Torsten]

If the underlying equation is a 3rd order differential equatiom, already your definition for f looks wrong.

Maybe you could state your problem first so that we have a chance to understand your code.

[kashif Rashid]

actually i want to make shooting method for 3rd order differential equation and my problem is

y'''-xy=(x^3-2x^2-5x-3)exp(x)

y(0)=0; y'(0)=1; y'(1)=-2.718

Answer 1

[Torsten]

y20 = 2;
sol = fsolve(@fun_shooting,y20);

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

[X,Y] = fun_ode(sol);
plot(X,Y(:,1))

function res = fun_shooting(y2)
  [X,Y] = fun_ode(y2);
  res = Y(end,2) + exp(1);
end

function [X,Y] = fun_ode(y2)
  f = @(x,y)[y(2);y(3);x*y(1)+exp(x)*(x^3-2*x^2-5*x-3)];
  y0 = [0;1;y2];
  xspan = [0 1];
  [X,Y] = ode45(f,xspan,y0);
end