# Preface

A function μ is called an integrating factor if and only if multiplication by it reduces the differential equation \( M(x,y)\,{\text d}x + N(x,y)\,{\text d} y =0 \) to an exact equation. Although Alexis Clairaut was the first to discover integrating factors, the fundamental conception of this technique iis due to Leonhard Euler, who set up classes of equations that admit integrating factors.

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## Glossary

# Integrating Factors

In many cases, a differential equation of first order

**integrating factor**if, upon multiplication the above equation by μ, we obtain an exact equation. In other words, μ is an integrating factor if and only if

The **integrating factor** method was introduced by the French mathematician, astronomer, and geophysicist Alexis
Claude Clairaut (1713--1765). He was a prominent Newtonian follower whose work helped to establish the validity of the principles and results that Sir
Iaac Newton had outlined in the Principia of 1687. Clairaut was one of the key
figures in the expedition to Lapland that helped to confirm Newton's theory for the figure of the Earth. In that
context, Clairaut worked out a mathematical result now known as
"Clairaut's theorem". He also tackled the
gravitational three-body problem, being the first to obtain a
satisfactory result for the
apsidal precession (rotation of the orbit of a celestial body) of
the Moon's orbit. Clairaut published some important work during the period 1733 to 1743. He wrote a paper in 1733 on
the calculus of variations, and in the same year he published on the geodesics of quadrics of rotation. The following
year Clairaut studied the differential equations now known as
Clairaut's differential equations, and introduced a
singular solution in addition to the general integral of the equations. In 1739 and 1740, he published further work
on the integral calculus, proving the existence of integrating factors for solving first order differential
equations. In 1742, Clairaut published an important work on dynamics. Clairaut was unmarried, and known for leading
an active social life.

Using simplified notations μ_{x} and μ_{y} for partial derivatives with respect to *x* and *y*, respectively, we rewrite the above partial differential equation as

*F*and

*G*are solutions of the algebraic equation

We start with two simple (but very important) classes when an integrating factor depends on only one variable.

- Suppose that an integrating factor depends only on the independent variable only. Then
\( \mu = \mu (x) \) satisfies the differential equation
\[ \frac{\partial}{\partial y} \left( \mu \,M(x,y)\right) = \frac{\partial}{\partial x} \left( \mu \,N(x,y)\right) \qquad \Longleftrightarrow \qquad \mu \, \frac{\partial M(x,y)}{\partial y} = \frac{{\text d} \mu}{{\text d} x} \, N(x,y) + \mu \, \frac{\partial N(x,y)}{\partial x} . \]Then we get the equation\[ \frac{1}{\mu} \,\frac{{\text d} \mu}{{\text d} x} = \frac{M_y - N_x}{N(x,y)} \qquad \mbox{must be a function of $x$ only} \]because the left-hand side is a function of
*x*. In this case, variable are separated and upon integration, we obtain\[ \mu (x) = \exp \left\{ \int \frac{M_y - N_x}{N(x,y)} \, {\text d} x \right\} . \] - Suppose that an integrating factor depends only on the dependent variable only. Then
\( \mu = \mu (y) \) satisfies the differential equation
\[ \frac{\partial}{\partial y} \left( \mu \,M(x,y)\right) = \frac{\partial}{\partial x} \left( \mu \,N(x,y)\right) \qquad \Longleftrightarrow \qquad \mu \, \frac{\partial M(x,y)}{\partial y} + \frac{{\text d} \mu}{{\text d} y} \, M(x,y) = \mu \, \frac{\partial N(x,y)}{\partial x} . \]Then we get the equation\[ \frac{1}{\mu} \,\frac{{\text d} \mu}{{\text d} y} = -\frac{M_y - N_x}{M(x,y)} \qquad \mbox{must be a function of $y$ only} \]because the left-hand side is a function of
*y*. In this case, variable are separated and upon integration, we obtain\[ \mu (y) = \exp \left\{ -\int \frac{M_y - N_x}{M(x,y)} \, {\text d} y \right\} . \]

**Example: **
Solve the equation

*Solution.* With \( M(x,y) = 3xy + y^2 , \quad N(x,y) = x^2 + xy , \) we see that
\( M_y = 3x + 2y \ne N_x = 2x +y . \) Therefore, the given equation is not exact. Since
the ratio

*x*, there exists an integrating factor \( \mu = \mu (x) = x . \) Multiplication by μ reduces the given differential equation to an exact one:

*h(y)*. Differentiating ψ with respect to

*y*and equating the result to

*N*, we get

*h(y)*is a constant, which we drop. Therefore, we obtain the general solution in implicit form:

*C*.

Note that the function

**Example: **
Solve the equation

*Solution.* With \( M(x,y) = 5x^4 y^4 + 2y\, e^{2x} + y^2 \cos (xy) , \quad
N(x,y) = 3x^5 y^3 + xy \, \cos (xy) , \) we see that \( M_y = 20x^4 y^3 + 2\, e^{2x} + 2y \,
\cos (xy) - xy^2 \sin (xy) \ne N_x = 15 x^4 y^3 + y\,\cos (xy) - xy^2 \sin (xy) . \) Therefore, the given
equation is not exact, but the ratio

*y*only. Therefore, there exists an integrating factor as a factor of

*y*:

*(y)*, we obtain an exact equation, which upon integrating, we obtain the general solution:

*C*is an arbitrary constant. ■

In some instances, an integrating factor of a differential equation can be found by inspection, a process based on ingenuity and experience. We present a list of integrating factors that may be helpful in solving differential equations.

With the | Use the | |
---|---|---|

combination | integrating factor | To obtain the exact differential |

\( y\,{\text d}x + x\,{\text d}y \) | \( \displaystyle \begin{split} \frac{1}{xy} \\ \frac{1}{(xy)^n} , \ n \ne 1 \end{split} \) | \( \displaystyle \begin{split} \frac{y\,{\text d}x + x\,{\text d}y}{xy} = {\text d}(\ln xy) \\ \frac{y\,{\text d}x + x\,{\text d}y}{(xy)^n} = -{\text d} \left( \frac{1}{(n-1) (xy)^{n-1}} \right) \end{split} \) |

\( x\,{\text d}x + y\,{\text d}y \) | \( \displaystyle \begin{split} \frac{1}{x^2 + y^2} \\ \frac{1}{(x^2 + y^2 )^n} , \ n\neq 1 \end{split} \) | \( \displaystyle \begin{split} \frac{x\,{\text d}x + y\,{\text d}y}{x^2 + y^2} = \frac{1}{2} \, {\text d}\left( \ln (x^2 + y^2 ) \right) \\ \frac{x\,{\text d}x + y\,{\text d}y}{(x^2 + y^2 )^n} = -\frac{1}{2} \,{\text d}\left( \frac{1}{(n-1) (x^2 + y^2 )^{n-1}} \right) \end{split} \) |

\( y\,{\text d}x - x\,{\text d}y \) | \( \displaystyle \begin{split} \frac{1}{y^2} \\ \frac{1}{x^2} \\ \frac{1}{xy} \\ \frac{1}{x^2 + y^2} \end{split} \) | \( \displaystyle \begin{split} \frac{y\,{\text d}x - x\,{\text d}y}{y^2} = {\text d}\left( \frac{x}{y} \right) \\ \frac{y\,{\text d}x - x\,{\text d}y}{x^2} = -{\text d}\left( \frac{y}{x} \right) \\ \frac{y\,{\text d}x - x\,{\text d}y}{xy} = {\text d}\left( \ln\frac{x}{y} \right) \\ \frac{y\,{\text d}x - x\,{\text d}y}{x^2 + y^2} = {\text d}\left( \arctan \frac{x}{y} \right) \end{split} \) |

**Example: **
Solve the equation

*Solution.* First, we rewrite the equation as

*(xy)*

^{-1}as an integrating factor. Multiplying by μ gives

*C*. ■

**Example: **
Solve the equation

*Solution.* We consider the group \( y\,{\text d}x - x\,{\text d}y , \) which is
not exact, but becomes so after division by *xy* (\( x\ne 0 \mbox{ and } y\ne
0 \) ). Then the given equation is reduced to

*x/y*. Therefore, we let

*z=x/y*and choose φ

*(z)*in such a way that \( \phi (z) \left( 3y^3 \,{\text d}x + xy^2 \,{\text d}y \right) \) is exact. Using relations \( \partial \phi /\partial x = \phi' (z) /y \) and \( \partial \phi /\partial y = -\phi' (z) x/ y^2 , \) we evaluate partial derivatives

*(z)*to be an integrating factor, we need to choose it so that these two partial derivatives are equal:

*C*is an arbitrary constant. ■

A function of
two variables *g(x,y)* is called homogeneous of degree *r* if
\( g(\lambda x , \lambda y ) = \lambda^r g(x,y) \) for any nonzero
constant λ and some real number *r* (possibly zero). If
*M(x,y)* and *N(x,y)* are homogeneous functions of the same
degree, then an integrating factor

**Example: **
Consider the differential equation

*(x,y)*leads to an exact differential equation \( \tilde{M} (x,y) \,{\text d}x + \tilde{N} (x,y)\,{\text d}y =0 \) with

*(x,y)*such that

*h(y)*. Equating ψ

_{y}, the partial derivative of ψ

*(x,y)*with respect to

*y*, to \( \tilde{N} (x,y) \) yields

*y=vx*, we obtain \( x\, v' + v = - v/(1+ v^2 ) . \) Separation of variables yields

*v=y/x*yields the same general solution as before: \( \psi (x,y) \equiv y^2 ( 2 x^2 + y^2 ) = C. \) ■

The integrating factor

**Example: **
The differential equation

*p*(

*z*) = z,

*q*(

*z*) =

*z*². Therefore, this equation can be reduced to an exact equation with the integrating factor \( \mu (x,y) = \frac{1}{xy [ p(xy) - q(xy) ]}; \) namely,

*(x,y)*yields

*(x,y)*. Integration gives

# Integrating factors of the form \( \mu = \mu (\omega (x,y)) \)

Suppose that for a given differential equation \( M(x,y)\,{\text d}x + N(x,y)\,{\text d}y =0 \)
there exists an integrating factor of the form \( \mu =
\mu (\omega (x,y)) \) for some function ω*(x,y)* of two variables.
From equation \( \left( \mu\,M \right)_y = \left( \mu\,N \right)_x , \) we obtain

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