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Solving same symbolic equation but getting different results?

Asked by Yilmaz Baris Erkan on 30 Jan 2018 at 20:38
Latest activity Commented on by John D'Errico
on 30 Jan 2018 at 23:02
if true
  % clear,clc
format short
syms dy v0 t theta g
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
eq1 = (dy == v0*t*sin(theta) - 1/2*g*t^2);
T1 = solve(eq1,t)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
dy = v0 * t *sin(theta) - 1/2*g*t^2;
T2 = solve(dy,t)
end

The code I am struggling to interpret is given above. I would expected the same result from both ways, however apperantly, it is not the case. Even though there might be a trivial distinction that I am missing, I would appreciate if anyone enlightens me about it.

Thank you in advance.

  1 Comment

Please describe what you want to do.

Do you want to integrate ‘eq1’ as a differential equation?

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1 Answer

Answer by John D'Errico
on 30 Jan 2018 at 22:35
Edited by John D'Errico
on 30 Jan 2018 at 22:38
 Accepted Answer

They are NOT the same equations!!!!!!!!

There is a difference. In the first equation, MATLAB is told that dy is a constant. The NAME of the equation is eq1. This is a quadratic equation with a constant term.

Now, look at the second equation.

You named it dy. dy is NOT seen as a constant here, but a container for the equation itself, much like eq1 was before. Importantly, see there is NO constant term in equation dy.

Note that even though you defined dy as a symbolic variable before, the line:

dy = v0 * t *sin(theta) - 1/2*g*t^2;

overwrites the value of dy. So when you try to "solve" dy, it solves dy by setting it to zero, and solving for t, thus dy==0.

What is the solution there? Clearly, that is either t=0, or t=2*v0*sin(theta)/g.

There is indeed a difference, and it is important to recognize.

  2 Comments

Thank you for showing me difference with detail. It helped me a lot.

Its kind of subtle, and I admit that they LOOK the same.

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