Definition and Discrimination Method of an Exact Differential Equation
Definition
The given differential equation
$\psi=\psi (x,y)$
is said to be an exact differential equation if there exists $\psi=\psi (x,y)$ that satisfies
$\psi (x,y)$.
Explanation
If the given differential equation is exact, it can be represented as a total differential with respect to $\psi (x,y)$.
$d\psi (x,y)=\dfrac{\partial \psi }{\partial x}dx + \dfrac{\partial \psi }{\partial y}dy$
Since $d\psi (x,y)=\dfrac{\partial \psi }{\partial x}dx + \dfrac{\partial \psi }{\partial y}dy$, it follows that $d\psi (x,y)=0$ is also true. Hence,
$d\psi (x,y)=0$
That is, the solution of the differential equation is not represented as a function in the form of $y=y(x)$, but instead as an implicit function in the form of $\psi (x,y)=C$. Meanwhile, whether the given differential equation is exact or not can be determined according to the following theorem.
Theorem
Let function $M,\ N,\ M_{y},\ N_{x}$ be continuous. The subscript denotes partial differentiation with respect to the indicated variable. Then, the differential equation
$y=y(x)$
is exact if and only if
$\psi (x,y)=C$.
$M,\ N,\ M_{y},\ N_{x}$
Proof
$(\implies)$
If $M(x,y)dx+N(x,y)dy=0$ is exact, by definition, there exists $\psi$ satisfying:
$(\implies)$
Taking partial derivatives with respect to $y, x$ yields: $M(x,y)dx+N(x,y)dy=0$
By the assumption of continuity, it follows that:
$\psi$
Therefore,
$y, x$
That is,
$(\impliedby)$
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$(\impliedby)$
Assume $M_{y}=N_{x}$. Assume there exists $\psi (x,y)$ satisfying:
$M_{y}=N_{x}$
Then, proving that $\psi (x,y)$ satisfies $\psi_{y}=N$ completes the proof. Integrating both sides of $\eqref{eq1}$ with respect to $x$,
$\psi (x,y)$
Since $\psi$ is a function of two variables with respect to $x,y$, note that the constant of integration is a function of $y$, denoted as $h(y)$, not just $C$. Differentiating $h(y)$ with respect to $x$ gives $0$. Now, differentiating both sides of $\eqref{eq2}$ with respect to $y$ again,
$\psi (x,y)$
Upon arranging the equation with respect to $h^{\prime}(y)$,
$\psi_{y}=N$
Observing this equation reveals that the left side is a function solely of $y$. Hence, the right side must also be, which implies that differentiating the right side with respect to $x$ yields $0$. Differentiating the right side with respect to $x$,
$\eqref{eq1}$
The third equality is valid under the assumption of $M_{y}=N_{x}$. Since it must hold that $N=N(x,y)$ and it is independent of $N$, the expression inside the parenthesis equals $0$. Therefore,
$x$
Thus, if $M_{y}=N_{x}$, there exists $\psi (x,y)$ satisfying $\psi_{x}=M \ \mathrm{and}\ \psi_{y}=N$, and the given differential equation is exact.
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