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3.8: Integrating Factors

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

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,

(3x+2y2)dx+2xydy=0

is not exact, since My(x,y)=4yNx(x,y)=2y in Equation ???. However, multiplying Equation ??? by x yields

(3x2+2xy2)dx+2x2ydy=0,

which is exact, since My(x,y)=Nx(x,y)=4xy in Equation ???. Solving Equation ??? by the procedure given in Section 2.5 yields the implicit solution

x3+x2y2=c.

A function μ=μ(x,y) is an integrating factor for Equation ??? if μ(x,y)M(x,y)dx+μ(x,y)N(x,y)dy=0 is exact. If we know an integrating factor μ for Equation ???, we can solve the exact equation Equation ??? by the method of Section 2.5. It would be nice if we could say that Equation ??? and Equation ??? always have the same solutions, but this isn’t so. For example, a solution y=y(x) of Equation ??? such that μ(x,y(x))=0 on some interval a<x<b could fail to be a solution of ??? (Exercise 3.9.1), while Equation ??? may have a solution y=y(x) such that μ(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 ??? and Equation ???. However, if μ(x,y) is defined and nonzero for all (x,y), Equation ??? and Equation ??? are equivalent; that is, they have the same solutions.

Finding Integrating Factors

By applying Theorem 3.8.2 (with M and N replaced by μM and μN), we see that Equation ??? is exact on an open rectangle R if μM, μN, (μM)y, and (μN)x are continuous and y(μM)=x(μN)or, equivalently,μyM+μMy=μxN+μNx on R. It’s better to rewrite the last equation as μ(MyNx)=μxNμyM, which reduces to the known result for exact equations; that is, if My=Nx then Equation ??? holds with μ=1, so Equation ??? is exact.

You may think Equation ??? is of little value, since it involves partial derivatives of the unknown integrating factor μ, and we haven’t studied methods for solving such equations. However, we’ll now show that Equation ??? 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, μ(x,y)=P(x)Q(y). We’re not saying that every equation Mdx+Ndy=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 μ(x,y)=P(x)Q(y), then μx(x,y)=P(x)Q(y) and μy(x,y)=P(x)Q(y), so Equation ??? becomes

P(x)Q(y)(MyNx)=P(x)Q(y)NP(x)Q(y)M, or, after dividing through by P(x)Q(y),

MyNx=P(x)P(x)NQ(y)Q(y)M. Now let p(x)=P(x)P(x)andq(y)=Q(y)Q(y), so Equation ??? becomes

MyNx=p(x)Nq(y)M.

We obtained Equation ??? by assuming that Mdx+Ndy=0 has an integrating factor μ(x,y)=P(x)Q(y). However, we can now view Equation ??? differently: If there are functions p=p(x) and q=q(y) that satisfy Equation ??? and we define

P(x)=±ep(x)dxandQ(y)=±eq(y)dy,

then reversing the steps that led from Equation ??? to Equation ??? shows that μ(x,y)=P(x)Q(y) is an integrating factor for Mdx+Ndy=0. In using this result, we take the constants of integration in Equation ??? 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 ??? 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 3.8.1

Let M, N, My, and Nx be continuous on an open rectangle R. Then:

(a) If (MyNx)/N is independent of y on R and we define p(x)=MyNxN then μ(x)=±ep(x)dx is an integrating factor for M(x,y)dx+N(x,y)dy=0 on R.

(b) If (NxMy)/M is independent of x on R and we define q(y)=NxMyM, then μ(y)=±eq(y)dy is an integrating factor for Equation ??? on R.

Proof

(a) If (MyNx)/N is independent of y, then Equation ??? holds with p=(MyNx)/N and q0. Therefore P(x)=±ep(x)dx andQ(y)=±eq(y)dy=±e0=±1, so Equation ??? is an integrating factor for Equation ??? on R.

(b) If (NxMy)/M is independent of x then eqrefeq:3.9.8 holds with p0 and q=(NxMy)/M, and a similar argument shows that Equation ??? is an integrating factor for Equation ??? on R.

The next two examples show how to apply Theorem 3.8.1.

Example 3.8.1

Find an integrating factor for the equation (2xy32x3y34xy2+2x)dx+(3x2y2+4y)dy=0 and solve the equation.

Solution

In Equation ??? M=2xy32x3y34xy2+2x, N=3x2y2+4y, and MyNx=(6xy26x3y28xy)6xy2=6x3y28xy, so Equation ??? isn’t exact. However, MyNxN=6x3y2+8xy3x2y2+4y=2x is independent of y, so Theorem 3.8.1 (a) applies with p(x)=2x. Since p(x)dx=2xdx=x2, μ(x)=ex2 is an integrating factor. Multiplying Equation ??? by μ yields the exact equation ex2(2xy32x3y34xy2+2x)dx+ex2(3x2y2+4y)dy=0.

To solve this equation, we must find a function F such that Fx(x,y)=ex2(2xy32x3y34xy2+2x) and Fy(x,y)=ex2(3x2y2+4y). Integrating Equation ??? with respect to y yields F(x,y)=ex2(x2y3+2y2)+ψ(x). Differentiating this with respect to x yields Fx(x,y)=ex2(2xy32x3y34xy2)+ψ(x). Comparing this with Equation ??? shows that ψ(x)=2xex2; therefore, we can let ψ(x)=ex2 in Equation ??? and conclude that ex2(y2(x2y+2)1)=c is an implicit solution of Equation ???. It is also an implicit solution of Equation ???.

Figure 3.8.1 shows a direction field and some integral curves for Equation ???

clipboard_e37bccd51f78c60aa2e6d41ad58458511.png
Figure 3.8.1: A direction field and integral curves for (2xy32x3y34xy2+2x)dx+(3x2y2+4y)dy=0

Example 3.8.2

Find an integrating factor for

2xy3dx+(3x2y2+x2y3+1)dy=0

and solve the equation.

Solution

In Equation ???,

M=2xy3,N=3x2y2+x2y3+1,

and

MyNx=6x2(6xy2+2xy3)=2xy3,

so Equation ??? isn't exact. Moreover,

MyNxN=2xy33x2y2+x2y2+1

is not independent of y, so Theorem 3.9.1(a) does not apply. However, Theorem 3.9.1(b) does apply, since

NxMyM=2xy32xy3=1

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

q(y)dy=dy=y,

μ(y)=ey is an integrating factor. Multiplying Equation ??? by μ yields the exact equation.

2xy3eydx+(3x2y2+x2y3+1)eydy=0.

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

Fx(x,y)=2xy3ey

and

Fy(x,y)=(3x2y2+x2y3+1)ey.

Integrating Equation ??? with respect to x yields

F(x,y)=x2y3ey+ϕ(y)

Differentiating this with respect to y yields

Fy=(3x2y2+x2y3)ey+ϕ(y)

and comparing this with Equation ??? shows that φ 0 (y) = e y . Therefore we set φ(y) = e y in Equation ??? and conclude that

(x2y3+1)ey=c

is an implicit solution of ???. It is also an implicit solution of ???. Figure 3.8.2 shows a direction field and some integral curves for ???.

clipboard_edba7aa19ca01c674c2ab6272b30d74ad.png
Figure 3.8.2: A direction field and integral curves for (2xy3eydx+(3x2y2+x2y3+1)eydy=0

Theorem 3.8.1 does not apply in the next example, but the more general argument that led to Theorem 3.8.1 provides an integrating factor.

Example 3.8.3

Find an integrating factor for

(3xy+6y2)dx+(2x2+9xy)dy=0

and solve the equation.

Solution

In Equation ???

M=3xy+6y2,N=2x2+9xy,

and

MyNx=(3x+12y)(4x+9y)=x+3y.

Therefore

MyNxM=x+3y3xy+6y2andNxMyN=x3y2x2+9xy

so Theorem 3.8.1 does not apply. Following the more general argument that led to Theorem 3.8.1, we look for functions p=p(x) and q=q(y) such that

MyNx=p(x)Nq(y)M;

that is,

x+3y=p(x)(2x2+9xy)q(y)(3xy+6y2).

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.

This will be an identity if

xp(x)=Aandyq(y)=B,

where A and B are constants such that

x+3y=A(2x+9y)B(3x+6y),

or, equivalently,

x+3y=(2A3B)x+(9A6B)y.

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

2A3B=19A6B=3

which has the solution A = 1, B = 1. Therefore Equation ??? implies that

p(x)=1xandq(y)=1y.

Since

p(x)dx=ln|x|andq(y)dy=ln|y|,

we can let P(x)=x and Q(y)=y; hence, µ(x,y)=xy is an integrating factor. Multiplying Equation ??? by µ yields the exact equation

(3x2y2+6xy3)dx+(2x3y+9x2y2)dy=0.

clipboard_ed79d6552dc5e4e92b44da72ed59746c5.png
Figure 3.8.3: A direction field and integral curves for (3xy+6y2)dx+(2x2+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

x3y2+3x2y3=c.

This is also an implicit solution of Equation ???. Since x ≡ 0 and y ≡ 0 satisfy Equation ???, you should check to see that x ≡ 0 and y ≡ 0 are also solutions of Equation ???. (Why is it necesary to check this?) Figure 3.8.3 shows a direction field and integral curves for Equation ???. See Exercise 3.9.28 for a general discussion of equations like Equation ???.

Example 3.8.4

The separable equation

ydx+(x+x6)dy=0

can be converted to the exact equation

dxx+x6+dyy=0

by multiplying through by the integrating factor

μ(x,y)=1y(x+x6).

However, to solve Equation ??? by the method of Section 2.5 we would have to evaluate the nasty integral

dxx+x6.

Instead, we solve Equation ??? explicitly for y by finding an integrating factor of the form µ(x,y)=xayb.

clipboard_e2d16b8955ad8c37ee1c25f09e4be726c.png
Figure 3.8.4: A direction field and integral curves for ydx+(x+x6)dy=0

Solution

In Equation ???

M=y, N=x+x6,

and

MyNx=1(1+6x5)=26x5.

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

MyNx=p(x)Nq(y)M;

that is,

26x5=p(x)(x+x6)+q(y)y.

The right side will contain the term 6x5 if p(x)=6/x. Then Equation ??? becomes

26x5=66x5+q(y)y,

so q(y)=4/y. Since

p(x)dx=6xdx=6ln|x|=ln1x6,

and

q(y)dy=4ydy=4ln|y|=lny4,

we can take P(x)=x6 and Q(y)=y4, which yields the integrating factor μ(x,y)=x6y4. Multiplying Equation ??? by μ yields the exact equation

y5x6dx+(y4x5+y4)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

(yx)5+y5=k.

Solving for y yields

y=k1/5x(1+x5)1/5,

which we rewrite as

y=cx(1+x5)1/5

by renaming the arbitrary constant. This is also a solution of Equation ???.

Figure 3.8.4 shows a direction field and some integral curves for Equation ???.


3.8: Integrating Factors is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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