Integrating Factors for Nonexact Equations, differential equations lesson

The equation $y\,dx + 2x\,dy = 0$ is not exact, because $M_y = 1$ while $N_x = 2$ . Multiplied by $x^{-1/2}$ on the half-plane $x > 0$ , it becomes exact, and the potential $2y\sqrt{x}$ is then constant along every solution in that half-plane. This lesson shows when a multiplier that depends on only one variable exists, how to compute it from a quotient of partial derivatives, and how to recover the solutions that live where the multiplier is undefined or zero.

Key terms

Nonexact equation

Definition Let $M$ and $N$ be continuously differentiable on an open region $R$ of the $xy$ -plane. The equation

M\,dx + N\,dy = 0

is nonexact on $R$ if $M_y \ne N_x$ at some point of $R$ . Then no potential exists on any open set containing that point, because a potential $F$ with $F_x = M$ and $F_y = N$ would be twice continuously differentiable, and its equal mixed partial derivatives would force $M_y = N_x$ there. An integrating factor on $R$ is a continuously differentiable function $\mu(x, y)$ with no zeros on $R$ such that

(\mu M)_y = (\mu N)_x

at every point of $R$ . Since $\mu \ne 0$ , a direction $(dx, dy)$ satisfies $\mu M\,dx + \mu N\,dy = 0$ exactly when it satisfies the original equation, so on $R$ the two equations have the same solution curves. In particular, a graph $y = y(x)$ in $R$ , with $y' = dy/dx$ , solves $M + Ny' = 0$ exactly when it solves $\mu M + \mu N y' = 0$ .

In plain words The left side of a nonexact equation is not the differential of any function. Multiplying both coefficients by the same nonzero function does not change which directions the equation allows, but it can turn the left side into the differential of a potential.

Example For $y\,dx - x\,dy = 0$ we have $M_y = 1$ and $N_x = -1$ , so the equation is nonexact on every region. On the half-plane $x > 0$ the factor $\mu = 1/x^2$ gives

\frac{y}{x^2}\,dx - \frac{1}{x}\,dy = 0,

which is exact with potential $F = -y/x$ , since $F_x = y/x^2$ and $F_y = -1/x$ . Its level curves, and so the solution curves in this half-plane, are the half-lines $y = kx$ with $x > 0$ . On the half-plane $y > 0$ the factor $1/y^2$ works as well, with potential $x/y$ , so an integrating factor is not unique.

Integrating factor depending on x

Definition Let $M$ and $N$ be continuously differentiable on an open rectangle $R = I \times J$ , where $I$ and $J$ are open intervals, and suppose $N \ne 0$ on $R$ . A continuously differentiable function $\mu(x)$ with no zeros on $I$ is an integrating factor on $R$ if and only if

\frac{\mu'(x)}{\mu(x)} = \frac{M_y - N_x}{N}

at every point of $R$ . The reason is that $(\mu M)_y = \mu M_y$ , while the product rule gives $(\mu N)_x = \mu' N + \mu N_x$ . Setting the two equal and dividing by $\mu N$ gives the condition. So such a factor exists exactly when the quotient on the right depends on $x$ alone, say equals $r(x)$ . The function $r$ is then continuous on $I$ , and one choice of factor is

\mu(x) = e^{\int r(x)\,dx},

with any antiderivative of $r$ in the exponent. Since a rectangle is simply connected, the multiplied equation then has a potential on all of $R$ .

In plain words Try a multiplier that depends on $x$ only. It works precisely when the quotient $(M_y - N_x)/N$ depends on $x$ only, and the multiplier is then the exponential of an antiderivative of that quotient.

Example For $y\,dx + 2x\,dy = 0$ on the half-plane $x > 0$ , the quotient is

\frac{M_y - N_x}{N} = \frac{1 - 2}{2x} = -\frac{1}{2x}.

An antiderivative is $-\tfrac{1}{2}\ln x$ , so $\mu = e^{-\frac{1}{2}\ln x} = x^{-1/2}$ . The multiplied equation

\frac{y}{\sqrt{x}}\,dx + 2\sqrt{x}\,dy = 0

has both cross derivatives equal to $1/\sqrt{x}$ , and $F = 2y\sqrt{x}$ is a potential, because $F_x = y/\sqrt{x}$ and $F_y = 2\sqrt{x}$ . The solutions satisfy $2y\sqrt{x} = C$ . The level $C = 4$ gives $y = 2/\sqrt{x}$ , and substituting $y' = -x^{-3/2}$ into the graph form $y + 2xy' = 0$ of the original equation gives $2x^{-1/2} - 2x^{-1/2} = 0$ .

Integrating factor depending on y

Definition Let $M$ and $N$ be continuously differentiable on an open rectangle $R = I \times J$ , and suppose $M \ne 0$ on $R$ . A continuously differentiable function $\mu(y)$ with no zeros on $J$ is an integrating factor on $R$ if and only if

\frac{\mu'(y)}{\mu(y)} = \frac{N_x - M_y}{M}

at every point of $R$ . This time the product rule acts on the other side, since $(\mu M)_y = \mu' M + \mu M_y$ and $(\mu N)_x = \mu N_x$ . So such a factor exists exactly when the quotient depends on $y$ alone, say equals $s(y)$ , which is then continuous on $J$ , and one choice is $\mu(y) = e^{\int s(y)\,dy}$ .

In plain words The test for a multiplier of $y$ alone mirrors the test for $x$ , but the difference in the numerator is reversed and the denominator is $M$ instead of $N$ . Mixing up the two versions is the most common slip.

Example For $2y\,dx + x\,dy = 0$ on the half-plane $y > 0$ , we have $M_y = 2$ and $N_x = 1$ , so

\frac{N_x - M_y}{M} = \frac{1 - 2}{2y} = -\frac{1}{2y},

and the factor is $\mu = y^{-1/2}$ . The multiplied equation $2\sqrt{y}\,dx + (x/\sqrt{y})\,dy = 0$ is exact with potential $F = 2x\sqrt{y}$ , because $F_x = 2\sqrt{y}$ and $F_y = x/\sqrt{y}$ . For $x > 0$ the level $2x\sqrt{y} = C$ with $C > 0$ is the curve $y = C^2/(4x^2)$ , and indeed every $y = k/x^2$ satisfies $2y + xy' = 0$ . In this equation the quotient for $x$ also depends on one variable, since $(M_y - N_x)/N = 1/x$ , so the factor $x$ works too, with potential $x^2 y$ .

Quotient tests

Definition For $M\,dx + N\,dy = 0$ with $M$ and $N$ continuously differentiable on an open rectangle, the quotient tests look for an integrating factor of one variable in four steps.

Compute $M_y - N_x$ . If it is zero on the rectangle, the equation is already exact there.
Where $N \ne 0$ , check whether $(M_y - N_x)/N$ depends on $x$ alone. If it does, its exponentiated antiderivative is a factor $\mu(x)$ .
Where $M \ne 0$ , check whether $(N_x - M_y)/M$ depends on $y$ alone. If it does, its exponentiated antiderivative is a factor $\mu(y)$ .
Check $(\mu M)_y = (\mu N)_x$ directly, construct a potential, and state the region where it is valid.

A failed test shows only that no factor of that one-variable kind exists on the region. Integrating factors that depend on both variables may still exist.

In plain words Compute the mismatch $M_y - N_x$ once, divide it by $N$ and by $-M$ , and look for a quotient that contains only one variable. The integrating factor is the exponential of an antiderivative of that quotient.

Example The linear equation $y' + y = x$ , with $y' = dy/dx$ , can be written as $(y - x)\,dx + dy = 0$ . Here $M_y - N_x = 1$ , so the quotient for $x$ is the constant $1$ , while the quotient for $y$ is $-1/(y - x)$ , which depends on both variables. The first test gives $\mu = e^x$ , and

e^x(y - x)\,dx + e^x\,dy = 0

has the potential $F = e^x(y - x + 1)$ . Setting $F = C$ gives $y = x - 1 + Ce^{-x}$ on the whole line. For every linear equation $y' + p(x)y = q(x)$ the quotient for $x$ is $p(x)$ , so the familiar factor $e^{\int p(x)\,dx}$ of a linear equation is a special case of this test.

Domain restrictions

Definition The multiplied equation $\mu M\,dx + \mu N\,dy = 0$ has the same solution curves as the original only on a region where $\mu$ is defined, finite and nonzero, and the quotient tests also assume $N \ne 0$ or $M \ne 0$ where they divide. Outside such a region two things can go wrong. Where $\mu$ is undefined or infinite, solutions of the original equation that reach that set are not described by the potential and must be found from the original equation. Where $\mu = 0$ , every direction satisfies the multiplied equation, so it gives no information about the slope there, and those points must also be checked in the original equation. A factor found under a division may satisfy $(\mu M)_y = (\mu N)_x$ on a larger set, but that has to be checked in the multiplied equation, not assumed.

In plain words Multiplying is safe only where the multiplier is a finite nonzero number. Wherever it blows up, vanishes or was found by dividing by zero, go back to the original equation and look for solutions there.

Example The equation $y\,dx + 2x\,dy = 0$ is defined on the whole plane, but the factor $x^{-1/2}$ exists only for $x > 0$ . The nonzero solutions $y = C/(2\sqrt{x})$ blow up as $x \to 0^+$ . The zero function, however, satisfies $y + 2xy' = 0$ for every real $x$ , so $y = 0$ is a solution on the whole line that crosses the excluded line $x = 0$ . On the half-plane $x < 0$ the factor is $(-x)^{-1/2}$ , the potential is $-2y\sqrt{-x}$ , and the solutions are $y = k/\sqrt{-x}$ . Finally, the vertical line $x = 0$ is a solution curve of the differential form, since along it $dx = 0$ and $N = 2x = 0$ , although it is not the graph of a function of $x$ .

Check yourself

Which quotient test applies to $y\,dx + (y^2 - x)\,dy = 0$ ? Find the integrating factor and a potential.

Show the answer

Here $M_y = 1$ and $N_x = -1$ , so the equation is not exact. The quotient for $x$ is $2/(y^2 - x)$ , which depends on both variables, so there is no factor of $x$ alone. The quotient for $y$ is

\frac{N_x - M_y}{M} = \frac{-1 - 1}{y} = -\frac{2}{y},

so on either half-plane $y > 0$ or $y < 0$ the factor is $\mu = e^{-2\ln\lvert y \rvert} = 1/y^2$ . The multiplied equation has coefficients $1/y$ and $1 - x/y^2$ , whose cross derivatives both equal $-1/y^2$ . Integrating $1/y$ in $x$ gives $F = x/y + h(y)$ , and $F_y = -x/y^2 + h'(y)$ must equal $1 - x/y^2$ , so $h(y) = y$ and

F = \frac{x}{y} + y.

The solution curves are $x/y + y = C$ , that is $x = Cy - y^2$ with $y \ne 0$ . The line $y = 0$ , which the factor excludes, also solves the original equation, since along it $M = 0$ and $dy = 0$ .

Solve $(2y^2 + 3x)\,dx + 2xy\,dy = 0$ with $y(1) = 1$ , and give the maximal interval of the solution $y(x)$ .

Show the answer

The mismatch is $M_y - N_x = 4y - 2y = 2y$ , and where $xy \ne 0$

\frac{M_y - N_x}{N} = \frac{2y}{2xy} = \frac{1}{x}.

On $x > 0$ this gives $\mu = e^{\ln x} = x$ . In fact $x$ makes the equation exact on the whole plane, since the multiplied coefficients $2xy^2 + 3x^2$ and $2x^2y$ both have cross derivative $4xy$ . A potential is $F = x^2y^2 + x^3$ , and $F(1, 1) = 2$ , so the solution satisfies $x^2y^2 + x^3 = 2$ . Since $y(1) = 1 > 0$ ,

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

This needs $x \ne 0$ and $2 - x^3 > 0$ , so the maximal interval is $(0, \sqrt[3]{2})$ , where $\sqrt[3]{2} \approx 1.26$ . As $x \to 0^+$ the solution blows up. At $x = \sqrt[3]{2}$ it reaches $y = 0$ , where $N = 2xy = 0$ and the level curve has a vertical tangent, so the graph cannot be continued.

A classmate finds that neither quotient test works for $(3y + 4xy^2)\,dx + (2x + 3x^2y)\,dy = 0$ and concludes that the equation has no integrating factor. Check that $\mu = x^2y$ is an integrating factor in each open quadrant, and explain what the tests actually showed.

Show the answer

Here $M_y - N_x = (3 + 8xy) - (2 + 6xy) = 1 + 2xy$ , and the two quotients

\begin{gathered} \frac{M_y - N_x}{N} = \frac{1 + 2xy}{x(2 + 3xy)}, \\[0.4em] \frac{N_x - M_y}{M} = -\frac{1 + 2xy}{y(3 + 4xy)} \end{gathered}

each depend on both variables. For example, the first equals $3/5$ at $(1, 1)$ and $5/8$ at $(1, 2)$ . So there is no integrating factor of $x$ alone or of $y$ alone, and that is all the tests show. With $\mu = x^2y$ the coefficients become $3x^2y^2 + 4x^3y^3$ and $2x^3y + 3x^4y^2$ , and both cross derivatives equal $6x^2y + 12x^3y^2$ . A potential is $F = x^3y^2 + x^4y^3$ . The factor vanishes on the coordinate axes, so those lines must be checked in the original equation. Along $x = 0$ we have $dx = 0$ and $N = 0$ , and along $y = 0$ we have $dy = 0$ and $M = 0$ , so both axes are solution curves of the original equation as well.

Integrating Factors for Nonexact Equations

Key terms

Nonexact equation

Integrating factor depending on x

Integrating factor depending on y

Quotient tests

Domain restrictions

Check yourself

How this lesson connects

Builds on