# A calculus of the absurd

#### 13.2 Integrating Factors

$$e^x$$ shows up a lot in differential equations, because it has properties that are helpful when we differentiate it. One way in which it helps us is in solving first-order linear differential equations, which are equations of the form

$\frac {dy}{dx} + p(x)y = q(x)$

This can be solved using the product rule. If we define a function $$f(x)$$, we can write by the product rule that the derivative of $$y e^{f(x)}$$ is

\begin{equation} \frac {dy}{dx} e^{f} + e^{f}\frac {df}{dx}y \end{equation}

This doesn’t immediately look like our equation, but if we multiply through by $$e^{f}$$, we get that

\begin{equation} \frac {dy}{dx} e ^ {f(x)} + p(x) e^{f(x)} y = q(x) e ^{f(x)} \end{equation}

What we can do here is write that the left hand side is equal to the derivative of $$ye^{f(x)}$$. This only works, however, if the derivative of $$f(x)$$ is equal to $$p(x)$$. 8888 This is because \begin {aligned} \frac {d}{dx} \left [ ye^{f(x)} \right ] &= \frac {d}{dx}{y} e^{f(x)} + y \frac {d}{dx} \left [ e^{f(x)} \right ] \\ &= \frac {dy}{dx} e^{f(x)} + y \frac {d}{dx} \left [f(x)\right ] e^{f(x)} \\ \end {aligned} And if \begin {aligned} f(x) = \int p(x) dx \end {aligned} then \begin {aligned} \frac {d}{dx} \left [ f(x) \right ] = p(x) \end {aligned} And thus $$\frac {d}{dx} \left [ ye^{f(x)} \right ] = \frac {dy}{dx} e ^ {f(x)} + p(x) e^{f(x)} y$$ which is just the left-hand side of the equation. If it is, we can write that

\begin{equation} \frac {d}{dx} \left [ ye^{f(x)} \right ] = q(x) e ^{f(x)} \end{equation}

And thus we can solve the equation by integrating.