Why are exact differential equations called so?

Question

As the question says, why are exact differential equations called so?

From Wikipedia, I got "The nomenclature of "exact differential equation" refers to the exact differential of a function". That leads me to ask why is an exact differential called so? Usually, the term 'exact' in the context of English refers to some quantity that is not approximated in any way. How does that definition fit here?

Answer 1

An exact differential has the property that its integral over a path is path-independent: it does not depend upon the taken path, but only upon the origin and end of the path.

In mechanics, the work of a force can be path-dependent (e.g. friction), or path-independent (e.g. gravity), in which case it is also called conservative. When it is path-independent, the force can be associated with a potential energy, and the work is only the (opposite of the) potential energy variation; i.e. the difference between potential energies at origin and end of the path. Work and potential energy (Wikipedia)

Then why the path-independent case was called "exact"? Probably because calculating the work involves no numerical approximation: it is the difference between two values of a potential energy, which has a known expression for usual forces such as gravity.

On the other hand, when there is a non-conservative force such as friction, its contribution must be calculated by an integral and may not have a closed expression, hence the computing should be numerical and give an inexact result.

See also https://physics.stackexchange.com/questions/603147/why-does-an-exact-differential-mean-a-force-is-conservative.

Answer 2

A differential $k$ - form $\omega$ is exact if there exists a $(k-1)$ - form $\alpha$ such that $\omega=d \alpha$. This (applied to your context) motivates the name 'exact'.

EDIT: to give an example, consider the equation: $$P(x,y) dx + Q(x,y) dy =0 $$ This equation is exact if there is some $U$ such that: $$\frac{\partial U}{\partial x} = P$$ and $$\frac{\partial U}{\partial y} = Q$$ Notice that: $$ dU = \frac{\partial U}{\partial x} dx +\frac{\partial U}{\partial y} dy = P dx +Qdy$$ That is, $U$ is an exact differential form.

Answer 3

An exact differential equation is defined as an equation which has a solution of the form: $$du(x,y)=P(x,y)dx+Q(x,y)dy$$ if the DE is defined as: $$P(x,y)dx+Q(x,y)dy=0$$ leading to the general solution of: $$u(x,y)=C$$ It may be called an exact equation because it is based on the requirements of continuous pds or that the value of the constant can be worked out easily so values given are "exact" rather than how PDEs are often solved using numerical methods which are effectively good approximations