
# 2.6: Integrating Factors

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In Section 2.5 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

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

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 $$a<b$$>[exer: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 [exer: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.

By applying Theorem [thmtype: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 ${\partial\over\partial y}(\mu M)={\partial\over\partial x} (\mu N) \mbox{\quad or, equivalently, \quad} \mu_yM+\mu M_y=\mu_xN+\mu N_x$ 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 $p(x)={P'(x)\over P(x)} \mbox{\quad and \quad} q(y)={Q'(y)\over Q(y)},$ 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

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

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.

[thmtype:2.6.1] Let $$M,$$ $$N,$$ $$M_y,$$ and $$N_x$$ be continuous on an open rectangle $$R.$$ Then$$:$$

If $$(M_y-N_x)/N$$ is independent of $$y$$ on $$R$$ and we define $p(x)={M_y-N_x\over N}$ 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 $$R.$$

If $$(N_x-M_y)/M$$ is independent of $$x$$ on $$R$$ and we define $q(y)={N_x-M_y\over M},$ 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 $$R.$$

(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 $P(x)=\pm e^{\int p(x)\,dx}\mbox{\quad and\quad}Q(y)=\pm e^{\int q(y)\,dy}=\pm e^0=\pm1,$ 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 [thmtype: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.

In Equation \ref{eq:2.6.13} $M=2xy^3-2x^3y^3-4xy^2+2x,\ N=3x^2y^2+4y,$ and $M_y-N_x=(6xy^2-6x^3y^2-8xy)-6xy^2=-6x^3y^2-8xy,$ so Equation \ref{eq:2.6.13} isn’t exact. However, ${M_y-N_x\over N}=-{6x^3y^2+8xy\over 3x^2y^2+4y}=-2x$ is independent of $$y$$, so Theorem [thmtype:2.6.1] a

applies with $$p(x)=-2x$$. Since $\int p (x)\,dx=-\int 2x\,dx=-x^2,$ $$\mu(x)=e^{-x^2}$$ 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 $F_x(x,y)=e^{-x^2}(2xy^3-2x^3y^3-4xy^2)+\psi'(x).$ 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 $e^{-x^2}\left(y^2(x^2y+2)-1\right)=c$ 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 [figure:2.6.1] shows a direction field and some integral curves for Equation \ref{eq:2.6.13}

In Equation \ref{eq:2.6.26}

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

and

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

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

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

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

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

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

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

and

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

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:2.6.26} by $$\mu$$ yields the exact equation

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

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

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

Solving for $$y$$ yields

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

which we rewrite as

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

by renaming the arbitrary constant. This is also a solution of Equation \ref{eq:2.6.26}.

Figure [figure:2.6.4} shows a direction field and some integral curves for Equation \ref{eq:2.6.26}.