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 $y' - \frac{y}{x} = ( x^2 ) ' - \frac{x^2}{x} = 2x - x = x$