Continuous Time: First-Order Linear Diff Eq

General form

(1)
\begin{align} \frac{dy}{dt}+u(t)y=w(t) \end{align}

For simplicity, u(t) and w(t) are constant here

(2)
\begin{align} \frac{dy}{dt}+ay=b \end{align}

Homogeneous Case


(3)
\begin{align} \frac{dy}{dt}+ay=0 \end{align}
(4)
\begin{align} \frac{dy}{dt} \frac{1}{y}=-a \end{align}

so

(5)
\begin{equation} y(t)=A{e}^{-at} \end{equation}

or

(6)
\begin{equation} y(t)=y(0){e}^{-at} \end{equation}

Proof

(7)
\begin{align} \ln y(t)=\ln (A{e}^{-at})=\ln A+\ln {e}^{-at}=\ln A-at \end{align}
(8)
\begin{align} \ln 'y(t)=\frac{dy}{dt}\frac {1}{y}=-a \end{align}

Nonhomogeneous Case


(9)
\begin{equation} {y}_{c}=A{e}^{-at} \end{equation}
(10)
\begin{align} {y}_{p}=\frac{b}{a} \end{align}
(11)
\begin{align} y(t)={y}_{c}+{y}_{p}=A{e}^{-at}+\frac{b}{a} \end{align}
(12)
\begin{align} \because y(0)=A+\frac{b}{a} \end{align}
(13)
\begin{align} A=y(0)-\frac{b}{a} \end{align}
(14)
\begin{align} \therefore y(t)=[y(0)-\frac{b}{a}]{e}^{-at}+\frac{b}{a} \end{align}
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