令 dx/dt=axdx / dt = a xdx/dt=ax，則其所有解之形式皆為 keatke^{at}keat 其中 k∈Rk \in \mathbb{R}k∈R Proof. [local-0] 令 u(t)u(t)u(t) 為一解，則 du/dt=u′(t)=au(t)du / dt = u'(t) = au(t)du/dt=u′(t)=au(t)。計算 ddt(u(t)e−at)=u′(t)e−at+u(t)(−ae−at)=au(t)e−at−au(t)e−at=0\begin{aligned} \frac{d}{d t}(u(t) e^{-at}) &= u'(t) e^{-at} + u(t)(-a e^{-at}) \\ &= au(t) e^{-at} - au(t) e^{-at} = 0 \end{aligned}dtd(u(t)e−at)=u′(t)e−at+u(t)(−ae−at)=au(t)e−at−au(t)e−at=0 因此 u(t)e−atu(t) e^{-at}u(t)e−at 是常數，可以寫成 u(t)eat=k∈R\frac{u(t)}{e^{at}} = k \in \mathbb{R}eatu(t)=k∈R 因此 u(t)=keatu(t) = k e^{at}u(t)=keat
令 u(t)u(t)u(t) 為一解，則 du/dt=u′(t)=au(t)du / dt = u'(t) = au(t)du/dt=u′(t)=au(t)。計算 ddt(u(t)e−at)=u′(t)e−at+u(t)(−ae−at)=au(t)e−at−au(t)e−at=0\begin{aligned} \frac{d}{d t}(u(t) e^{-at}) &= u'(t) e^{-at} + u(t)(-a e^{-at}) \\ &= au(t) e^{-at} - au(t) e^{-at} = 0 \end{aligned}dtd(u(t)e−at)=u′(t)e−at+u(t)(−ae−at)=au(t)e−at−au(t)e−at=0 因此 u(t)e−atu(t) e^{-at}u(t)e−at 是常數，可以寫成 u(t)eat=k∈R\frac{u(t)}{e^{at}} = k \in \mathbb{R}eatu(t)=k∈R 因此 u(t)=keatu(t) = k e^{at}u(t)=keat