Deriving the Solution to First-Order Linear Differential Equations

$\newcommand{\dif}{\mathrm{d}}$ 

A first-order linear differential equation is one of many irritating nuisances encountered in calculus. I thus present to you its rancid definition:
    A first-order linear diffential equation is any differential equation of the form 

$$\frac{\dif x}{\dif t} + p(t)x = q(t)$$

Which has the plain and simple solution 

$$x = e^{-\int p(t) \dif t} \int e^{\int p(t) \dif t}q(t)\dif t$$

If you too are confused at this, then you've come to the right place.

First, notice that a first-order linear is very similar to the form $\frac{\dif u}{\dif t}v+u\frac{\dif v}{\dif t}$, which by the product rule would simply equal $\frac{\dif }{\dif t}(uv)$. Our goal is therefore to put the DE into this form.

The key step is to invent a new function, called $f(t)$, and multiply the first-order linear DE by it:

$$\frac{\dif x}{\dif t}f(t) + f(t)p(t)x = f(t)q(t)$$

And if we define $f(t)p(t)$ equal to $\frac{\dif f}{\dif t}$, the whole thing can simplify with the product rule, just as we expected: 

$$\frac{\dif x}{\dif t}f(t) + \frac{\dif f}{\dif t}x = f(t)q(t)$$

$$\frac{\dif }{\dif t}(x \cdot f(t)) = f(t)q(t)$$

Which is much more manageable. 

$$x \cdot f(t) = \int f(t)q(t) \dif t$$

$$x = \frac{1}{f(t)} \int f(t)q(t) \dif t$$

And you have the solution, assuming you're fine with not explicitly defining $f$. If that is not the case, then let's find $f$.

The only condition we gave was

$$\frac{\dif f}{\dif t}=f(t)p(t)$$

Rearanging and integrating:

$$\int  \frac{1}{f(t)}\dif f = \int p(t) \dif t$$  

$$\ln(f(t)) = \int p(t) \dif t$$

$$f(t) = e^{\int p(t) \dif t}$$

And finally substituting it into the solution:

$$x = e^{-\int p(t) \dif t} \int e^{\int p(t) \dif t}q(t)\dif t$$

$$\mathbb{Q.E.D.}$$