Preface

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This is a tutorial made solely for the purpose of education and it was designed for students taking Applied Math 0330. It is primarily for students who have very little experience or have never used Mathematica before and would like to learn more of the basics for this computer algebra system. As a friendly reminder, don't forget to clear variables in use and/or the kernel.

Finally, the commands in this tutorial are all written in bold black font, while Mathematica output is in normal font. This means that you can copy and paste all commands into Mathematica, change the parameters and run them. You, as the user, are free to use the scripts for your needs to learn the Mathematica program, and have the right to distribute this tutorial and refer to this tutorial as long as this tutorial is accredited appropriately.

Integrating Factors

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In many cases, a differential equation of first order can be converted into an exact equation by multiplying through appropriate function. Correspondingly, a function \( \mu = \mu (x,y) \) is called an 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 This is especially useful in thermodynamics where temperature becomes the integrating factor that makes entropy an exact differential.

The integrating factor method was introduced by the French mathematician, astronomer, and geophysicist Alexis Claude Clairaut (1713--1765). He was a prominent Newtonian whose work helped to establish the validity of the principles and results that Sir Isaac 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 gave 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

Upon division by μ, we get Actually, the conversion of a differential equation into an exact equation using an integrating factor is extremely general. Unfortunately, there is no systematic way to solve the above partial differential equation with respect to μ. There are known particular classes when it is possible, and we consider some of them. It is also sometimes convenient to reduce the partial differential equation for logarithm of μ. Let Since \( \frac{\partial}{\partial x}\, \ln \mu = G(x,y) \) and \( \frac{\partial}{\partial y}\, \ln \mu = F(x,y) , \) we find where 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
  \[ \mu (y) = \exp \left\{ -\int \frac{M_y - N_x}{M(x,y)} \, {\text d} y \right\} . \]

An integrating factor is not unique, as the following example shows.

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

is a function of x, there exists an integrating factor \( \mu = \mu (x) = x . \) Multiplication by μ reduces the given differential equation to an exact one: Indeed, with new coefficients \( M(x,y) = 3x^2 y + x\,y^2 , \quad N(x,y) = x^3 + x^2 y , \) we have \( M_y = 3x^2 + 2xy = N_x = 3x^2 + 2xy . \) Therefore, the given equation is exact. Then there exists a potential function ψ such that Integrating the former, we get \( \psi (x,y) = x^3 y + \frac{1}{2}\, x^2 y^2 + h(y) \) for some (unknown) function h(y). Differentiating ψ with respect to y and equating the result to N, we get So h(y) is a constant, which we drop. Therefore, we obtain the general solution in implicit form: for some constant C.

Note that the function

can also be used as an integrating factor.

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

is a function of y only. Therefore, there exists an integrating factor as a factor of y: Multiplying by μ(y), we obtain an exact equation, which upon integrating, we obtain the general solution: where 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

The table suggests that we try μ = (xy)^-1 as an integrating factor. Multiplying by μ gives and the general solution is obtained by just integration: \( x^4 + \ln \, xy = C , \) with some constant 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

The last term is now exact and will remain exact when multiplied by any function of 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 For φ(z) to be an integrating factor, we need to choose it so that these two partial derivatives are equal: Solving this equation for φ, we get the integrating factor \( \phi (z) =z^2 = x^2 /y^2 . \) Multiplying by it, we obtain the exact equation: \[ \left( 3x^2 y + \frac{x}{y^2} \right) {\text d}x + \left( x^3 - \frac{x^2}{y^3} \right) {\text d}y =0 \qquad \Longrightarrow \qquad {\text d} \left( x^3 y + \frac{x^2}{2y^2} \right) =0 . \]