Math

Integrating Factors

Integrating factors allow us to solve equations of the form \(y' + P(x)y = Q(x)\)

Intuition

The derivative of \(e^{R(x)}\) is \(R'(x) e^{R(x)}\) . This can be used to cancel out terms when we have anything of the form \(y' + R'(x)y\), and we can do by setting \(P(x) = R'(x)\) . By playing around with this idea the methodology below was developed.

General Solution

We know that \(y' + P(x)y = Q(x)\). By want \(R'(x) = P(x)\), so we define \(R(x) = \int P(x) dx\) to get \(y' + R'(x)y = Q(x)\). By multiplying all terms of the equation by \(e^{R(x)}\) we get

\[\begin{align} Q(x) e^{R(x)} &= y'e^{R(x)} + yR'(x)e^{R(x)} \\ &= y'e^{R(x)} + y(e^{R(x)})' \\ &= (ye^{R(x)})' \end{align}\]

By integrating both sides and then rearranging for \(y\), we get \(ye^{R(x)} = \int Q(x) e^{R(x)} dx \Leftrightarrow y = e^{-R(x)} \int Q(x) e^{R(x)}dx\), with \(R(x) = \int P(x) dx\) as initially defined.

Example

Many examples require harder integrals than the one that follows, however this example should give you an idea of how to proceed.

\[y' - \frac{y}{x} = x\]

We have that \(P(x) = -\frac{1}{x}\), so \(R(x) = \int P(x) dx = - \int \frac{1}{x} dx = -\ln{x}\). Hence our Integrating Factor is \(e^{-\ln{x}} = \frac{1}{x}\), so we can rewrite our original equation by multiplying it by \(\frac{1}{x}\) as

\[\begin{align} \frac{y'}{x} - \frac{y}{x}\frac{1}{x} &= x \frac{1}{x} \\ \frac{y'}{x} - \frac{y}{x^2} &= 1 \\ \frac{y'}{x} - y \left ( \frac{1}{x} \right ) ' &= 1 \\ \left ( \frac{y}{x} \right ) ' &= 1 \end{align}\]

Integrating both sides with respect to \(x\) we get

\[\frac{y}{x} = \int 1 dx = x \Leftrightarrow y = x^2\]

Confirming it within the original equation, we indeed get \(y' - \frac{y}{x} = ( x^2 ) ' - \frac{x^2}{x} = 2x - x = x\)