how to solve equation

when

x=1
y=x.^2+3 /(x-1)

and when

x=2:100, y=x.^3+55

how do you solve this in matlab using vectorization method

Answer 1

[Image Analyst]

Try this:

% When x=1,  y=x.^2+3 /(x-1) and when x=2:100, y=x.^3+55,
x = 1 : 100;
% In general
y = x .^ 3 + 55;
% Special case for x = 1
y(1) = 1 .^ 2 + 3 ./ (1 - 1); % y will be infinity of course.
% Plot non-infinite results.
plot(x(2:end), y(2:end), 'b-', 'LineWidth', 2);
grid on;
fontSize = 15;
title('y = x .^ 3 + 55', 'FontSize', fontSize, 'Interpreter', 'none');
xlabel('x', 'FontSize', fontSize);
ylabel('y', 'FontSize', fontSize);

[Ls]

Sorry In the case of y(1), the equation is x.^2+3/(x+1)

[Image Analyst]

@Ls, then the line of code would of course be

y(1) = 1 .^ 2 + 3 ./ (1 + 1);

Answer 2

[David Hill]

y=zeros(size(x));
y(x==1)=3;
y(x~=1)=x(x~=1).^3+55;

[Walter Roberson]

You overlooked the /(x-1) when x = 1 ...

[David Hill]

He changed the question.

[Image Analyst]

Yes. Originally it said

"when x=1, y=x.^2+3 /(x-1) and when x=2:100, y=x.^3+55,"

as I captured in the comment in my code. Then in his comment to me (after your and my Answers were posted) he changed it to x.^2+3/(x+1). And I added a comment to update it to the new formula.

[Ls]

Thank you everyone as x-1 does not give proper value i changed the question. Anyway thank you again.