# its not taking the values of y when y value is ranging from 0.25 to 0.45 with an interval of 0.05,but it is able to take the value when y value is constant. and its coming in some matrix but with constant values.

2 views (last 30 days)
ellora on 23 Jan 2014
Commented: ellora on 26 Jan 2014
Y1 = 7.7054
Y2 = 9.6584
I(3)= -0.0043
I(4)=0.4141
n=4 ;
m=2;
y=0.25:0.05:0.45;
for j=1:m
for i=1:n
if (j==1)
ud(i+1,1)=y;
taub(i+1,1)=y;
ud(i+1,2)=(I(3)*exp(Y2*y)+I(4)*exp(-Y2*y)+k2)^0.5;
taub(i+1,2)=(f*ro*ud(i+1,2)^2)/8;
end
end
end
ud
taub
plot(ud(i,1),ud(i,2))
sir here y will be my first column and ud will be my second column,bt ud is dependent on y,y is varying from 0.25 to 0.45 by having an interval of 0.05. here i just want to change the value of y in each iteration.
ellora on 25 Jan 2014
table does nt mean taub,and i said 'find' that means i want to ve my data in a tabular format. that too the table should contain 2 columns,one for the value of y and the 2nd column for ud.there is nothing related to taub(its a different variable).
ellora on 26 Jan 2014
it worked.thx

Jos (10584) on 25 Jan 2014
ud(i+1,1)=y;
the left hand expression "ud(i+1,1)" is a single element, whereas the right hand expression "y" is a list of numbers. This will not work.
What should ud look like when i equals 1?
ellora on 26 Jan 2014
yeah i got it, i used ud(i+1,1)=y(i),then it worked.thx

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