Exact Differential Equations, differential equations lesson

Let $x$ be the independent variable and write $y' = dy/dx$ . The solutions of $y' = -x/y$ lie on the circles $x^2 + y^2 = C$ , so the quantity $x^2 + y^2$ stays constant along every solution. Many first-order equations hide a conserved quantity of this kind, a function of both variables whose differential is the equation itself. This lesson shows how to recognize such an exact equation, how to build that function one variable at a time, and why a hole in the domain can prevent one function from working on the whole region.

Key terms

Exact differential

Definition Let $F(x, y)$ have continuous first partial derivatives on an open set $D$ in the plane. The partial derivative $F_x = \partial F/\partial x$ is the derivative in $x$ with $y$ held fixed, and $F_y = \partial F/\partial y$ is the derivative in $y$ with $x$ held fixed. The total differential of $F$ is

dF = F_x\,dx + F_y\,dy.

An expression $M\,dx + N\,dy$ is an exact differential on $D$ if it equals $dF$ for such a function $F$ , which means $F_x = M$ and $F_y = N$ at every point of $D$ . Along a differentiable graph $y = y(x)$ inside $D$ , with $y' = dy/dx$ , the chain rule gives

\begin{aligned} &\frac{d}{dx}F\bigl(x, y(x)\bigr) \\[0.4em] &= F_x + F_y\,y', \end{aligned}

with both partial derivatives evaluated at $\bigl(x, y(x)\bigr)$ .

In plain words An exact differential records how one function $F$ changes when $x$ and $y$ both change a little. Its two coefficients are the rates of change of $F$ in the two coordinate directions.

Example For $F(x, y) = xy^2 + \sin x$ , holding $y$ fixed gives $F_x = y^2 + \cos x$ , and holding $x$ fixed gives $F_y = 2xy$ . So

(y^2 + \cos x)\,dx + 2xy\,dy

is an exact differential on the whole plane, namely $dF$ .

Exact equation

Definition Let $M$ and $N$ be continuous functions of $(x, y)$ on an open set $D$ . The equation

M\,dx + N\,dy = 0

is exact on $D$ if $M\,dx + N\,dy$ is an exact differential there. With $x$ as the independent variable and $y' = dy/dx$ , a solution is a differentiable function $y$ on an open interval whose graph lies in $D$ and which satisfies $M\bigl(x, y(x)\bigr) + N\bigl(x, y(x)\bigr)\,y'(x) = 0$ . For an exact equation with $dF = M\,dx + N\,dy$ , the chain rule shows that $y$ is a solution exactly when $F\bigl(x, y(x)\bigr)$ is constant on the interval. So every solution lies on a level curve $F(x, y) = C$ . Conversely, if $(x_0, y_0)$ is a point of $D$ with $N(x_0, y_0) \ne 0$ , the implicit function theorem shows that the part of the level curve near $(x_0, y_0)$ is the graph of a differentiable solution with $y(x_0) = y_0$ , and its slope is $y' = -M/N$ .

In plain words An exact equation says that the change of some function $F$ along the solution is zero. So $F$ stays constant along every solution, and the solutions follow its level curves.

Example Where $y \ne 0$ , the equation $y' = -x/y$ can be written $x\,dx + y\,dy = 0$ , which is $dF = 0$ for $F = \tfrac{1}{2}(x^2 + y^2)$ . Its solutions lie on circles centered at the origin. Through $(0, 2)$ the level curve is $x^2 + y^2 = 4$ , and since $N = y = 2 \ne 0$ there, the solution near that point is the upper half $y = \sqrt{4 - x^2}$ . It continues on $(-2, 2)$ and ends at $x = \pm 2$ , where $N = y = 0$ and the circle turns vertical.

Potential function

Definition A potential function for $M\,dx + N\,dy$ on an open set $D$ is a function $F$ with continuous first partial derivatives such that $F_x = M$ and $F_y = N$ on $D$ . On a connected set any two potentials differ by a constant, so they have the same level curves. When the equation is exact on a rectangle or on the whole plane, a potential is built in three steps.

Integrate $M$ in $x$ with $y$ held fixed, and add an unknown function $h(y)$ in place of the constant of integration.
Differentiate the result in $y$ and set it equal to $N$ . Exactness guarantees that the resulting equation for $h'(y)$ involves $y$ alone.
Integrate $h'(y)$ , then check both $F_x = M$ and $F_y = N$ .

In plain words Recover $F$ from its two partial derivatives one variable at a time. The function $h(y)$ holds the part of $F$ that the first integration cannot see, because its derivative in $x$ is zero.

Example Take $M = 2xy + 3x^2$ and $N = x^2 + 4y$ , polynomials on the whole plane. Integrating $M$ in $x$ gives

F = x^2y + x^3 + h(y).

Then $F_y = x^2 + h'(y)$ must equal $x^2 + 4y$ , so $h'(y) = 4y$ and $h(y) = 2y^2$ , up to an additive constant that can be dropped. The potential is

F(x, y) = x^2y + x^3 + 2y^2,

and its partial derivatives are $2xy + 3x^2$ and $x^2 + 4y$ , as required. The solution through $(0, 1)$ lies on the level curve where $F = 2$ ,

x^2y + x^3 + 2y^2 = 2.

Since $N(0, 1) = 4 \ne 0$ , this relation defines a differentiable solution near $x = 0$ , and its slope $y' = -M/N$ is $0/4 = 0$ there, so the tangent at the initial point is horizontal. Solving the relation as a quadratic in $y$ and keeping the root with $y(0) = 1$ shows that along this solution $N = \sqrt{x^4 - 8x^3 + 16}$ . That is positive for $x < r$ , where $r \approx 1.339$ is the smallest positive root of $x^4 - 8x^3 + 16$ , so the solution interval is $(-\infty, r)$ . At $x = r$ we get $N = 0$ , the slope formula fails and the level curve has a vertical tangent.

Test for exactness

Definition Let $M$ and $N$ have continuous first partial derivatives on an open set $D$ . If $M\,dx + N\,dy = 0$ is exact on $D$ , then

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \quad \text{on } D.

The reason is that a potential $F$ then has continuous mixed partial derivatives $F_{xy} = M_y$ and $F_{yx} = N_x$ , and continuous mixed partial derivatives are equal. Conversely, if $M_y = N_x$ everywhere on $D$ and $D$ is an open rectangle, an open disk or more generally an open simply connected set, the equation is exact on $D$ . If $M_y \ne N_x$ at even one point of $D$ , the equation is not exact on $D$ , and by continuity it is not exact on any open set containing that point.

In plain words Before looking for $F$ , compare $M_y$ with $N_x$ . If they differ at some point, no potential exists near that point. If they agree throughout $D$ , a potential exists at least on every rectangle inside $D$ .

Example For the polynomial example above, $M_y = 2x$ and $N_x = 2x$ , so the test passes on the whole plane, which agrees with the potential already found. For $xy\,dx + x^2\,dy = 0$ the test gives $M_y = x$ and $N_x = 2x$ . They agree only on the line $x = 0$ , and every open set contains points off that line, so this equation is not exact on any open set. Not exact does not mean unsolvable. On the half-plane $x > 0$ , dividing by $x$ gives $y\,dx + x\,dy = 0$ , which is $d(xy) = 0$ , so the solutions there are $y = C/x$ . Multipliers of this kind are the integrating factors of the next lesson.

Local exactness

Definition The equation $M\,dx + N\,dy = 0$ is locally exact on an open set $D$ if every point of $D$ lies in an open rectangle inside $D$ on which the equation is exact. If $M$ and $N$ have continuous first partial derivatives on $D$ and $M_y = N_x$ on $D$ , the equation is locally exact on $D$ . On an open rectangle $R$ inside $D$ that contains $(x_0, y_0)$ , a potential is

\begin{aligned} F(x, y) &= \int_{x_0}^{x} M(s, y_0)\,ds \\[0.4em] &\quad + \int_{y_0}^{y} N(x, s)\,ds, \end{aligned}

the integral of the form along a horizontal segment and then a vertical segment, both inside $R$ . Local potentials on overlapping rectangles differ by constants, and those constants need not fit together around the whole of $D$ . So a locally exact equation need not be exact on $D$ .

In plain words Equal mixed partial derivatives always give a potential on small pieces of the domain. Whether the pieces join into one potential depends on the shape of the domain.

Example On the punctured plane, the plane without the origin, consider the angular form

\omega = \frac{-y\,dx + x\,dy}{x^2 + y^2}.

Its coefficients satisfy

M_y = N_x = \frac{y^2 - x^2}{(x^2 + y^2)^2}.

On the half-planes $x > 0$ and $x < 0$ the function $\arctan(y/x)$ is a potential, and on the half-planes $y > 0$ and $y < 0$ the function $-\arctan(x/y)$ is one. Every point other than the origin lies in one of these half-planes, so $\omega = 0$ is locally exact on the punctured plane. In the first quadrant both potentials apply, and they differ by the constant $\pi/2$ .

Simply connected domain

Definition A connected open set $D$ in the plane is simply connected if every closed curve in $D$ can be shrunk continuously to a point without leaving $D$ . For sets in the plane this means that $D$ has no holes. Open disks, rectangles, half-planes and the whole plane are simply connected, while the punctured plane and an annulus are not. If $D$ is open and simply connected, and $M$ and $N$ have continuous first partial derivatives with $M_y = N_x$ throughout $D$ , then $M\,dx + N\,dy = 0$ is exact on $D$ , with one potential defined on all of $D$ . Without that condition the conclusion can fail. The integral of an exact differential $dF$ around any closed curve in $D$ is $0$ , because $F$ returns to its starting value, so a form with a nonzero integral around some closed curve has no potential on $D$ .

In plain words A loop around a hole can carry a nonzero total change, which no single valued potential allows. In a region without holes every loop shrinks to a point, and this obstruction cannot occur.

Example Go once around the unit circle with $x = \cos\theta$ and $y = \sin\theta$ , so that $dx = -\sin\theta\,d\theta$ , $dy = \cos\theta\,d\theta$ and $x^2 + y^2 = 1$ . The numerator becomes $(\sin^2\theta + \cos^2\theta)\,d\theta = d\theta$ , so the angular form becomes $d\theta$ and

\oint \omega = \int_0^{2\pi} d\theta = 2\pi.

So $\omega = 0$ is not exact on the punctured plane, even though the test passes at every point there. On the plane with the nonpositive $x$ -axis removed, which is simply connected, the polar angle $\theta$ with $-\pi < \theta < \pi$ is a potential, and the solution curves there are the open rays from the origin that lie in this region.

Check yourself

Show that $(e^y + 2x)\,dx + (xe^y - 1)\,dy = 0$ is exact on the plane and find its solutions in implicit form.

Show the answer

The coefficients have continuous partial derivatives on the plane, which is simply connected, and $M_y = e^y = N_x$ , so the equation is exact. Integrating $M$ in $x$ gives $F = xe^y + x^2 + h(y)$ . Then $F_y = xe^y + h'(y)$ must equal $xe^y - 1$ , so $h'(y) = -1$ and $h(y) = -y$ . The solutions satisfy

xe^y + x^2 - y = C.

As a check, $F_x = e^y + 2x$ and $F_y = xe^y - 1$ .

Solve $(2x + y)\,dx + (x + 2y)\,dy = 0$ with $y(0) = 1$ , write the solution explicitly and give its solution interval.

Show the answer

Both coefficients are polynomials and $M_y = 1 = N_x$ on the plane, so the equation is exact. Integrating $M$ in $x$ gives $x^2 + xy + h(y)$ , and matching $F_y = x + h'(y)$ with $N = x + 2y$ gives $h'(y) = 2y$ , so $F = x^2 + xy + y^2$ . The point $(0, 1)$ gives $F = 1$ , so the solution lies on the ellipse $x^2 + xy + y^2 = 1$ . As a quadratic in $y$ this is $y^2 + xy + (x^2 - 1) = 0$ , and the root with $y(0) = 1$ is

y = \frac{-x + \sqrt{4 - 3x^2}}{2}.

Along it $N = x + 2y = \sqrt{4 - 3x^2}$ , which is positive exactly when $\lvert x \rvert < 2/\sqrt{3}$ . The solution interval is $(-2/\sqrt{3}, 2/\sqrt{3})$ , about $(-1.1547, 1.1547)$ . At the endpoints $N = 0$ and the ellipse has a vertical tangent, so the graph cannot be continued as a function of $x$ .

The equations $\dfrac{x\,dx + y\,dy}{x^2 + y^2} = 0$ and $\dfrac{-y\,dx + x\,dy}{x^2 + y^2} = 0$ both pass the test for exactness on the punctured plane. Which of them is exact there, and what does the answer say about the role of simple connectivity?

Show the answer

The first is exact on the punctured plane, with potential $F = \tfrac{1}{2}\ln(x^2 + y^2)$ , whose partial derivatives are $x/(x^2 + y^2)$ and $y/(x^2 + y^2)$ . Its solutions lie on the circles $x^2 + y^2 = C$ with $C > 0$ . The second is not exact there, because its integral around the unit circle is $2\pi$ , while the integral of $dF$ around a closed curve is $0$ .

The test passes for both, and the punctured plane is not simply connected, so the theorem decides neither case. Simple connectivity is a sufficient condition for the test to give one potential on the whole domain, not a necessary one.

Exact Differential Equations

Key terms

Exact differential

Exact equation

Potential function

Test for exactness

Local exactness

Simply connected domain

Check yourself

How this lesson connects

Builds on

Used later in