Kindly help me to solve this problem Thanks a lots!!! Non linear shooting method

% NONLINEAR SHOOTING ALGORITHM 11.2
%
% To approximate the solution of the nonlinear boundary-value problem
%
% Y'' = F(X,Y,Y'), A<=X<=B, Y(A) = ALPHA, Y(B) = BETA:
%
%
% INPUT: Endpoints A,B; boundary conditions ALPHA, BETA; number of
% subintervals N; tolerance TOL; maximum number of iterations M.
%
% OUTPUT: Approximations W(1,I) TO Y(X(I)); W(2,I) TO Y'(X(I))
% for each I=0,1,...,N or a message that the maximum
% number of iterations was exceeded.
syms('OK', 'A', 'B', 'ALPHA', 'BETA', 'TK', 'AA', 'N');
syms('TOL', 'NN', 'FLAG', 'NAME', 'OUP', 'H', 'K', 'W1');
syms('W2', 'U1', 'U2', 'I', 'X', 'T', 'K11', 'K12', 'K21');
syms('K22', 'K31', 'K32', 'K41', 'K42', 'J', 's', 'x', 'y', 'z');
TRUE = 1;
FALSE = 0;
fprintf(1,'This is the Nonlinear Shooting Method.\n');
fprintf(1,'Input the function F(X,Y,Z) in terms of x, y, z.\n');
fprintf(1,'followed by the partial of F with respect to y on the \n');
fprintf(1,'next line followed by the partial of F with respect to \n');
fprintf(1,'z or y-prime on the next line. \n');
fprintf(1,'For example: (32+2*x^3-y*z)/8 \n');
fprintf(1,' -z/8 \n');
fprintf(1,' -y/8 \n');
F = @(x,y,z) (y'cos(x)-y*log(y));
FY = @(x,y,z) (-y'sin(x)-log(y)-y);
FYP = @(x,y,z) (1);
OK = FALSE;
while OK == FALSE
fprintf(1,'Input left and right endpoints on separate lines.\n');
A = input(' ');
B = input(' ');
if A >= B
fprintf(1,'Left endpoint must be less than right endpoint.\n');
else OK = TRUE;
end;
end;
fprintf(1,'Input Y(%.10e).\n', A);
ALPHA = input(' ');
fprintf(1,'Input Y(%.10e).\n', B);
BETA = input(' ');
TK = (BETA-ALPHA)/(B-A);
fprintf(1,'TK = %.8e\n', TK);
fprintf(1,'Input new TK? Enter Y or N.\n');
AA = input(' ',';s');
if AA == 'Y' | AA == 'y'
fprintf(1,'input new TK\n');
TK = input(' ');
end;
OK = FALSE;
while OK == FALSE

fprintf(1,'Input an integer > 1 for the number of subintervals.\n');
N = input(' ');
if N <= 1
fprintf(1,'Number must exceed 1.\n');
else
OK = TRUE;
end;
end;
OK = FALSE;
while OK == FALSE
fprintf(1,'Input Tolerance.\n');
TOL = input(' ');
if TOL <= 0
fprintf(1,'Tolerance must be positive.\n');
else
OK = TRUE;
end;
end;
OK = FALSE;
while OK == FALSE
fprintf(1,'Input maximum number of iterations.\n');
NN = input(' ');
if NN <= 0
fprintf(1,'Must be positive integer.\n');
else
OK = TRUE;
end;
end;
if OK == TRUE
fprintf(1,'Choice of output method:\n');
fprintf(1,'1. Output to screen\n');
fprintf(1,'2. Output to text File\n');
fprintf(1,'Please enter 1 or 2.\n');
FLAG = input(' ');
if FLAG == 2
fprintf(1,'Input the file name in the form - drive:\\name.ext\n');
fprintf(1,'for example A:\\OUTPUT.DTA\n');
NAME = input(' ','s');
OUP = fopen(NAME,'wt');
else
OUP = 1;
end;
fprintf(OUP, 'NONLINEAR SHOOTING METHOD\n\n');
fprintf(OUP, ' I X(I) W1(I) W2(I)\n');
% STEP 1
W1 = zeros(1,N+1);
W2 = zeros(1,N+1);
H = (B-A)/N;
K = 1;
% TK already computed
OK = FALSE;
% STEP 2
while K <= NN &amp; OK == FALSE
% STEP 3
W1(1) = ALPHA;
W2(1) = TK;
U1 = 0 ;

U2 = 1;
% STEP 4
% Rung-Kutta method for systems is used in STEPS 5 and 6
for I = 1 : N
% STEP 5
X = A+(I-1)*H;
T = X+0.5*H;
% STEP 6
K11 = H*W2(I);
K12 = H*F(X,W1(I),W2(I));
K21 = H*(W2(I)+0.5*K12);
K22 = H*F(T,W1(I)+0.5*K11,W2(I)+0.5*K12);
K31 = H*(W2(I)+0.5*K22);
K32 = H*F(T,W1(I)+0.5*K21,W2(I)+0.5*K22);
K41 = H*(W2(I)+K32);
K42 = H*F(X+H,W1(I)+K31,W2(I)+K32);
W1(I+1) = W1(I)+(K11+2*(K21+K31)+K41)/6;
W2(I+1) = W2(I)+(K12+2*(K22+K32)+K42)/6;
K11 = H*U2;
K12 = H*(FY(X,W1(I),W2(I))*U1+FYP(X,W1(I),W2(I))*U2);
K21 = H*(U2+0.5*K12);
K22 = H*(FY(T,W1(I),W2(I))*(U1+0.5*K11)+FYP(T,W1(I),W2(I))*(U2+0.5*K21));
K31 = H*(U2+0.5*K22);
K32 = H*(FY(T,W1(I),W2(I))*(U1+0.5*K21)+FYP(T,W1(I),W2(I))*(U2+0.5*K22));
K41 = H*(U2+K32);
K42 = H*(FY(X+H,W1(I),W2(I))*(U1+K31)+FYP(X+H,W1(I),W2(I))*(U2+K32));
U1 = U1+(K11+2*(K21+K31)+K41)/6;
U2 = U2+(K12+2*(K22+K32)+K42)/6;
end;
% STEP 7
% test for accuracy
if abs(W1(N+1)-BETA) < TOL
% STEP 8
I = 0;
fprintf(OUP, '%3d %13.8f %13.8f %13.8f\n', I, A, ALPHA, TK);
for I = 1 : N
J = I+1;
X = A+I*H;
fprintf(OUP, '%3d %13.8f %13.8f %13.8f\n', I, X, W1(J), W2(J));
end;
fprintf(OUP, 'Convergence in %d iterations\n', K);
fprintf(OUP, ' t = %14.7e\n', TK);
% STEP 9
OK = TRUE;
else
% STEP 10
% Newton&#39;s method applied to improve TK
TK = TK-(W1(N+1)-BETA)/U1;
K = K+1;
end;
end;
% STEP 11
% method failed
if OK == FALSE
fprintf(OUP, 'Method failed after %d iterations\n', NN);
end;
end;

if OUP ~= 1
fclose(OUP);
fprintf(1,'Output file %s created successfully \n',NAME);
end;

Error :

F = @(x,y,z) (y'cos(x)-y*log(y));

↑

Error: Unexpected MATLAB expression.

Answer 1

[Alan Stevens]

You don't have multiplication signs between y' and cos and y' and sin:

F = @(x,y,z) (y'cos(x)-y*log(y));
FY = @(x,y,z) (-y'sin(x)-log(y)-y);

[Jin Hao Yen]

That's mean I straight away type the equation without y'?

[Alan Stevens]

You seem to have typed y' instead of y*

[Alan Stevens]

Or, if y is a vector and y' intentional, then you should have y'*.

You also have a lot of unneccessary semicolons (though these won't prevent the program from running). Look at the orange lines on the right of the editor. Hover your mouse over them.