Exact differential equations

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Consider the differential equation of the form

\[P(x,y)dx + Q(x,y)dy = 0 \tag{Eq:exact}\] 

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

\[\begin{eqnarray*} F_x(x,y) &=& P(x,y),\\ F_y(x,y) &=& Q(x,y). \end{eqnarray*}\]

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

Remark. Recall that

\[\begin{eqnarray} F_x(x,y) &=& \frac{\partial F}{\partial x}(x,y),\\ F_y(x,y) &=& \frac{\partial F}{\partial y}(x,y). \end{eqnarray}\]

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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

\[P(x,y)dx + Q(x,y)dy = 0,\tag{Eq:exactt}\]

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

\[\begin{eqnarray*} \frac{d}{dx}F(x, y(x)) &=& F_x(x,y) + F_y(x,y)\frac{dy}{dx}\\ &=& P(x,y) + Q(x,y)\frac{dy}{dx} = 0 \end{eqnarray*}\]

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

\[F(x,y) = \int_a^xP(u,b)du + \int_b^yQ(x,v)dv\tag{Eq:poten}\]

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 \(F_y(x,y) = Q(x,y)\).

\[\begin{eqnarray*} \frac{\partial}{\partial x}F(x,y) &=& \frac{d}{d x}\int_a^xP(u,b)du + \frac{\partial }{\partial x}\int_b^yQ(x,v)dv\\ &=& P(x,b) + \int_b^yQ_x(x,v)dv\\ &=& P(x,b) + \int_b^yP_y(x,v)dv\\ &=& P(x,b) + P(x,y) - P(x,b)\\ &=& P(x,y). \end{eqnarray*}\]

Similarly,

\[\begin{eqnarray}\frac{\partial}{\partial y}F(x,y) &=& \frac{\partial}{\partial y}\int_a^xP(u,b)du + \frac{\partial }{\partial y}\int_b^yQ(x,v)dv\\ &=& Q(x,y). \end{eqnarray}\]

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Example.  Let's solve

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

Let

\[\begin{eqnarray*} P(x,y) &=& 2x\sin(xy) + (x^2y + y^4)\cos(xy),\\ Q(x,y) &=& 3y^2\sin(xy) + (x^3 + xy^3)\cos(xy), \end{eqnarray*}\]

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

\[P\cdot \mu_y - Q\cdot \mu_x + \mu\cdot(P_y - Q_x) = 0.\tag{Eq:mucond}\]

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 \[\mu(x) = \exp\left[\int R(x) dx\right]\] is an integrating factor.
  2. If \(S(y) = -\frac{1}{P}(P_y - Q_x)\) is a function of \(y\) alone, then \[\mu(y) = \exp\left[\int S(y) dy\right]\] is an integrating factor.
Example. Let's solve
\[(x+2y)dx + dy = 0.\label{eq:inexact}\]
Let \(P = x + 2y\) and \(Q = 1\). Since \(P_y = 2 \neq 0 = Q_x\), this is not an exact differential equation. Noting \(R = \frac{1}{Q}(P_y - Q_x) = 2\), let \(\mu(x) = e^{\int R dx} = e^{2x}\). 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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