> [!definition] > > $ > a(t)\frac{d^2y}{dt^2} + b(t)\frac{dy}{dt} + c(t) = 0 > $ > Let $(L)$ be a second order [[Ordinary Differential Equation|ODE]]. Then all solutions of $(L)$ can be expressed as [[Linear Combination|linear combinations]] of two functions, which also takes care of the constants. > [!theorem] Strategy > > Suppose that $y_1(t)$ solves the ODE, then let $y(t) = u(t)y_1(t)$ where $y$ is also assumed to solve $(L)$, then we have > $ > y = uy_1 \quad \frac{dy}{dt}=\frac{du}{dt}y_1 + u\frac{dy_1}{dt} > $ > and > $ > \frac{d^2y}{dt^2} = \frac{d^2u}{dt^2}y_1 + 2\frac{d_1y}{dt}\frac{du}{dt} + u\frac{d^2y_1}{dt^2} > $ > Then we can write > $ > \begin{align*} > 0 &= a(t)\frac{d^2y}{dt^2} + b(t)\frac{dy}{dt} + c(t) \\ > &= a\paren{\frac{d^2u}{dt^2}y_1 + 2\frac{d_1y}{dt}\frac{du}{dt} + u\frac{d^2y_1}{dt^2}} \\ > &+b\paren{\frac{du}{dt}y_1 + u\frac{dy_1}{dt}} + cuy_1 \\ > &= ay_1\frac{d^2u}{dt^2} + \paren{2a\frac{dy_1}{dt} + by_1}\frac{du}{dt} \\ > &+\paren{a\frac{d^2y_1}{dt^2} + b\frac{dy_1}{dt} + cy_1}u \\ > &= ay_1\frac{d^2u}{dt^2} + \paren{2a\frac{dy_1}{dt} + by_1}\frac{du}{dt} \\ > &+\cancel{\paren{a\frac{d^2y_1}{dt^2} + b\frac{dy_1}{dt} + cy_1}}u \\ > &= ay_1\frac{d^2u}{dt^2} + \paren{2a\frac{dy_1}{dt} + by_1}\frac{du}{dt} > \end{align*} > $ > Let $v = \frac{du}{dt}$, then we have created > $ > 0 = ay_1 \frac{dv}{dt} + \paren{2a\frac{dy_1}{dt}+by_1}v > $ > a [[First Order Inexact ODE|first order ODE]].