Applied Laplace Transforms

Big Idea

The Laplace transform can be used to solve initial value problems since

$\mathcal L\{y'(t)\} = sY(s) - y(0),\quad \mathcal L\{y''(t)\} = s^2Y(s) - sy(0) - y'(0).$

For a derivation of these results, check out this post.

Questions

Question 1. Solve the initial value problem

$\left\{ \begin{aligned}y'' + y &= \delta(t-2), \\ y(0) &= 1, \\ y'(0) &= 1. \end{aligned} \right.$

(Click for Solution)

Solution. Taking Laplace transforms on both sides and applying linearity,

$\mathcal L \{ y'' \} + \mathcal L \{ y \} = \mathcal L \{ \delta(t-2) \}.$

Evaluating each term and denoting $Y \equiv Y(s) = \mathcal L\{y(t) \} \equiv \mathcal L\{y(t)\}$ ,

$\begin{aligned} \mathcal L\{ \delta(t-2) \} &= e^{-2s}, \\ \mathcal L\{ y'' \} &= s^2 Y(s) - sy(0) - y'(0) \\ &= s^2 Y(s) - s - 1. \end{aligned}$

Consolidating,

$\begin{aligned} (s^2 Y(s) - s - 1) + Y(s) &= e^{-2s} \\ (s^2 + 1) Y(s) &= s+1 + e^{-2s} \\Y(s) &= \frac s{s^2 + 1} + \frac 1{s^2 + 1} + e^{-2s} \cdot \frac 1{s^2 + 1}. \end{aligned}$

Taking inverse Laplace transforms and applying linearity,

$\begin{aligned} y(t) = \mathcal L^{-1} \{Y(s) \} &= \mathcal L^{-1} \left\{ \frac s{s^2 + 1^2} \right\} + \mathcal L^{-1} \left\{ \frac 1{s^2 + 1^2} \right\} + \mathcal L^{-1} \left\{ e^{-2s} \cdot \frac 1{s^2 + 1} \right\} \\ &= \cos(t) + \sin(t) + \mathcal L^{-1} \left\{ e^{-2s} \cdot \frac 1{s^2 + 1} \right\}.\end{aligned}$

Denoting $F(s) = 1/(s^2+1)$ yields $f(t) = \mathcal L^{-1} \{F(s)\} = \sin(t)$ . By the shift theorems,

$\begin{aligned} \mathcal L^{-1} \{ e^{-2s} F(s) \} &= f(t-2) U (t-2) \\ &= \sin(t-2) U(t-2).\end{aligned}$

Consolidating, we obtain the final answer

$y(t) = \cos(t) + \sin(t) + \sin(t-2) U(t-2).$

Question 2. Solve the initial value problem

$\left\{ \begin{aligned}y'' + y &= 1 + U(t-2), \\ y(0) &= 1, \\ y'(0) &= 1. \end{aligned} \right.$

(Click for Solution)

Solution. Taking Laplace transforms on both sides and applying linearity,

$\mathcal L \{ y'' \} + \mathcal L \{ y \} = \mathcal L\{1\} + \mathcal L \{ U(t-2) \}.$

Evaluating each term and denoting $Y \equiv Y(s) = \mathcal L\{y(t) \} \equiv \mathcal L\{y(t)\}$ ,

$\begin{aligned} \mathcal L\{1\} &= \frac 1s, \\ \mathcal L\{ U(t-2) \} &= \frac{ e^{-2s} }s, \\ \mathcal L\{ y'' \} &= s^2 Y(s) - sy(0) - y'(0) \\ &= s^2 Y(s) - s - 1. \end{aligned}$

Consolidating,

$\begin{aligned} (s^2 Y(s) - s - 1) + Y(s) &= \frac 1s + \frac{ e^{-2s} }s\\ (s^2 + 1) Y(s) &= s+1 + \frac 1s + \frac{ e^{-2s} }s \\Y(s) &= \frac s{s^2 + 1} + \frac 1{s^2 + 1} + \frac 1{s(s^2 + 1)} + e^{-2s} \cdot \frac 1{s( s^2 + 1)} \\ &= \frac {s^2}{s(s^2 + 1)} + \frac 1{s(s^2 + 1)} + \frac 1{s^2 + 1} + e^{-2s} \cdot \frac 1{s( s^2 + 1)} \\ &= \frac {s^2+1}{s(s^2 + 1)} + \frac 1{s^2 + 1} + e^{-2s} \cdot \frac 1{s( s^2 + 1)} \\ &= \frac {1}{s} + \frac 1{s^2 + 1} + e^{-2s} \cdot \frac 1{s( s^2 + 1)}. \end{aligned}$

Taking inverse Laplace transforms and applying linearity,

$\begin{aligned} y(t) = \mathcal L^{-1} \{Y(s) \} &= \mathcal L^{-1} \left\{ \frac 1{s} \right\} + \mathcal L^{-1} \left\{ \frac {1}{s^2 + 1^2} \right\} + \mathcal L^{-1} \left\{ e^{-2s} \cdot \frac 1{s(s^2 + 1)} \right\} \\ &= 1 + \sin(t) + \mathcal L^{-1} \left\{ e^{-2s} \cdot \frac 1{s(s^2 + 1)} \right\}.\end{aligned}$