Question

Fun DE time!

Given\[f(x)=\begin{cases}
-1 & \text{ if } 0\leq t<1 \\
1 & \text{ if } 0\geq1
\end{cases}\]Find\[\mathcal{L}\{f(x)\}\]

Answer 1

Drats, I don't remember.

Answer 2

\[\mathcal{L}\{f(t)\}=\int_{0}^{1}e^{-st}(-1)dt+\int_{1}^{\infty}e^{-st}(1)dt.\]

Answer 3

Let u(t) = 1, x >= 0
0, x < 0

the unit step function.

Then your function f(x) = -2u(t) + 3u(t-1)
and the Laplace transform is just the linear combination of these

L{f(x)} = -2 L{u(t)} + 3 L{u(t-1)}
= -2/s + 3e^(-s)/s

Answer 4

The unit step function...

... I somehow got\[\mathcal{L}\{f(t)\}=\frac{2}{s}e^{-s}-\frac{1}{s}.\]I'll recheck.

Answer 5

You got the right answer.

Answer 6

oh, sorry, I wrote down f incorrectly

f = -u(t) + 2u(t-1)

So
L(f(t)) = -1/s + 2e^(-s)/s

the same thing

Answer 7

Hate it when if forget things like
\[-2 \neq -1\]