> [!NOTE] Theorem > Let $\frac{d}{dt} x(t) = f(x(t))g(t), \quad x(t_{0})=x_{0}$with $t\in(\alpha,\beta)\subset \mathbb{R},$ $f:\mathbb{R}\to \mathbb{R}$ and $g:(\alpha,\beta)\to \mathbb{R}$ be an [[Initial Value Problem for Scalar Ordinary Differential Equation|initial value problem]] for a [[Separable Differential Equation|separable differential equation]] that has a unique [[Solution to Scalar Ordinary Differential Equation|solution]]. > > If $f(x_{0})=0,$ then $x(t)=x_{0},$ for all $t\in (\alpha,\beta),$ is the solution to the equation. > > If $f(x_{0})\neq 0,$ then $f(x(t))\neq 0$ for all $t\in(\alpha,\beta).$ **Proof**: If $f(x_{0})=0$ then the function $x(t)=x_{0}$ satisfies both $\frac{d}{dt}x(t)=0$ and $f(x(t))g(t)=f(x_{0})g(t)=0$ for all $t$. Now suppose $f(x_{0})\neq 0.$ Assume there is a time $t_{1}\in(\alpha,\beta)$ such that $f(x(t_{1}))=0.$ Let $y$ denote the solution to the equation but with the initial condition $y(t_{1})=x(t_{1}).$ Note that $t_{0}$ is arbitrary, and that the solution is unique. Thus $x=y.$ However, as $f(y(t_{1}))=f(x(t_{1}))=0,$ the previous result yields that $y(t)=x(t_{1})$ for all $t\in(\alpha,\beta).$ But $f(y(t_{0}))=0\neq f(x(t_{0}))$ which contradicts $y=x.$ Therefore $f(x(t))\neq 0$ for all $t\in (\alpha,\beta).$ # Applications **Consequences**: See [[Implicit Solution to Initial Value Problem for Separable Equation]].