Tutorial Week 8
From UBCMATH WIKI
Worksheet Questions
You can print the PDF. This worksheet is about Heaviside functions and solving IVPs with the Laplace transform.
- Write an expression for the function $f(t)$ shown below using Heaviside functions. In your final answer, all terms should be in the form $u_c(t) g(t-c)$ for some $g$, such that the Laplace transform is easy to compute.
- Use Laplace transform to solve the following initial-value problem
\begin{equation}
\left\{
\begin{array}{lr}
y'' + 4y' + 13y = 0 \\
y(0)=1\\
y'(0)=4
\end{array}
\right.
\end{equation}
Solutions
1. $f(t)=2(t-1)u_1(t)-2(t-2)u_2(t)-u_2(t)-(t-3)u_3(t)+(t-4)u_4(t)$ (each term in $f(t)$ 2 points , total 10 points)
- If the answer is close to the correct one: 1 point for slopes (+/- sign), 1 point for correct $c$.
- If it is far from correct, 2 points for using Heaviside in the correct way somewhere in their answer, 2 points for demonstrating sign (+/-) understanding.
- If they have the correct answer but not in the form $u_c(t)g(t-c)$, they will lose 2 points. one such answer can be in the form $f(t)=2(t-1)(u_1(t)-u_2(t))+(u_2(t)-u_3(t))+(u_3(t)-u_4(t))(4-t)$
$Y(s)= \dfrac{s+8}{s^2+4s+13}$ (2 points for correct numerator and 2 points for correct denominator) $= \dfrac{s+8}{(s+2)^2+3^2}$ (2 points for completing the square (recognizing and doing it correctly)) $= \dfrac{s+2}{(s+2)^2+3^2}+2\dfrac{3}{(s+2)^2+3^2}$ (1 point) $= e^{-2t} (cos(3t)+2 sin(3t))$ (each term 1 point).
2. (total 10 points):
