linear first-order equation
This is the workhorse equation of first-order theory — the one with a guaranteed, formulaic solution every single time. Unlike the separable case (which is a lucky structural accident) or the nonlinear cases (which usually have no closed form), the linear first-order equation is always solvable in closed form by a fixed recipe. If you can put your equation into its shape, you are essentially done.
Its standard form is dy/dx + P(x)*y = Q(x): the unknown y appears alone to the first power, multiplied only by a function of x, and the right side Q(x) (the forcing or source term) depends only on x. The solution is delivered by the integrating factor mu(x) = e^(integral of P(x) dx). Multiplying through turns the left side into d/dx(mu*y), so one integration gives y = (1/mu)*(integral of mu*Q dx + C). When Q = 0 the equation is homogeneous and the answer is just y = C/mu = C*e^(-integral P), an exponential; a nonzero Q adds the particular response driven by the source. That split — homogeneous part plus particular part — is the seed of the whole linear theory to come.
These equations model an enormous range of first-order linear systems: an RC or RL electrical circuit, a mixing tank with inflow and outflow, radioactive decay with a steady feed, Newtonian cooling with a varying ambient. The reliability is the point — given the standard form, the integrating factor formula never fails, which is exactly why so much effort (Bernoulli and Riccati substitutions, for instance) goes into transforming awkward nonlinear equations into this trustworthy linear shape.
An RC circuit charging from a battery obeys dq/dt + q/(R*C) = V/R. With P = 1/(R*C), the integrating factor is e^(t/(R*C)); the solution is q(t) = C_0*V*(1 - e^(-t/(R*C))), the familiar exponential approach to full charge.
Put the equation in standard form, build mu, integrate once — a guaranteed solution.
The formula assumes the equation is genuinely in standard form, with the coefficient of dy/dx normalized to 1. If your equation reads a(x)*y' + b(x)*y = c(x), you must first divide through by a(x) — and watch for points where a(x) = 0, since those can be genuine singular points where the solution misbehaves.