need help with problem

what is wrong in this code i dont get the correct answer

xleft=0; 
xright=(9*pi)/4;
N=20;
x=linspace(xleft,xright,N);
x=x';
dx=(xright-xleft)/(N-1);
yleft=1;
yright=1;

a=1;
b=-2;
c=-2;
% M is the matrix when multiplied by y gives y'' y' and y
M2=(diag(-2*ones(N,1),0)+diag(ones(N-1,1),-1)+diag(ones(N-1,1),1))/dx^2;
M1=(diag(ones(N-1,1),1)-diag(ones(N-1,1),-1))/(2*dx);
M0=eye(N);
M=a*M2+b*M1+c*M0;
%treating boundaries separate
M(1,:)=[1,zeros(1,N-1)];
M(end,:)=[zeros(1,N-1),1];

k=(sin(x))/((cos(x))+2);
k(1) = yleft; 
k(end) = yright; 
y=M\k

Answer 1

[Matt J]

is my code correct i dont get the correct answer

Then by definition, it is not correct...

[dulanga]

what is wrong in the code any help?

[Matt J]

k=(sin(x))./((cos(x))+2);
k(1) = yleft; 
k(end) = yright; 
y=M\k(:)

[dulanga]

thanks a lot Matt

[dulanga]

what changes would you make if say the boundary chnaged from y(0)=1 to y'(0)=1?

[Matt J]

That would be the same as saying

M1(1,:)*k(:)=1

would it not?

[dulanga]

yes but what if the function is y''= smthing in this case there is no m1 for instantance this question

i think the M matrix and b has to chnage altoghter but i am not sure how ?

[Matt J]

Why is there no M1? The only thing that has changed is b=c=0.

[dulanga]

yes so when it goes to M equation since b is zero M1 doesnt matter anymore ?

There is a M1 yes since b*M1 and b is zero M1 holds no impact on M then ?

Answer 2

[Muhammad Ibrahim]

k=(sin(x))/((cos(x))+2);

should be

k=(sin(x))./((cos(x))+2);

That small dot after sin(x) can make a big difference.

Also. read the question carefully, you might be keying the maximum value when it maybe asking minimum value.If you still have problems just ask again in this forum maybe I can help out.

[dulanga]

How would u solve this ?