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# 3.8: Integrating Factors

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In Section 3.8 we saw that if $$M$$, $$N$$, $$M_y$$ and $$N_x$$ are continuous and $$M_y=N_x$$ on an open rectangle $$R$$ then

$\label{eq:3.9.1} M(x,y)\,dx+N(x,y)\,dy=0$

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

$\label{eq:3.9.2} (3x+2y^2)\,dx+2xy\,dy=0$

is not exact, since $$M_y(x,y)=4y\ne N_x(x,y)=2y$$ in Equation \ref{eq:3.9.2}. However, multiplying Equation \ref{eq:3.9.2} by $$x$$ yields

$\label{eq:3.9.3} (3x^2+2xy^2)\,dx+2x^2y\,dy=0,$

which is exact, since $$M_y(x,y)=N_x(x,y)=4xy$$ in Equation \ref{eq:3.9.3}. Solving Equation \ref{eq:3.9.3} 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:3.9.1} if $\label{eq:3.9.4} \mu(x,y)M (x,y)\,dx+\mu(x,y)N (x,y)\,dy=0$ is exact. If we know an integrating factor $$\mu$$ for Equation \ref{eq:3.9.1}, we can solve the exact equation Equation \ref{eq:3.9.4} by the method of Section 2.5. It would be nice if we could say that Equation \ref{eq:3.9.1} and Equation \ref{eq:3.9.4} always have the same solutions, but this isn’t so. For example, a solution $$y=y(x)$$ of Equation \ref{eq:3.9.4} such that $$\mu(x,y(x))=0$$ on some interval $$a<x<b$$ could fail to be a solution of \ref{eq:3.9.1} (Exercise 3.9.1), while Equation \ref{eq:3.9.1} may have a solution $$y=y(x)$$ such that $$\mu(x,y(x))$$ isn’t even defined (Exercise 3.9.2). Similar comments apply if $$y$$ is the independent variable and $$x$$ is the dependent variable in Equation \ref{eq:3.9.1} and Equation \ref{eq:3.9.4}. However, if $$\mu(x,y)$$ is defined and nonzero for all $$(x,y)$$, Equation \ref{eq:3.9.1} and Equation \ref{eq:3.9.4} are equivalent; that is, they have the same solutions.

## Finding Integrating Factors

By applying Theorem 3.8.2 (with $$M$$ and $$N$$ replaced by $$\mu M$$ and $$\mu N$$), we see that Equation \ref{eq:3.9.4} is exact on an open rectangle $$R$$ if $$\mu M$$, $$\mu N$$, $$(\mu M)_y$$, and $$(\mu N)_x$$ are continuous and ${\partial\over\partial y}(\mu M)={\partial\over\partial x} (\mu N) \quad \text{or, equivalently,} \quad \mu_yM+\mu M_y=\mu_xN+\mu N_x\nonumber$ on $$R$$. It’s better to rewrite the last equation as $\label{eq:3.9.5} \mu(M_y-N_x)=\mu_xN-\mu_yM,$ which reduces to the known result for exact equations; that is, if $$M_y=N_x$$ then Equation \ref{eq:3.9.5} holds with $$\mu=1$$, so Equation \ref{eq:3.9.1} is exact.

You may think Equation \ref{eq:3.9.5} 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:3.9.5} 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:3.9.5} becomes

$\label{eq:3.9.6} P(x)Q(y)(M_y-N_x)=P'(x)Q(y)N-P(x)Q'(y)M,$ or, after dividing through by $$P(x)Q(y)$$,

$\label{eq:3.9.7} M_y-N_x={P'(x)\over P(x)}N-{Q'(y)\over Q(y)}M.$ 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:3.9.7} becomes

$\label{eq:3.9.8} M_y-N_x=p(x)N-q(y)M.$

We obtained Equation \ref{eq:3.9.8} 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:3.9.7} differently: If there are functions $$p=p(x)$$ and $$q=q(y)$$ that satisfy Equation \ref{eq:3.9.8} and we define

$\label{eq:3.9.9} P(x)=\pm e^{\int p(x)\,dx}\quad \text{and} \quad Q(y)=\pm e^{\int q(y)\,dy},$

then reversing the steps that led from Equation \ref{eq:3.9.6} to Equation \ref{eq:3.9.8} 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:3.9.9} 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:3.9.8} 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 $$\PageIndex{1}$$

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

(a) 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 $\label{eq:3.9.10} \mu(x)=\pm e^{\int p(x)\,dx}$ is an integrating factor for $\label{eq:3.9.11} M(x,y)\,dx+N(x,y)\,dy=0$ on $$R.$$

(b) 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 $\label{eq:3.9.12} \mu(y)=\pm e^{\int q(y)\,dy}$ is an integrating factor for Equation \ref{eq:3.9.11} on $$R.$$

Proof

(a) If $$(M_y-N_x)/N$$ is independent of $$y$$, then Equation \ref{eq:3.9.8} 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:3.9.10} is an integrating factor for Equation \ref{eq:3.9.11} on $$R$$.

(b) If $$(N_x-M_y)/M$$ is independent of $$x$$ then eqrefeq:3.9.8 holds with $$p\equiv0$$ and $$q=(N_x-M_y)/M$$, and a similar argument shows that Equation \ref{eq:3.9.12} is an integrating factor for Equation \ref{eq:3.9.11} on $$R$$.

The next two examples show how to apply Theorem $$\PageIndex{1}$$.

Example $$\PageIndex{1}$$

Find an integrating factor for the equation $\label{eq:3.9.13} (2xy^3-2x^3y^3-4xy^2+2x)\,dx+(3x^2y^2+4y)\,dy=0$ and solve the equation.

Solution

In Equation \ref{eq:3.9.13} $M=2xy^3-2x^3y^3-4xy^2+2x,\ N=3x^2y^2+4y,\nonumber$ and $M_y-N_x=(6xy^2-6x^3y^2-8xy)-6xy^2=-6x^3y^2-8xy,\nonumber$ so Equation \ref{eq:3.9.13} 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 $$\PageIndex{1}$$ (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:3.9.13} by $$\mu$$ yields the exact equation $\label{eq:3.9.14} e^{-x^2}(2xy^3-2x^3y^3-4xy^2+2x)\,dx+ e^{-x^2}(3x^2y^2+4y)\,dy=0.$

To solve this equation, we must find a function $$F$$ such that $\label{eq:3.9.15} F_x(x,y)=e^{-x^2}(2xy^3-2x^3y^3-4xy^2+2x)$ and $\label{eq:3.9.16} F_y(x,y)=e^{-x^2}(3x^2y^2+4y).$ Integrating Equation \ref{eq:3.9.16} with respect to $$y$$ yields $\label{eq:3.9.17} F(x,y)=e^{-x^2}(x^2y^3+2y^2)+\psi(x).$ 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:3.9.15} shows that $$\psi'(x)= 2xe^{-x^2}$$; therefore, we can let $$\psi(x)=-e^{-x^2}$$ in Equation \ref{eq:3.9.17} and conclude that $e^{-x^2}\left(y^2(x^2y+2)-1\right)=c\nonumber$ is an implicit solution of Equation \ref{eq:3.9.14}. It is also an implicit solution of Equation \ref{eq:3.9.13}.

Figure $$\PageIndex{1}$$ shows a direction field and some integral curves for Equation \ref{eq:3.9.13}

Figure $$\PageIndex{1}$$: A direction field and integral curves for $$(2xy^3-2x^3y^3-4xy^2+2x)\,dx+(3x^2y^2+4y)\,dy=0$$

Example $$\PageIndex{2}$$

Find an integrating factor for

$\label{eq:3.9.30} 2xy^{3}dx+(3x^{2}y^{2}+x^{2}y^{3}+1)dy=0$

and solve the equation.

Solution

In Equation \ref{eq:3.9.30},

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

and

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

so Equation \ref{eq:3.9.30} 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 3.9.1(a) does not apply. However, Theorem 3.9.1(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:3.9.30} by $$\mu$$ yields the exact equation.

$\label{eq:3.9.36} 2xy^{3}e^{y}dx+(3x^{2}y^{2}+x^{2}y^{3}+1)e^{y}dy=0.$

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

$\label{eq:3.9.37} F_x (x,y)=2xy^{3}e^{y}$

and

$\label{eq:3.9.38} F_{y}(x,y)=(3x^{2}y^{2}+x^{2}y^{3}+1)e^{y}.$

Integrating Equation \ref{eq:3.9.37} with respect to $$x$$ yields

$\label{eq:3.9.39} F (x,y)=x^{2} y^{3} e^{y} + \phi (y)$

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:3.9.38} shows that φ 0 (y) = e y . Therefore we set φ(y) = e y in Equation \ref{eq:3.9.39} and conclude that

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

is an implicit solution of \ref{eq:3.9.36}. It is also an implicit solution of \ref{eq:3.9.30}. Figure $$\PageIndex{2}$$ shows a direction field and some integral curves for \ref{eq:3.9.30}.

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

Theorem $$\PageIndex{1}$$ does not apply in the next example, but the more general argument that led to Theorem $$\PageIndex{1}$$  provides an integrating factor.

Example $$\PageIndex{3}$$

Find an integrating factor for

$\label{eq:3.9.42} (3xy+6y^{2})dx+(2x^{2} +9xy)dy=0$

and solve the equation.

Solution

In Equation \ref{eq:3.9.42}

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

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 $$\PageIndex{1}$$ does not apply. Following the more general argument that led to Theorem $$\PageIndex{1}$$, 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

$\label{eq:3.9.49} xp(x)=A\quad\text{and}yq(y)=B,$

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 \\ 9A-6B &=3 \end{aligned}\nonumber

which has the solution A = 1, B = 1. Therefore Equation \ref{eq:3.9.49} 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:3.9.42} by $$µ$$ yields the exact equation

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

Figure $$\PageIndex{3}$$: 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

$\label{eq:3.9.55} x^{3}y^{2}+3x^{2}y^{3}=c.$

This is also an implicit solution of Equation \ref{eq:3.9.42}. Since x ≡ 0 and y ≡ 0 satisfy Equation \ref{eq:3.9.55}, you should check to see that x ≡ 0 and y ≡ 0 are also solutions of Equation \ref{eq:3.9.42}. (Why is it necesary to check this?) Figure $$\PageIndex{3}$$ shows a direction field and integral curves for Equation \ref{eq:3.9.42}. See Exercise 3.9.28 for a general discussion of equations like Equation \ref{eq:3.9.42}.

Example $$\PageIndex{4}$$

The separable equation

$\label{eq:3.9.56} -ydx+(x+x^{6})dy=0$

can be converted to the exact equation

$\label{eq:3.9.57} -\frac{dx}{x+x^{6}}+\frac{dy}{y}=0$

by multiplying through by the integrating factor

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

However, to solve Equation \ref{eq:3.9.57} 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:3.9.56} explicitly for $$y$$ by finding an integrating factor of the form $$µ(x, y) = x^{a}y^{b}$$ .

Figure $$\PageIndex{4}$$: A direction field and integral curves for $$-ydx+(x+x^{6})dy=0$$

Solution

In Equation \ref{eq:3.9.56}

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

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,

$\label{eq:3.9.28} -2-6x^5=p(x)(x+x^6)+q(y)y.$

The right side will contain the term $$-6x^5$$ if $$p(x)=-6/x$$. Then Equation \ref{eq:3.9.28} 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:3.9.56} 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:3.9.56}.

Figure $$\PageIndex{4}$$ shows a direction field and some integral curves for Equation \ref{eq:3.9.56}.