Integrating Factor

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 we multiply a differential equation with 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 \( \frac{dy}{dx} \) 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 with 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 \( Q(x) = x \). 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 with 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 as shown below. 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 with the integrating factor, the LHS always becomes the derivative of the 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 the integrating factor \( M(x) \) on both sides of the differential equation \eqref{if-eq-diff}. 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 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 the integrating factor on both sides of the differential equation 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, I crafted 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.

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