Prove: Theorem. Let M(x, y), N(x, y), ∂M/∂y and ∂N/∂x be continuous functions of x and y. Then a necessary and sufficient condition that the differential equation
1) M(x, y) dx + N(x, y) dy = 0
be exact is
Proof.
Part 1. We first prove that if 1) is exact then necessarily ∂M/∂y = ∂N/∂x. If 1) is exact then there exists a function u such that
where
2) M = ∂u/∂x , N = ∂u/∂y
Consequently, from 2) we get
From calculus we know
provided the partial derivatives are continuous. Thus ∂M/∂y = ∂N/∂x.
Part 2. We will now prove that if ∂M/∂y = ∂N/∂x, 1) will necessarily be exact i.e. that ∂M/∂y = ∂N/∂x is a sufficient condition for 1) to be exact. To do this it is sufficient to show that if ∂M/∂y = ∂N/∂x we can produce a function u(x, y) such that
3) du = M dx + N dy
Let Φ(x, y) be a function for which
It is obtained by integrating M dx with respect to x, keeping y constant i.e.
(where S M ∂x denotes integration with respect to x, keeping y constant). Let us now take the partial of both sides of 4) with respect to y:
Now we know from calculus that
so 6) can be written
and because ∂M/∂y = ∂N/∂x 7) can be written as
Let us integrate both sides of this last equation with respect to x, holding y fixed. The “arbitrary constant” will be any function of y. We will call it B'(y) for ease in referring to its primitive B(y). Integrating 8) with respect to x gives
9) ∂Φ/∂y = N + B'(y)
We will now draw a function out of a magic hat and claim that it will meet the condition of 3) above. It is the function
10) u = Φ(x, y) - B(y)
From 10) we get
Substituting 9) into11) we get
du = M dx + [N + B'(y) ] dy - B'(y) dy
or
du = M dx + N dy
References.
1. Earl D. Rainville. Elementary Differential Equations.