Integrating Factor (Differential Equations)
Integrating Factor - Susam Pal


By Susam Pal on 10 Nov 2021
Introduction

  One of the many techniques for solving ordinary differential
  equations involves using an integrating factor.  An
  integrating factor is a function that a differential equation is
  multiplied by to simplify it and make it integrable.  It almost
  appears to work like magic!

The Method

  Let us first see how the integrating factor method works.  In this
  post, we will work with linear first-order ordinary differential
  equations of type

  \[
    \frac{dy}{dx} + y P(x) = Q(x)
  \]

  to discuss, reason about, and illustrate this method.  We will also
  often use the Leibniz's notation \( dy/dx \) and the Lagrange's
  notation \( y'(x) \) or simply \( y' \) interchangeably as is
  typical in calculus.  They all mean the same thing: the derivative
  of the function \( y \) with respect to \( x.  \) Thus the above
  differential equation may also be written as

  \[
    y' + y P(x) = Q(x).
  \]

  Given a differential equation of this form, we first find an
  integrating factor \( M(x) \) using the formula

  \[
    M(x) = e^{\int P(x) \, dx}.
  \]

  Then we multiply both sides of the differential equation by this
  integrating factor.  Now remarkably, the left-hand side (LHS)
  reduces to a single term consisting only of a derivative.  As a
  result, we can get rid of that derivative by integrating both sides
  of the equation and we then proceed to obtain a solution.

An Example

  Here is an example that demonstrates the method of using an
  integrating factor.  Let us say we want to solve the differential
  equation

  \[
    y' + y \left( \frac{x + 1}{x} \right) = \frac{1}{x}.
  \]

  Indeed this is in the form \( y' + y P(x) = Q(x) \) with \( P(x) =
  (x + 1)/x \) and \( Q(x) = 1/x.  \) We first obtain the integrating
  factor

  \[
    M(x)
    = e^{\int P(x) \, dx}
    = e^{\int (x + 1)/x \, dx}
    = e^{\int (1 + 1/x) \, dx}
    = e^{x + \ln x}
    = x e^x.
  \]

  Now we multiply both sides of the differential equation by this
  integrating factor and get

  \[
    y' x e^x + y (x + 1) e^x = e^x.
  \]

  The LHS can now be simplified to \( \frac{d}{dx} (y x e^x).  \) This
  can be verified using the product rule for derivatives.  This
  simplification of the LHS is the remarkable feature of this method.
  Therefore the above equation can be written as

  \[
    \frac{d}{dx} (y x e^x) = e^x.
  \]

  Note that the expression on the LHS is a product of the function \(
  y \) and the integrating factor \( x e^x.  \) We will discuss this
  observation in more detail a little later.  Let us first complete
  solving this differential equation.  Since the LHS is now a single
  term that consists of a derivative, obtaining a solution now simply
  involves integrating both sides with respect to \( x.  \)
  Integrating both sides we get

  \[
    y x e^x = e^x + C
  \]

  where \( C \) is the constant of integration.  Finally, we divide
  both sides by the integrating factor \( x e^x \) to get

  \[
    y = \frac{1}{x} + \frac{C}{x e^x}.
  \]

  We have now obtained a solution for the differential equation.  If
  we review the steps above, we will find that after multiplying both
  sides of the given differential equation by the integrating factor,
  the differential equation becomes significantly simpler and
  integrable.  In fact, after multiplying both sides of the given
  differential equation by the integrating factor, the LHS always
  becomes the derivative of the product of the function \( y \) and
  the integrating factor.  We will now see why this is so.

An Interesting Relationship

  Consider once again the linear first-order differential equation

  \begin{equation}
    \label{if-eq-diff}
    y' + yP(x) = Q(x).
  \end{equation}

  We first find the integrating factor

  \begin{equation}
    \label{if-eq-integrating-factor}
    M(x) = e^{\int P(x)\, dx}.
  \end{equation}

  The integrating factor obtained like this satisfies an interesting
  relationship:

  \begin{equation}
    \label{if-eq-property}
    M'(x) = M(x) P(x).
  \end{equation}

  We can prove this relationship easily by differentiating both sides
  of \eqref{if-eq-integrating-factor} as follows:

  \[
    M'(x)
    = \frac{d}{dx} \left( e^{\int P(x)\, dx} \right)
    = e^{\int P(x)\, dx} \frac{d}{dx} \left( \int P(x)\, dx \right)
    = M(x) P(x).
  \]

  Note that we use the chain rule to work out the derivative above.
  This beautiful result is due to how the derivative of the
  exponential function works.  When we apply the chain rule to obtain
  the derivative of \( e^{f(x)} \) we get

  \[
    \frac{d}{dx} e^{f(x)} = e^{f(x)} f'(x).
  \]

  This nice property of the exponential function leads to the
  interesting relationship in \eqref{if-eq-property}.

Simplification of LHS

  Now let us multiply both sides of the differential equation
  \eqref{if-eq-diff} by the integrating factor \( M(x) \) By doing so,
  we get

  \[
    y' M(x) + y P(x) M(x) = Q(x) M(x).
  \]

  But from \eqref{if-eq-property} we know that \( P(x) M(x) = M'(x),
  \) so the above equation can be written as

  \[
    y' M(x) + y M'(x) = Q(x) M(x).
  \]

  Look what we have got on the LHS!  We have the expansion of \(
  \frac{d}{dx}(yM(x)) \) on the LHS.  By product rule of
  differentiation, we have \( \frac{d}{dx}(yM(x)) = y' M(x) + y M'(x).
  \) Therefore the above equation can be written as

  \[
    \frac{d}{dx}(yM(x)) = Q(x) M(x).
  \]

  The "magic" has occurred here!  Multiplying both sides of the
  differential equation by the integrating factor has led us to an
  equation that has got a single derivative only on the LHS.  As a
  result, finding the solution is now a simple matter of integrating
  both sides, i.e.,

  \[
    y M(x) = \int Q(x) M(x) \, dx.
  \]

  Thus

  \[
    y = \frac{1}{M(x)} \int Q(x) M(x) \, dx.
  \]

  Note that the result of indefinite integral on the RHS will contain
  the constant of integration, which we will denote as \( C, \) so the
  final solution looks like

  \begin{equation}
    \label{if-eq-general-solution}
    y = \frac{1}{M(x)} \int Q(x) M(x) \, dx + \frac{C}{M(x)}.
  \end{equation}

Illustration

  Let us illustrate the method and its magic with a very simple
  differential equation:

  \[
    y' + \frac{y}{x} = x.
  \]

  First we note that this equation is in the form \( y' + yP(x) = Q(x)
  \) with \( P(x) = 1/x \) and \( Q(x) = x.  \) We then find the
  integrating factor

  \[
  M(x)
  = e^{\int P(x) \, dx}
  = e^{\int \frac{1}{x} \, dx}
  = e^{\ln x}
  = x.
  \]

  Then we multiply both sides of the differential equation by the
  integrating factor to get

  \[
    y'x + y = x^2.
  \]

  Now indeed the LHS can be written down as a single derivative as
  shown below:

  \[
    \frac{d}{dx} yx = x^2.
  \]

  Note that the LHS is the derivative of the product of \( y \) and
  the integrating factor \( x.  \) This is exactly what we discussed
  in the previous section.  We integrate both sides of the above
  equation to get

  \[
    yx = \frac{x^3}{3} + C.
  \]

  Finally we divide both sides by the integrating factor \( x \) to
  get

  \[
    y = \frac{x^2}{3} + \frac{C}{x}.
  \]

  We have arrived at the solution \( y(x) \) for the differential
  equation.

Conclusion

  In this post, we used very simple and convenient differential
  equations that led to nice closed-form solutions.  In practice,
  differential equations can be quite complicated and may not always
  lead to closed-form solutions.  In such cases, we leave the result
  in the form of an expression that contains an unsolved integral.
  Such solutions may resemble the form shown in
  \eqref{if-eq-general-solution}.


  The method of using integrating factors to solve differential
  equations can also be extended to linear higher-order differential
  equations.  That is something we did not discuss in this post.
  However, I hope that the intuition gained from understanding how and
  why this method works for linear first-order differential equations
  will be useful while studying such extensions of this method.


