📂Odinary Differential Equations

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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See Also

• Solving Exact Differential Equations