Section E.2 Laplace Transforms

\begin{gather*} \small \os{\vphantom{m}}{ y'' + 5y' + 6y = 0}\\ \small y(0) = 2,\ y'(0) = -1\\ \small \os{\large 🔻 🔻 🔻 🔻 🔻}{\text{Differential Equation}}\\ \\ \small \us{\large 🔻 🔻 🔻 🔻 🔻}{\text{Solution}}\\ \small y(t) = 5e^{-2t} - 3e^{-3t} \end{gather*}

\begin{gather*} \small \underrightarrow{\text{1️⃣ Forward}}\\ \small \text{Apply}\ \laplacesym\\ \\ \\ \\ \small \text{Apply}\ \laplacesym^{-1}\\ \small \overleftarrow{\text{3️⃣ Backward}} \end{gather*}

\begin{align*} \amp\small\DLBa s^2Y - 2s + 1 + 5sY - 10 + 6Y = 0\\ \amp\small\qquad\qquad {\Big\downarrow}\quad\text{Solve for}\ Y\\ \amp\small\DLBa Y(s) = \frac{2s + 9}{s^2 + 5s + 6}\\ \amp\small\qquad\qquad {\Big\downarrow}\ \ \text{Prepare for Inverse}\\ \amp\small\DLBa Y(s) = \frac{5}{s + 2} - \frac{3}{s + 3} \end{align*}

\begin{gather*} \small \os{\vphantom{m}}{ y'' - 2y' + y = e^{t}}\\ \small y(0) = 0,\ y'(0) = 1\\ \small \os{\large 🔺 🔺 🔺 🔺 🔺}{\text{Differential Equation}}\\ \\ \\ \small \us{\large 🔻 🔻 🔻 🔻 🔻}{\text{Solution}}\\ \small y(t) = t e^{t} + \frac12 t^2 e^{t} \end{gather*}

\begin{gather*} \small \underrightarrow{\text{1️⃣ Forward}}\\ \small \text{Apply}\ \laplacesym\\ \\ \\ \\ \\ \small \text{Apply}\ \laplacesym^{-1}\\ \small \overleftarrow{\text{3️⃣ Backward}} \end{gather*}

\begin{align*} \amp\small\DLBa s^2Y - sy(0) - y'(0)\\ \amp\small\DLBa \qquad\qquad - 2(sY - y(0)) + Y = \frac{1}{s-1}\\ \amp\small\qquad\qquad {\Big\downarrow}\quad\text{Solve for}\ Y\\ \amp\small\DLBa Y(s) = \frac{1}{(s-1)^3} + \frac{1}{(s-1)^2}\\ \amp\small\qquad\qquad {\Big\downarrow}\ \ \text{Prepare for Inverse}\\ \amp\small\DLBa Y(s)\ \text{already matches} \end{align*}

\begin{gather*} \small \os{\vphantom{m}}{ y'' + y = t^2 }\\ \small y(0) = 1,\ y'(0) = 0\\ \small \os{\large 🔻 🔻 🔻 🔻 🔻}{\text{Differential Equation}}\\ \\ \small \us{\large 🔻 🔻 🔻 🔻 🔻}{\text{Solution}}\\ \small y(t) = 3\cos(t) + t^2 - 2 \end{gather*}

\begin{gather*} \small \underrightarrow{\text{1️⃣ Forward}}\\ \small \text{Apply}\ \laplacesym\\ \\ \\ \\ \small \text{Apply}\ \laplacesym^{-1}\\ \small \overleftarrow{\text{3️⃣ Backward}} \end{gather*}

\begin{align*} \amp\small\DLBa s^2Y - sy(0) - y'(0) + Y = \frac{2}{s^3}\\ \amp\small\qquad\qquad {\Big\downarrow}\quad\text{Solve for}\ Y\\ \amp\small\DLBa Y(s) = \frac{s}{s^2 + 1} + \frac{2}{s^3(s^2 + 1)}\\ \amp\small\qquad\qquad {\Big\downarrow}\ \ \text{Prepare for Inverse}\\ \amp\small\DLBa Y(s) = \frac{3s}{s^2 + 1} - \frac{2}{s} + \frac{2}{s^3} \end{align*}

\begin{gather*} \small \os{\vphantom{m}}{ y'' + 6y' + 9y = 0 }\\ \small y(0) = 3,\ y'(0) = -3\\ \small \os{\large 🔺 🔺 🔺 🔺 🔺}{\text{Differential Equation}}\\ \\ \small \us{\large 🔻 🔻 🔻 🔻 🔻}{\text{Solution}}\\ \small y(t) = (3 - 24t)e^{-3t} \end{gather*}

\begin{gather*} \small \underrightarrow{\text{1️⃣ Forward}}\\ \small \text{Apply}\ \laplacesym\\ \\ \\ \\ \small \text{Apply}\ \laplacesym^{-1}\\ \small \overleftarrow{\text{3️⃣ Backward}} \end{gather*}

\begin{align*} \amp\small\DLBa s^2Y - 3s + 3 + 6sY - 18 + 9Y = 0\\ \amp\small\qquad\qquad {\Big\downarrow}\quad\text{Solve for}\ Y\\ \amp\small\DLBa Y(s) = \frac{3s - 15}{(s + 3)^2}\\ \amp\small\qquad\qquad {\Big\downarrow}\ \ \text{Prepare for Inverse}\\ \amp\small\DLBa Y(s) = \frac{3}{s + 3} - \frac{24}{(s + 3)^2} \end{align*}

\begin{gather*} \small \os{\vphantom{m}}{ y''' + y'' - y' - y = 0 }\\ \small y(0) = 1,\ y'(0) = 0,\ y''(0) = -1\\ \small \os{\large 🔺 🔺 🔺 🔺 🔺}{\text{Differential Equation}}\\ \\ \small \us{\large 🔻 🔻 🔻 🔻 🔻}{\text{Solution}}\\ \small y(t) = -\frac12e^{t} - \frac12e^{-t} \end{gather*}

\begin{gather*} \small \underrightarrow{\text{1️⃣ Forward}}\\ \small \text{Apply}\ \laplacesym\\ \\ \\ \\ \small \text{Apply}\ \laplacesym^{-1}\\ \small \overleftarrow{\text{3️⃣ Backward}} \end{gather*}

\begin{align*} \amp\small\DLBa s^3Y + s^2 + 1 + s^2Y + s\\ \amp\small\DLBa \qquad\qquad - sY - 1 - Y = 0\\ \amp\small\qquad\qquad {\Big\downarrow}\quad\text{Solve for}\ Y\\ \amp\small\DLBa Y(s) = \frac{-s^2 - s}{s^3 + s^2 - s - 1}\\ \amp\small\qquad\qquad {\Big\downarrow}\ \ \text{Prepare for Inverse}\\ \amp\small\DLBa Y(s) = -\frac12\frac{1}{s - 1} - \frac12\frac{1}{s + 1} \end{align*}

\begin{gather*} \small \os{\vphantom{m}}{ y'' + 2y = e^{t} \cos(3t) }\\ \small y(0) = 0,\ y'(0) = 0\\ \small \os{\large 🔺 🔺 🔺 🔺 🔺}{\text{Differential Equation}}\\ \\ \small \us{\large 🔻 🔻 🔻 🔻 🔻}{\text{Solution}}\\ \small y(t) = e^{t}\cos(3t) - \cos(\sqrt{2}t) \end{gather*}

\begin{gather*} \small \underrightarrow{\text{1️⃣ Forward}}\\ \small \text{Apply}\ \laplacesym\\ \\ \\ \\ \small \text{Apply}\ \laplacesym^{-1}\\ \small \overleftarrow{\text{3️⃣ Backward}} \end{gather*}

\begin{align*} \amp\small\DLBa s^2 Y + 2Y = \frac{s - 1}{(s - 1)^2 + 9}\\ \amp\small\qquad\qquad {\Big\downarrow}\quad\text{Solve for}\ Y\\ \amp\small\DLBa Y(s) = \frac{s - 1}{(s^2 + 2)[(s - 1)^2 + 9]}\\ \amp\small\qquad\qquad {\Big\downarrow}\ \ \text{Prepare for Inverse}\\ \amp\small\DLBa Y(s) = \frac{1}{(s - 1)^2 + 9} - \frac{s}{s^2 + 2} \end{align*}