Method of finite differences

Hello, i have to solve this with a method of finite differences

x'' = -2x'(t) – 2x(t) + e^(-t) + sin(2t)

t ϵ [0,4]; x(0) = 0,6; x(4) = -0,1

And here is my code in Matlab

clear all;
al=0.6; bt=-0.1; h=0.01;
t=0:h:4;
n=length(t)-2; 
for i=1:n
	if i~=1, a(i,i-1)=-1+h;end
	a(i,i)=2-2*(h^2);
	if i~=n, a(i,i+1)=-1-h;end
	f(i)=-(h^2)*(exp(-t))-(h^2)*(sin(2*t(i)));
end
f(1)=f(1)-al*(-1+h);
f(n)=f(n)-bt*(-1+h);
x=a\f';
x1=[al,y',bt];
figure(1),plot(t,x1)

when i run this i get this error: Unable to perform assignment because the indices on the left side are not compatible with the size of the right side. Error in differences (line 9) f(i)=-(h^2)*(exp(-t))-(h^2)*(sin(2*t(i)));

And is my solution of the task without matlab:

x'' + 2x'(t) + 2x(t) = e^(-t) + sin(2t)

• a = 1; b = 2; c=2; d = e^(-t) + sin(2t)

Ai,i-1 = -ai + (h/2)bi =-1 + (h/2).2=-1+h

Ai,i = 2ai – h2ci = 2.1 – 2h^2 = 2 - 2h^2

Ai,i+1 = -ai - (h/2)bi = -1 – (h/2)2 = -1 - h

fi = -h^2di = -h^2(e^(-t)+ sin(2t))

so my questions are why im getting this error and are my coeficients right.

Thanks in advance. :)

[Adam]

t is a vector, f(i) is a scalar. You should pre-declare f before the loop, but even so f(i) would still be a scalar. I imagine you probably just want to use t(i) instead of t. You do this later on on the same line, but not for the first usage.

As for the seconod question, I can't answer that with a glance at the code!