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Solving same symbolic equation but getting different results?

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Yilmaz Baris Erkan
Yilmaz Baris Erkan 2018 年 1 月 30 日
コメント済み: John D'Errico 2018 年 1 月 30 日
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 件のコメント
Star Strider
Star Strider 2018 年 1 月 30 日
Please describe what you want to do.
Do you want to integrate ‘eq1’ as a differential equation?

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採用された回答

John D'Errico
John D'Errico 2018 年 1 月 30 日
編集済み: John D'Errico 2018 年 1 月 30 日
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 件のコメント
John D'Errico
John D'Errico 2018 年 1 月 30 日
Its kind of subtle, and I admit that they LOOK the same.

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