Asked by Yilmaz Baris Erkan
on 30 Jan 2018 at 20:38

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.

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.

Yilmaz Baris Erkan
on 30 Jan 2018 at 22:49

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

John D'Errico
on 30 Jan 2018 at 23:02

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

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## 1 Comment

## Star Strider (view profile)

Direct link to this comment:https://www.mathworks.com/matlabcentral/answers/379848-solving-same-symbolic-equation-but-getting-different-results#comment_530198

Please describe what you want to do.

Do you want to integrate

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