2.6: Integrating Factors

In the previous section on Exact Equations we saw that if $M$, $N$, $M_y$ and $N_x$ are continuous and $M_y=N_x$ on an open rectangle $R$ then

$$\label{eq:2.6.1} M(x,y)\,dx+N(x,y)\,dy=0$$

is exact on $R$. Sometimes an equation that isn’t exact can be made exact by multiplying it by an appropriate function. For example,

$$\label{eq:2.6.2} (3x+2y^2)\,dx+2xy\,dy=0$$

is not exact, since $M_y(x,y)=4y\ne N_x(x,y)=2y$ in Equation $\ref{eq:2.6.2}$. However, multiplying Equation $\ref{eq:2.6.2}$ by $x$ yields

$$\label{eq:2.6.3} (3x^2+2xy^2)\,dx+2x^2y\,dy=0,$$

which is exact, since $M_y(x,y)=N_x(x,y)=4xy$ in Equation $\ref{eq:2.6.3}$. Solving Equation $\ref{eq:2.6.3}$ by the procedure given in Section 2.5 yields the implicit solution

$$x^3+x^2y^2=c.\nonumber$$

A function $\mu=\mu(x,y)$ is an integrating factor for Equation $\ref{eq:2.6.1}$ if $$\label{eq:2.6.4} \mu(x,y)M (x,y)\,dx+\mu(x,y)N (x,y)\,dy=0$$ is exact. If we know an integrating factor $\mu$ for Equation $\ref{eq:2.6.1}$, we can solve the exact equation Equation $\ref{eq:2.6.4}$ by the method of Section 2.5. It would be nice if we could say that Equation $\ref{eq:2.6.1}$ and Equation $\ref{eq:2.6.4}$ always have the same solutions, but this isn’t so. For example, a solution $y=y(x)$ of Equation $\ref{eq:2.6.4}$ such that $\mu(x,y(x))=0$ on some interval $a<x<b$ could fail to be a solution of $\ref{eq:2.6.1}$ (Exercise 2.6.1), while Equation $\ref{eq:2.6.1}$ may have a solution $y=y(x)$ such that $\mu(x,y(x))$ isn’t even defined (Exercise 2.6.2). Similar comments apply if $y$ is the independent variable and $x$ is the dependent variable in Equation $\ref{eq:2.6.1}$ and Equation $\ref{eq:2.6.4}$. However, if $\mu(x,y)$ is defined and nonzero for all $(x,y)$, Equation $\ref{eq:2.6.1}$ and Equation $\ref{eq:2.6.4}$ are equivalent; that is, they have the same solutions.

Finding Integrating Factors

By applying Theorem 2.5.2 (with $M$ and $N$ replaced by $\mu M$ and $\mu N$), we see that Equation $\ref{eq:2.6.4}$ is exact on an open rectangle $R$ if $\mu M$, $\mu N$, $(\mu M)_y$, and $(\mu N)_x$ are continuous and

$${\partial\over\partial y}(\mu M)={\partial\over\partial x} (\mu N)\nonumber$$

or, equivalently

$$\mu_yM+\mu M_y=\mu_xN+\mu N_x\nonumber$$

on $R$. It’s better to rewrite the last equation as $$\label{eq:2.6.5} \mu(M_y-N_x)=\mu_xN-\mu_yM,$$ which reduces to the known result for exact equations; that is, if $M_y=N_x$ then Equation $\ref{eq:2.6.5}$ holds with $\mu=1$, so Equation $\ref{eq:2.6.1}$ is exact.

You may think Equation $\ref{eq:2.6.5}$ is of little value, since it involves partial derivatives of the unknown integrating factor $\mu$, and we haven’t studied methods for solving such equations. However, we’ll now show that Equation $\ref{eq:2.6.5}$ is useful if we restrict our search to integrating factors that are products of a function of $x$ and a function of $y$; that is, $\mu(x,y)=P(x)Q(y)$. We’re not saying that every equation $M\,dx+N\,dy=0$ has an integrating factor of this form; rather, we are saying that some equations have such integrating factors.We’llnow develop a way to determine whether a given equation has such an integrating factor, and a method for finding the integrating factor in this case.

If $\mu(x,y)=P(x)Q(y)$, then $\mu_x(x,y)=P'(x)Q(y)$ and $\mu_y(x,y)=P(x)Q'(y)$, so Equation $\ref{eq:2.6.5}$ becomes

$$\label{eq:2.6.6} P(x)Q(y)(M_y-N_x)=P'(x)Q(y)N-P(x)Q'(y)M,$$

or, after dividing through by $P(x)Q(y)$,

$$\label{eq:2.6.7} M_y-N_x={P'(x)\over P(x)}N-{Q'(y)\over Q(y)}M.$$

Now let

$$p(x)={P'(x)\over P(x)} \quad \text{and} \quad q(y)={Q'(y)\over Q(y)},\nonumber$$

so Equation $\ref{eq:2.6.7}$ becomes

$$\label{eq:2.6.8} M_y-N_x=p(x)N-q(y)M.$$

We obtained Equation $\ref{eq:2.6.8}$ by assuming that $M\,dx+N\,dy=0$ has an integrating factor $\mu(x,y)=P(x)Q(y)$. However, we can now view Equation $\ref{eq:2.6.7}$ differently: If there are functions $p=p(x)$ and $q=q(y)$ that satisfy Equation $\ref{eq:2.6.8}$ and we define

$$\label{eq:2.6.9} P(x)=\pm e^{\int p(x)\,dx}\quad \text{and} \quad Q(y)=\pm e^{\int q(y)\,dy},$$

then reversing the steps that led from Equation $\ref{eq:2.6.6}$ to Equation $\ref{eq:2.6.8}$ shows that $\mu(x,y)=P(x)Q(y)$ is an integrating factor for $M\,dx+N\,dy=0$. In using this result, we take the constants of integration in Equation $\ref{eq:2.6.9}$ to be zero and choose the signs conveniently so the integrating factor has the simplest form.

There’s no simple general method for ascertaining whether functions $p=p(x)$ and $q=q(y)$ satisfying Equation $\ref{eq:2.6.8}$ exist. However, the next theorem gives simple sufficient conditions for the given equation to have an integrating factor that depends on only one of the independent variables $x$ and $y$, and for finding an integrating factor in this case.

Theorem 2.6.1

Let $M,$ $N,$ $M_y,$ and $N_x$ be continuous on an open rectangle $R.$ Then:

1. If $(M_y-N_x)/N$ is independent of $y$ on $R$ and we define $$p(x)={M_y-N_x\over N}\nonumber$$ then $$\label{eq:2.6.10} \mu(x)=\pm e^{\int p(x)\,dx}$$ is an integrating factor for $$\label{eq:2.6.11} M(x,y)\,dx+N(x,y)\,dy=0$$ on $R.$
2. If $(N_x-M_y)/M$ is independent of $x$ on $R$ and we define $$q(y)={N_x-M_y\over M},\nonumber$$ then $$\label{eq:2.6.12} \mu(y)=\pm e^{\int q(y)\,dy}$$ is an integrating factor for Equation $\ref{eq:2.6.11}$ on $R.$

Proof

(a) If $(M_y-N_x)/N$ is independent of $y$, then Equation $\ref{eq:2.6.8}$ holds with $p=(M_y-N_x)/N$ and $q\equiv0$. Therefore $$P(x)=\pm e^{\int p(x)\,dx}\quad\text{ and}\quad Q(y)=\pm e^{\int q(y)\,dy}=\pm e^0=\pm1,\nonumber$$ so Equation $\ref{eq:2.6.10}$ is an integrating factor for Equation $\ref{eq:2.6.11}$ on $R$.

(b) If $(N_x-M_y)/M$ is independent of $x$ then eqrefeq:2.6.8 holds with $p\equiv0$ and $q=(N_x-M_y)/M$, and a similar argument shows that Equation $\ref{eq:2.6.12}$ is an integrating factor for Equation $\ref{eq:2.6.11}$ on $R$.

The next two examples show how to apply Theorem 2.6.1 .

Example 2.6.1

Find an integrating factor for the equation $$\label{eq:2.6.13} (2xy^3-2x^3y^3-4xy^2+2x)\,dx+(3x^2y^2+4y)\,dy=0$$ and solve the equation.

Solution

In Equation $\ref{eq:2.6.13}$ $$M=2xy^3-2x^3y^3-4xy^2+2x,\ N=3x^2y^2+4y,\nonumber$$ and $$M_y-N_x=(6xy^2-6x^3y^2-8xy)-6xy^2=-6x^3y^2-8xy,\nonumber$$ so Equation $\ref{eq:2.6.13}$ isn’t exact. However, $${M_y-N_x\over N}=-{6x^3y^2+8xy\over 3x^2y^2+4y}=-2x\nonumber$$ is independent of $y$, so Theorem 2.6.1 (a) applies with $p(x)=-2x$. Since $$\int p (x)\,dx=-\int 2x\,dx=-x^2,\nonumber$$ $\mu(x)=e^{-x^2}$ is an integrating factor. Multiplying Equation $\ref{eq:2.6.13}$ by $\mu$ yields the exact equation $$\label{eq:2.6.14} e^{-x^2}(2xy^3-2x^3y^3-4xy^2+2x)\,dx+ e^{-x^2}(3x^2y^2+4y)\,dy=0.$$

To solve this equation, we must find a function $F$ such that $$\label{eq:2.6.15} F_x(x,y)=e^{-x^2}(2xy^3-2x^3y^3-4xy^2+2x)$$ and $$\label{eq:2.6.16} F_y(x,y)=e^{-x^2}(3x^2y^2+4y).$$ Integrating Equation $\ref{eq:2.6.16}$ with respect to $y$ yields $$\label{eq:2.6.17} F(x,y)=e^{-x^2}(x^2y^3+2y^2)+\psi(x).$$ Differentiating this with respect to $x$ yields $$F_x(x,y)=e^{-x^2}(2xy^3-2x^3y^3-4xy^2)+\psi'(x).\nonumber$$ Comparing this with Equation $\ref{eq:2.6.15}$ shows that $\psi'(x)= 2xe^{-x^2}$; therefore, we can let $\psi(x)=-e^{-x^2}$ in Equation $\ref{eq:2.6.17}$ and conclude that $$e^{-x^2}\left(y^2(x^2y+2)-1\right)=c\nonumber$$ is an implicit solution of Equation $\ref{eq:2.6.14}$. It is also an implicit solution of Equation $\ref{eq:2.6.13}$.

Figure 2.6.1 shows a direction field and some integral curves for Equation $\ref{eq:2.6.13}$

Figure 2.6.1 : A direction field and integral curves for $(2xy^3-2x^3y^3-4xy^2+2x)\,dx+(3x^2y^2+4y)\,dy=0$

Example 2.6.2

Find an integrating factor for

$$\label{eq:2.6.30} 2xy^{3}dx+(3x^{2}y^{2}+x^{2}y^{3}+1)dy=0$$

and solve the equation.

Solution

In Equation $\ref{eq:2.6.30}$,

$$M=2xy^{3},\quad N=3x^{2}y^{2}+x^{2}y^{3}+1,\nonumber$$

and

$$M_{y}-N_{x}=6x^{2}-(6xy^{2}+2xy^{3})=-2xy^{3},\nonumber$$

so Equation $\ref{eq:2.6.30}$ isn't exact. Moreover,

$$\frac{M_y-N_x}{N}=-\frac{2xy^3}{3x^2y^2+x^2y^2+1}\nonumber$$

is not independent of $y$, so Theorem 2.6.1(a) does not apply. However, Theorem 2.6.1(b) does apply, since

$$\frac{N_x-M_y}{M}=\frac{2xy^3}{2xy^3}=1\nonumber$$

is not independent of $x$, so we can take $q(y)=1$. Since

$$\int q(y)dy=\int dy=y,\nonumber$$

$\mu (y)=e^{y}$ is an integrating factor. Multiplying Equation $\ref{eq:2.6.30}$ by $\mu$ yields the exact equation.

$$\label{eq:2.6.36} 2xy^{3}e^{y}dx+(3x^{2}y^{2}+x^{2}y^{3}+1)e^{y}dy=0.$$

To solve this equation, we must find a function $F$ such that

$$\label{eq:2.6.37} F_x (x,y)=2xy^{3}e^{y}$$

and

$$\label{eq:2.6.38} F_{y}(x,y)=(3x^{2}y^{2}+x^{2}y^{3}+1)e^{y}.$$

Integrating Equation $\ref{eq:2.6.37}$ with respect to $x$ yields

$$\label{eq:2.6.39} F (x,y)=x^{2} y^{3} e^{y} + \phi (y)$$

Differentiating this with respect to $y$ yields

$$F_{y}= (3x^{2} y^{2} + x^{2} y^{3}) e^{y} + \phi ' (y)\nonumber$$

and comparing this with Equation $\ref{eq:2.6.38}$ shows that φ 0 (y) = e y . Therefore we set φ(y) = e y in Equation $\ref{eq:2.6.39}$ and conclude that

$$(x^{2}y^{3}+1)e^{y}=c\nonumber$$

is an implicit solution of $\ref{eq:2.6.36}$. It is also an implicit solution of $\ref{eq:2.6.30}$. Figure 2.6.2 shows a direction field and some integral curves for $\ref{eq:2.6.30}$.

Figure 2.6.2 : A direction field and integral curves for $(2xy^3e^ydx+(3x^2y^2+x^2y^3+1)e^ydy=0$

Theorem 2.6.1 does not apply in the next example, but the more general argument that led to Theorem 2.6.1 provides an integrating factor.

Example 2.6.3

Find an integrating factor for

$$\label{eq:2.6.42} (3xy+6y^{2})dx+(2x^{2} +9xy)dy=0$$

and solve the equation.

Solution

In Equation $\ref{eq:2.6.42}$

$$M=3xy+6y^2, \quad N=2x^2+9xy,\nonumber$$

and

$$M_y -N_x =(3x+12y)-(4x+9y)=-x+3y.\nonumber$$

Therefore

$$\frac{M_y - N_x}{M}=\frac{-x+3y}{3xy+6y^2}\quad\text{and}\quad\frac{N_x - M_y}{N}=\frac{x-3y}{2x^2 +9xy}\nonumber$$

so Theorem 2.6.1 does not apply. Following the more general argument that led to Theorem 2.6.1 , we look for functions $p = p(x)$ and $q = q(y)$ such that

$$M_y - N_x = p(x)N-q(y)M;\nonumber$$

that is,

$$-x+3y=p(x)(2x^2+9xy)-q(y)(3xy+6y^2).\nonumber$$

Since the left side contains only first degree terms in $x$ and $y$, we rewrite this equation as

$$xp(x)(2x+9y)-yq(y)(3x+6y)=-x+3y.\nonumber$$

This will be an identity if

$$\label{eq:2.6.49} xp(x)=A\quad\text{and}yq(y)=B,$$

where $A$ and $B$ are constants such that

$$-x+3y=A(2x+9y)-B(3x+6y),\nonumber$$

or, equivalently,

$$-x+3y=(2A-3B)x+(9A-6B)y.\nonumber$$

Equating the coefficients of x and y on both sides shows that the last equation holds for all $(x, y)$ if

$$\begin{aligned} 2A-3B &=-1 \\[4pt] 9A-6B &=3 \end{aligned}\nonumber$$

which has the solution A = 1, B = 1. Therefore Equation $\ref{eq:2.6.49}$ implies that

$$p(x)=\frac{1}{x}\quad\text{and}\quad q(y)=\frac{1}{y}.\nonumber$$

Since

$$\int p(x)dx=\ln |x|\quad\text{and}\quad\int q(y)dy=\ln |y|,\nonumber$$

we can let $P(x) = x$ and $Q(y) = y$; hence, $µ(x, y) = xy$ is an integrating factor. Multiplying Equation $\ref{eq:2.6.42}$ by $µ$ yields the exact equation

$$(3x^{2}y^{2}+6xy^{3})dx + (2x^{3}y+9x^{2}y^{2})dy=0.\nonumber$$

Figure 2.6.3 : A direction field and integral curves for $(3xy+6y^{2})dx + (2x^{2}+9xy)dy=0$

We leave it to you to use the method of Section 2.5 to show that this equation has the implicit solution

$$\label{eq:2.6.55} x^{3}y^{2}+3x^{2}y^{3}=c.$$

This is also an implicit solution of Equation $\ref{eq:2.6.42}$. Since x ≡ 0 and y ≡ 0 satisfy Equation $\ref{eq:2.6.55}$, you should check to see that x ≡ 0 and y ≡ 0 are also solutions of Equation $\ref{eq:2.6.42}$. (Why is it necesary to check this?) Figure 2.6.3 shows a direction field and integral curves for Equation $\ref{eq:2.6.42}$. See Exercise 2.6.28 for a general discussion of equations like Equation $\ref{eq:2.6.42}$.

Example 2.6.4

The separable equation

$$\label{eq:2.6.56} -ydx+(x+x^{6})dy=0$$

can be converted to the exact equation

$$\label{eq:2.6.57} -\frac{dx}{x+x^{6}}+\frac{dy}{y}=0$$

by multiplying through by the integrating factor

$$\mu (x,y)=\frac{1}{y(x+x^{6})}.\nonumber$$

However, to solve Equation $\ref{eq:2.6.57}$ by the method of Section 2.5 we would have to evaluate the nasty integral

$$\int\frac{dx}{x+x^{6}}.\nonumber$$

Instead, we solve Equation $\ref{eq:2.6.56}$ explicitly for $y$ by finding an integrating factor of the form $µ(x, y) = x^{a}y^{b}$.

Solution

In Equation $\ref{eq:2.6.56}$

$$M=-y,\ N=x+x^6,\nonumber$$

and

$$M_y-N_x=-1-(1+6x^5)=-2-6x^5.\nonumber$$

We look for functions $p=p(x)$ and $q=q(y)$ such that

$$M_y-N_x=p(x)N-q(y)M;\nonumber$$

that is,

$$\label{eq:2.6.28} -2-6x^5=p(x)(x+x^6)+q(y)y.$$

The right side will contain the term $-6x^5$ if $p(x)=-6/x$. Then Equation $\ref{eq:2.6.28}$ becomes

$$-2-6x^5=-6-6x^5+q(y)y,\nonumber$$

so $q(y)=4/y$. Since

$$\int p(x)\,dx=-\int{6\over x}\,dx=-6\ln|x|=\ln{1\over x^6},\nonumber$$

and

$$\int q(y)\,dy=\int{4\over y}\,dy=4\ln |y|=\ln{y^4},\nonumber$$

we can take $P(x)=x^{-6}$ and $Q(y)=y^4$, which yields the integrating factor $\mu(x,y)=x^{-6}y^4$. Multiplying Equation $\ref{eq:2.6.56}$ by $\mu$ yields the exact equation

$$-{y^5\over x^6}\,dx+\left({y^4\over x^5}+y^4\right) \,dy=0.\nonumber$$

We leave it to you to use the method of the Section 2.5 to show that this equation has the implicit solution

$$\left({y\over x}\right)^5+y^5=k.\nonumber$$

Solving for $y$ yields

$$y=k^{1/5}x(1+x^5)^{-1/5},\nonumber$$

which we rewrite as

$$y=cx(1+x^5)^{-1/5}\nonumber$$

by renaming the arbitrary constant. This is also a solution of Equation $\ref{eq:2.6.56}$.

Figure 2.6.4 shows a direction field and some integral curves for Equation $\ref{eq:2.6.56}$.

Figure 2.6.4 : A direction field and integral curves for $-ydx+(x+x^{6})dy=0$