Integrating Factors

Author: John J Weber III, PhD Corresponding Textbook Sections:

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Integrating Factors

Definition: Integrating Factor

If the equation M(x,y)dx+N(x,y)dy=0 is not exact, but the equation μ(x,y)M(x,y)dx+μ(x,y)N(x,y)dy=0 for μ(x,y)0 is exact, then the function μ(x,y) is an integrating factor of the equation.

Derivation

We know from the Test for Exactness that μ(x,y)M(x,y)dx+μ(x,y)N(x,y)dy=0 is an exact equation if and only if [μ(x,y)M(x,y)]y=[μ(x,y)N(x,y)]x

Using the product rule for derivatives:

M(x,y)μ(x,y)y+μ(x,y)M(x,y)y=N(x,y)μ(x,y)x+μ(x,y)N(x,y)x

Rewriting to isolate for μ(x,y):

M(x,y)μ(x,y)yN(x,y)μ(x,y)x=(N(x,y)xM(x,y)y)μ(x,y)

Solving for μ(x)

Suppose the integrating is a function of only x, i.e., μ(x,y)=μ(x):

M(x,y)μ(x)yN(x,y)μ(x)x=(N(x,y)xM(x,y)y)μ(x)

We know from Calculus III: μ(x)y=0 and μ(x)x=dμ(x)dx. Thus,

0N(x,y)dμ(x)dx=(N(x,y)xM(x,y)y)μ(x)

Solve for dμ(x)dx:

dμ(x)dx=(N(x,y)xM(x,y)yN(x,y))μ(x)

Rewriting the expression in the parenthesis:

dμ(x)dx=(M(x,y)yN(x,y)xN(x,y))μ(x)

which is separable if and only if (M(x,y)yN(x,y)xN(x,y)) is a function of only x. That is,

dμ(x)μ(x)=(M(x,y)yN(x,y)xN(x,y))dx

and

μ(x)=e(M(x,y)yN(x,y)xN(x,y))dx

Solving for μ(y)

Suppose the integrating is a function of only x, i.e., μ(x,y)=μ(y):

M(x,y)μ(y)yN(x,y)μ(y)x=(N(x,y)xM(x,y)y)μ(y)

We know from Calculus III: μ(y)x=0 and μ(y)y=dμ(y)dy. Thus,

M(x,y)dμ(y)dy0=(N(x,y)xM(x,y)y)μ(y)

Solve for dμ(y)dy:

dμ(y)dy=(N(x,y)xM(x,y)yM(x,y))μ(y)

which is separable if and only if (N(x,y)xM(x,y)yM(x,y)) is a function of only y. That is,

dμ(y)μ(y)=(N(x,y)xM(x,y)yM(x,y))dy

and

μ(y)=e(N(x,y)xM(x,y)yM(x,y))dy

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Last Modified: Monday, 6 September 2020 13:33 EDT