Solving same symbolic equation but getting different results?

Question

Yilmaz Baris Erkan 2018-1-30

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/379848-solving-same-symbolic-equation-but-getting-different-results

评论： John D'Errico 2018-1-30

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.

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Star Strider 2018-1-30

Please describe what you want to do.

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

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Answer 1

John D'Errico 2018-1-30

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/379848-solving-same-symbolic-equation-but-getting-different-results#answer_302612

编辑：John D'Errico 2018-1-30

在 MATLAB Online 中打开

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.