Question

Use Laplace transforms to solve the initial value problem y'+y=f(t),
where f(t)=2 (0≤t<2)
=-6( t≥2)
, and y(0)=7.

Answer 1

\[y'+y=f(t)\\
f(t)=\begin{cases}2&\text{for }0\le t<2\\-6&\text{for }t\ge2\end{cases}\]
Take the Laplace transform of both sides:
\[s\mathcal{L}\{y\}-y(0)+\mathcal{L}\{y\}=\int_0^\infty f(t)e^{-st}~dt\\
(s+1)\mathcal{L}\{y\}=\left(2\int_0^2e^{-st}~dt-6\int_2^\infty e^{-st}~dt\right)+y(0)\\
(s+1)\mathcal{L}\{y\}=\left(-\frac{2}{s}(e^{-2s}-1)+\frac{6}{s}(0-e^{-2s}) \right)+7\\
(s+1)\mathcal{L}\{y\}=-\frac{8}{s}e^{-2s}+\frac{2}{s}+7\\
\mathcal{L}\{y\}=\color{red}{-\frac{8}{s(s+1)}e^{-2s}}+\color{blue}{\frac{2}{s(s+1)}}+\color{green}{\frac{7}{s+1}}\\
y=\mathcal{L}^{-1}\left\{\color{red}{-\frac{8}{s(s+1)}e^{-2s}}+\color{blue}{\frac{2}{s(s+1)}}+\color{green}{\frac{7}{s+1}}\right\}\]
Important table formula to use: \(\mathcal{L}^{-1}\left\{e^{-cs}F(s)\right\}=\theta(t-c)~f(t-c)\).
\[\begin{align*}\mathcal{L}^{-1}\left\{\color{red}{-\frac{8}{s(s+1)}e^{-2s}}\right\}&=-8~\theta(t-2)~\mathcal{L}^{-1}\left\{\frac{1}{s(s+1)}\right\}\\
&=-8~\theta(t-2)~\mathcal{L}^{-1}\left\{\frac{1}{s}-\frac{1}{s+1}\right\}\\
&=-8~\theta(t-2)~\left(\mathcal{L}^{-1}\left\{\frac{1}{s}\right\}-\mathcal{L}^{-1}\left\{\frac{1}{s+1}\right\}\right)\\
&=-8~\theta(t-2)~\left(1-e^{-\bf\color{red}{(t-2)}}\right)
\end{align*}\]
\[\begin{align*}\mathcal{L}^{-1}\left\{\color{blue}{\frac{2}{s(s+1)}}\right\}&=2\mathcal{L}^{-1}\left\{\frac{1}{s(s+1)}\right\}\\
&=2(1-e^{-t})
\end{align*}\]
\[\begin{align*}\mathcal{L}^{-1}\left\{\color{green}{\frac{7}{s+1}}\right\}&=7\mathcal{L}^{-1}\left\{\frac{1}{s+1}\right\}\\
&=7e^{-t}
\end{align*}\]