I think a good way to preface this section is to recall the product rule from single variable calculus.
$$\boxed{\dfrac{d}{dx}\left(f(x)g(x)\right) = f'(x)g(x) + f(x)g'(x)}$$
With that in mind, let's look at an example:
$$\dfrac{dy}{dx} - 3y = e^{2x}, y(0) = 3$$
This isn't exactly the product rule, but it can be. Let's multiply by a function, $\rho$.
$$\rho\dfrac{dy}{dx} -3\rho{}y = \rho{}e^{2x}$$
So, on the left side, the y is differentiated, so on the right, the $\rho$ is differentiated. We can say
$$\dfrac{d\rho}{dx} = -3\rho \to \rho = e^{-3x} + C$$
We can pick our own function, so we just set C = 0. So, what we were trying to do is combine the left side of the equation into the derivative of one term. This is what it becomes:
$$\dfrac{d}{dx}(e^{-3x}y) = e^{-x}$$
Now we just integrate, and we're done. Here, we can't forget the constant term. Before, we were allowed to, because we just had to find any function that satisfy the product rule.
$$e^{-3x}y = -e^{-x} + C \to \boxed{y = Ce^{3x} - e^{-x}}$$

# Example

$$y' = x^2 - y$$
$$y' + y = x^2$$
We now multiply by $\rho$
$$\rho{}y' + \rho y = \rho x^2$$
$$\rho' = \rho \to \rho = e^{x}$$
$$\dfrac{d}{dx}\left(e^xy\right) = x^2e^x$$
$$y = e^{-x}\left(\int x^2e^x\,\mathrm{d}x\right)$$
Because this post isn't intended to be about integration methods, I just plugged the integral into WolframAlpha. If you want to do it out, it requires integration by parts.
$$y = e^{-x}\left((x^2 - 2x + 2)e^x + C\right)$$
$$y = x^2 - 2x + 2 + Ce^{-x}$$

# Existence and Uniqueness

Given this equation:
$$\dfrac{dy}{dx} + P(x)y = Q(x), y(x_0) = y_0$$
A unique solution exists given P(x) and Q(x) are continuous in a region around $x_0$

David Witten