Exact differential equations

Consider the differential equation of the form

$$\begin{array}{}\text{(Eq:exact)}& P(x,y)dx+Q(x,y)dy=0\end{array}$$ 

where $P(x,y)$ and $Q(x,y)$ are some bivariate functions. In this differential equation, the variables $x$ and $y$ have equal status (neither is an independent nor dependent variable). Thus, the solution should also be given as some equation of $x$ and $y$. If necessary, $y$ may be interpreted as an implicit function of $x$.

The above differential equation (Eq:exact) is said to be an exact or total differential equation if there exists a function $F(x,y)$ of class $C^1$ such that

$$\begin{array}{rcl}{F}_x(x,y)& =& P(x,y),\\ F_y(x,y)& =& Q(x,y).\end{array}$$

This function $F(x,y)$ is called a potential function.

Remark. Recall that

$$\begin{array}{rcl}{F}_x(x,y)& =& \frac{\partial F}{\partial x}(x,y),\\ F_y(x,y)& =& \frac{\partial F}{\partial y}(x,y).\end{array}$$

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As the following theorem shows, exact differential equations are solvable.

Theorem (Solution of an exact differential equation)

Given the exact differential equation

$$\begin{array}{}\text{(Eq:exactt)}& P(x,y)dx+Q(x,y)dy=0,\end{array}$$

if there exists a function $F(x,y)$ such that $F_x(x,y)=P(x,y)$ and $F_y(x,y)=Q(x,y)$, then the solution is $F(x,y)=C$ (constant).

Proof. Suppose $F_x(x,y)=P(x,y)$ and $F_y(x,y)=Q(x,y)$. If $F(x,y)=C$, then differentiating both sides with respect to $x$ yields

$$F_x(x,y)+F_y(x,y)\frac{dy}{dx}=0.$$

Hence $F(x,y)=C$ is indeed a solution of (Eq:exactt). Conversely, if the solution of (Eq:exactt) is given by $y=y(x)$, for example, then

$$\begin{array}{rcl}\frac{d}{dx}F(x,y(x))& =& F_x(x,y)+F_y(x,y)\frac{dy}{dx}\\ & =& P(x,y)+Q(x,y)\frac{dy}{dx}=0\end{array}$$

so that $F(x,y)=C$ (constant). ■

Hence, solving an exact differential equation boils down to finding the potential function. But how can we find it?

Theorem (Condition for exact differential equations)

The differential equation

$$P(x,y)dx+Q(x,y)dy=0$$

with $P$ and $Q$ of class $C^1$ is exact if and only if

$$P_y(x,y)=Q_x(x,y).$$

In this case, the potential function $F(x,y)$ is given by

$$\begin{array}{}\text{(Eq:poten)}& F(x,y)={\int }_a^{x}P(u,b)du+{\int }_b^{y}Q(x,v)dv\end{array}$$

where $(a,b)$ is an arbitrary point in the domain of $P(x,y)$ and $Q(x,y)$.

Proof. If $P(x,y)dx+Q(x,y)dy=0$ is an exact differential equation, then 

$$P_y(x,y)=F_{xy}(x,y)=F_{yx}(x,y)=Q_x(x,y)$$

as $P$ and $Q$ are of class $C^1$, which implies that the potential $F(x,y)$ is of class $C^2$.

Conversely, suppose that $P_y(x,y)=Q_x(x,y)$ holds. Let us define $F(x,y)$ as in (Eq:poten). It suffices to show that $F_x(x,y)=P(x,y)$ and $F_y(x,y)=Q(x,y)$.

$$\begin{array}{rcl}\frac{\partial }{\partial x}F(x,y)& =& \frac{d}{dx}{\int }_a^{x}P(u,b)du+\frac{\partial }{\partial x}{\int }_b^{y}Q(x,v)dv\\ & =& P(x,b)+{\int }_b^y{Q}_x(x,v)dv\\ & =& P(x,b)+{\int }_b^y{P}_y(x,v)dv\\ & =& P(x,b)+P(x,y)-P(x,b)\\ & =& P(x,y).\end{array}$$

Similarly,

$$\begin{array}{rcl}\frac{\partial }{\partial y}F(x,y)& =& \frac{\partial }{\partial y}{\int }_a^{x}P(u,b)du+\frac{\partial }{\partial y}{\int }_b^{y}Q(x,v)dv\\ & =& Q(x,y).\end{array}$$

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Example.  Let's solve

$$\{2x\sin (xy)+(x^{2}y+y^4)\cos (xy)\}dx+\{3y^2\sin (xy)+(x^3+x{y}^3)\cos (xy)\}dy=0.$$

Let

$$\begin{array}{rcl}P(x,y)& =& 2x\sin (xy)+(x^{2}y+y^4)\cos (xy),\\ Q(x,y)& =& 3y^2\sin (xy)+(x^3+x{y}^3)\cos (xy),\end{array}$$

and we have

$$P(x,y)dx+Q(x,y)dy=0.$$

Let us define

$$F(x,y)=(x^2+y^3)\sin (xy).$$

Then, $F_x(x,y)=P(x,y)$ and $F_y(x,y)=Q(x,y)$. Thus, the solution is

$$(x^2+y^3)\sin (xy)=C\text{(constant)}.$$

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Integrating factor

Consider the differential equation $Pdx+Qdy=0$ which may not be exact. In such a case, there may (or may not) exist a function $\mu (x,y)$ such that $\mu Pdx+\mu Qdy=0$ is an exact differential equation. Such a function $\mu (x,y)$ is called an integrating factor.

In general, finding an integrating factor is not easy. Let us find a condition an integrating factor should satisfy. By the above Theorem (Condition for exact differential equations), an integrating factor $\mu =\mu (x,y)$ must satisfy $(\mu P{)}_y=(\mu Q{)}_x$, that is,

$$\begin{array}{}\text{(Eq:mucond)}& P\cdot {\mu }_y-Q\cdot {\mu }_x+\mu \cdot (P_y-Q_x)=0.\end{array}$$

However, solving this equation for $\mu$ is difficult in general. Therefore, we consider a few special cases.

1. If $R(x)=\frac{1}{Q}(P_y-Q_x)$ is a function of $x$ alone, then $$\mu (x)=\exp [\int R(x)dx]$$ is an integrating factor.
2. If $S(y)=-\frac{1}{P}(P_y-Q_x)$ is a function of $y$ alone, then $$\mu (y)=\exp [\int S(y)dy]$$ is an integrating factor.

Example. Let's solve

$$(x+2y)dx+dy=0.$$

Let $P=x+2y$ and $Q=1$. Since $P_y=2\ne 0=Q_x$, this is not an exact differential equation. Noting $R=\frac{1}{Q}(P_y-Q_x)=2$, let $\mu (x)=e^{\int Rdx}=e^{2x}$. Then, with $\mu P=e^{2x}(x+2y)$ and $\mu Q=e^{2x}$, we have $(\mu P{)}_y=2e^{2x}=(\mu Q{)}_x$. Hence

$$(\mu P)dx+(\mu Q)dy=0$$

is exact. Define

$$F(x,y)=\frac{1}{4}{e}^{2x}(2x+4y-1),$$

and this satisfies $F_x=\mu P$ and $F_y=\mu Q$. Thus, the solution is

$$e^{2x}(2x+4y-1)=C\text{ (constant)}.$$

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