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I understand that if a constraint equation given on a differential form is exact, that means it is also holonomic since I can find a solution. But there are other types of differential equations, like separable and linear, such that I can find an equation in the form of a holonomic constraint. Why is it that being exact is a condition, rather than just having a solution of the type $f(x,y) = 0$?

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    $\begingroup$ $f(x,y)=0$ looks like a holonomic constraint. $\endgroup$ – Qmechanic Nov 26 '17 at 11:47
  • $\begingroup$ Well, yeah... I meant that. $\endgroup$ – Ana Luiza Ferrari Nov 27 '17 at 20:10
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  1. Here I would like to mention the notion of a semi-holonomic constraint $$ \sum_{j=1}^n a_j(q,t)~ \mathrm{d}q^j + a_t(q,t)~\mathrm{d}t~=~0, \tag{1'}$$ which puts an inexact differential equal to zero.

  2. It is possible to incorporate semi-holonomic constraints into Lagrange equations, cf. e.g. my Phys.SE answer here.

  3. Note that eq. (1') does not mean that we demand that the $n+1$ co-vector components $$ a_t(q,t)~= a_1(q,t)~= \ldots ~=~a_n(q,t)~=~0 \qquad\qquad (\longleftarrow \text{Wrong!} ) \tag{2'}$$ of a co-vector (1') should be zero. Perhaps this potential misunderstanding (2') is secretly the core of OP's question?

  4. Rather eq. (1') means that $$ \sum_{j=1}^n a_j(q,t)~ \dot{q}^j + a_t(q,t)~=~0. \tag{3'}$$

  5. In particular, a holonomic constraint $$f(q,t)~=~0 \tag{0}$$ can be put on the above semi-holonomic form $$ \mathrm{d}f~=~\sum_{j=1}^n \frac{\partial f}{\partial q^j}~ \mathrm{d}q^j + \frac{\partial f}{\partial t}~\mathrm{d}t~=~0 .\tag{1}$$ Explicitly, eq. (1) means that the total time derivative $$ \frac{df}{dt}~\equiv~ \sum_{j=1}^n \frac{\partial f}{\partial q^j}~ \dot{q}^j + \frac{\partial f}{\partial t}~=~0\tag{3}$$ is zero.

References:

  1. H. Goldstein, Classical Mechanics, Section 2.4.
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