Consider the following differential equation \begin{equation*} \frac{dy}{dx} = q(x) - p(x)y, \end{equation*} in general, this equation is not separable. There are a few circumstances that will make it separable such as if \(q(x) = p(x)\) and some others, but that is not the point. In general, we can't solve it using separation of variables.

However, if we move the term involving \(y\) to the left, then the left side almost looks as if it could be the derivative of a product e.g. \([uv]' = u'v + uv'\text{.}\) \begin{equation} \underbrace{y'}_{u'} \underbrace{\underline{\phantom{m}}}_{v} + \underbrace{y}_{u} \underbrace{p(x)}_{v'} = q(x) \label{eqn-first-order-linear}\tag{2.3.1} \end{equation} If the left side is to be the derivative of a product, then we need to fill the blank with some function of \(x\text{,}\) say \(I(x)\text{.}\) But if we multiply one term in the equation by some factor, then we must multiply all terms in the equation by the same factor. If we can determine \(I(x)\text{,}\) then upon multiplying the entire equation by that function we would be able to solve the equation by integrating both sides, and solving for \(y\text{.}\)

Here is what equation (2.3.1) looks like if we multiply both sides by some as yet unknown function, \(I(x)\text{:}\) \begin{equation*} \underbrace{y'}_{u'} \underbrace{I(x)}_{v} + \underbrace{y}_{u} \underbrace{p(x)I(x)}_{v'} = q(x)I(x) \end{equation*} Whatever, this function \(I\) is, it clearly must satisfy: \begin{equation} I' = p(x)I \quad \text{ or, equivalently } \quad \frac{dI}{dx} = p(x)I. \label{eqn-integrating-factor-condition}\tag{2.3.2} \end{equation} This is because \(v=I(x)\) and \(v'=I'=p(x)I(x)\text{.}\) Equation (2.3.2) is separable and has solution: \begin{gather} \int \frac{dI}{I} = \int p(x) \, dx\notag\\ \ln \abs{I} = \int p(x) \, dx\notag\\ I = e^{\int p(x) \, dx}\label{eqn-integrating-factor}\tag{2.3.3} \end{gather}

Since, we multiply equation (2.3.1) by \(I(x)\) in order to be able to integrate, we call, \(I(x)\) an integrating factor. Upon multiplying both sides of the equation by \(I(x)\text{,}\) the left side is indeed the derivative of a product: \begin{equation*} [y\cdot I]' = y'e^{\int p(x) \, dx} + yp(x)e^{\int p(x) \, dx}. \end{equation*}

This method allows us to solve linear equations where the right hand side is not zero! That is, it will allow us to solve equations of the form: \begin{equation*} y' + p(x)y = q(x). \end{equation*} An example will illustrate the method more clearly.