MATLAB Answers


Solving same symbolic equation but getting different results?

Asked by Yilmaz Baris Erkan on 30 Jan 2018
Latest activity Commented on by John D'Errico
on 30 Jan 2018
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)
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
Edited by John D'Errico
on 30 Jan 2018
 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.


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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