Math 555: Differential Equations I

Lecture Companion

§2.6: Exact Equations and Integrating Factors

Consider a first order differential equation of the form

$$ M(x,y)\, dx + N(x,y)\,dy = 0. \tag{1} $$
Equation $(1)$ is said to be exact if and only if there is a function $\varphi(x,y)$ for which $\partial_x \varphi = M$ and $\partial_y \varphi = N$. Such a $\varphi$ will exist if and only if

$$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}. \tag{2} $$
In this case, assuming $y=y(x)$ is an implicit function of $x$, then equation $(1)$ may be rewritten as

$$ \frac{\partial \varphi}{\partial x} + \frac{\partial \varphi}{\partial y}\frac{dy}{dx} = \frac{d}{dx}\Big[\varphi(x,y(x))\Big] = 0. \tag{*} $$
Integrating, we obtain the solution

$$ \varphi(x,y) = C, \tag{3} $$
where

$$ \varphi(x,y) = \int M(x,y)\,dx = \int N(x,y)\, dy. \tag{4} $$
For readers with a Calculus III background, an exact differential equation is closely related to a conservative vector field. The solution $\varphi$ of the differential equation is a potential function of the vector field. Solution curves of the differential equation are equipotential lines of the vector field.

If the functions $M$ and $N$ of equation $(1)$ do not satisfy the criterion of equation $(2)$, then we must try to find an integrating factor, $\mu$.

An acceptable integrating factor is a function $\mu = \mu(x,y)$ that is never equal to $0$ (think: $\mu > 0$) and makes the equation

$$ \mu(x,y)M(x,y)\, dx + \mu(x,y)N(x,y)\,dy = 0 \tag{5} $$
exact. Equation $(5)$ being exact requires that

$$ \frac{\partial}{\partial y} \Big[ \mu(x,y)M(x,y)\Big] = \frac{\partial}{\partial x} \Big[ \mu(x,y)N(x,y)\Big]. \tag{6} $$
If $\mu$ is solely a function of $x$, and not $y$, then $\mu$ will satisfy the differential equation

$$ \frac{1}{\mu} d\mu = \frac{\partial_y M - \partial_x N}{N} dx \tag{8} $$
where the quotient on the righthand side must be a function of $x$ alone. Similarly, if $\mu$ is solely a function of $y$, then it will satisfy the differential equation

$$ \frac{1}{\mu} d\mu = \frac{\partial_y M - \partial_x N}{-M} dy \tag{9} $$
where the righthand side is a function of only $y$. Equations $(8)$ and $(9)$ give two special cases for choosing $\mu$, but in general it may be difficult to find an acceptable integrating factor.

We consider some special cases and look at some examples below.

Example 1. Suppose $M(x,y) = M(x)$ and $N(x,y) = N(y)$. Then equation $(2)$ is trivially satisfied, so that equation $(1)$ is exact. But in this case equation $(1)$ may be rewritten as

$$ \frac{dy}{dx} = \frac{-M(x)}{N(y)}; $$
a separable equation. So, all separable FODE are exact.

Example 2. Suppose $M(x,y) = p(x)y - q(x)$ and $N(x,y) = 1$. Then equation $(1)$ may be rewritten as

$$ \frac{dy}{dx} + p(x)y = q(x); $$
a linear equation. In general, linear equations are not exact. As an exercise, work through the procedure described above to determine $\mu$ as a function of $x$ alone. You should get the same equation for $\mu$ that we derived for linear DE,

$$ \mu(x) = e^{\int p(x)\,dx}. $$

Example 3. Consider the differential equation

$$ 2x + y^2 + 2xy\,y' = 0. $$
This DE is neither separable nor linear. Comparing with equation $(1)$, $M(x,y) = 2x + y^2$ and $N(x,y) = 2xy$. The partial derivatives are

$$ \frac{\partial M}{\partial y} = 2y,\ \ \ \frac{\partial N}{\partial x} = 2y, $$
so the equation is exact. Integrating we obtain

$$ \varphi = \int 2x + y^2\, dx = x^2 + xy^2 + C_1(y), $$
where $C_1$ could be an unknown function of $y$. Similarly,

$$ \varphi = \int 2xy\, dy = xy^2 + C_2(x), $$
where $C_2$ could be an unknown function of $x$. These two integrals must agree--they are both equal to $\varphi$. Therefore $C_2(x)$ must be set equal to $x^2$. The function $C_1(y)$ may be set to $0$. Therefore, we have obtained that

$$ \varphi = x^2 + xy^2 $$
is a function satisfying $\partial_x \varphi = M$ and $\partial_y \varphi = N$. Hence the solution to the DE is

$$ x^2 + xy^2 = C. $$
We plot the slope field and some solution curves below.

Example 4. Consider the differential equation

$$ (3xy-y^2)\, dx + x(x-y)\,dy= 0. $$
This equation is not exact since

$$ \frac{\partial M}{\partial y} = 3x - 2y,\ \ \ \frac{\partial N}{\partial x} = 2x -y. $$
Notice that $\partial_y M - \partial_x N = (x - y)$, so that

$$ \frac{\partial_y M - \partial_x N}{N} = \frac{1}{x} $$
is a function solely of $x$. Therefore an acceptable integrating factor for this equation is

$$ \mu(x) = e^{\int \frac{1}{x}\, dx} = x. $$
Using this integrating factor, the solution is

$$ \varphi = x^3y - \tfrac{1}{2}x^2y^2 = C. $$
The details are left as an exercise.

Again, we plot the slope field together with some solution curves.

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