> [!theorem] Strategy > > Let $(L)$ be a first order linear [[Ordinary Differential Equation|ODE]], > $ > a_1(t)\frac{dy}{dt} + a_0(t)y = g(t) \Rightarrow > \frac{dy}{dt} + \frac{a_0}{a_1}y = \frac{g}{a_0} > $ > Then using the [[Product Rule|product rule]], introduce an integrating factor and write > $ > u\frac{dy}{dt} + puy = uq > $ > then choose $u(t)$ such that $\frac{du}{dt}=pu$ (solve as [[Separable ODE|separable ODE]]). In which case, > $ > \frac{d}{dt}uy = uq \Rightarrow > y = \frac{C}{u} + \frac{1}{u}\int uq dt > $ > Where we can simply pick $u = e^{\int pdt}$.