Question

Posting along as I solve a problem, using the LaTeX eqn editor; solving an IVP with heaviside and laplace. Don't need help yet.

Answer 1

k just tag me wen u need help

Answer 2

Use the Laplace Transform to solve the given Initial Value Problem:


...right now I'm just having trouble figuring out how to make a large, one-sided curly bracket to describe a piecewise-defined function. Anybody know how to do this? @wio @Kainui

Answer 3

Can you give an example?

Answer 4

One sec, can do.

Answer 5

The bracket on the lefthand side, right here: http://puu.sh/d0yCA/860b16f591.png

Answer 6

looks like it is recursive.

Answer 7

If so, I think you'd want to make it explicit first.

Answer 8

I don't formally know LaTeX, so I have no idea what that means in this context; From picking around StackExchange, I found someone posting that \begin{cases} is the way to do it, but I have no clue how to do that within the OpenStudy editor, which I'm not sure (but I'm guessing) is different from a "real" LaTeX editor.

Answer 9

\[
\begin{cases}
c(a_i) =a_{i+1}& 1\leq i < l \\
c(a_l)=a_1
\end{cases}
\]
```
\begin{cases}
c(a_i) =a_{i+1}& 1\leq i < l \\
c(a_l)=a_1
\end{cases}
```

Answer 10

OH, okay, now I know why it wasn't working, I didn't put the \end{cases}.

Answer 11

Actually, no, I don't know, I still don't get something generated in the equation editor when I do that.

Answer 12

One second, lemme screencap what I'm doing.

Answer 13

```
\[
\begin{cases}
c(a_i) =a_{i+1}& 1\leq i < l \\
c(a_l)=a_1
\end{cases}
\]
```

Answer 14

Oh, okay, is that an OpenStudy thing or what?

Answer 15

Inline formula \(x=2\).
Multiline formula: \[
y = mx+b
\]
```
Inline formula \(x=2\).
Multiline formula: \[
y = mx+b
\]
```

Answer 16

This is a MathJax thing, which OpenStudy uses.

Answer 17

I'm using the equation editor and awkwardly writing stuff in LaTeX *inside* the editor, and then noticing additional stuff that I *did not type* being generated in the body of the post prior to posting, it's usually those external brackets and stuff-yeah, is that what is going on?

Answer 18

Maybe. I dunno. I just type it out manually and rely on the preview.

Answer 19

Alright, well, either way:

Use the Laplace Transform to solve the given Initial Value Problem:

\[y'+y=f(t),\]where f(t) is
\[
\begin{cases}
0, \ 0 \le t <1 \\ 5, \ t \ge 1
\end{cases}
\]

Answer 20

First off, putting the right-hand function in the terms of Unit Step Functions:

Answer 21

\[f(t)=(0)-(0)\mathcal{U}(t-1)+5\mathcal{U}(t-1)\] (Using the general formula for converting a piecewise-defined function to linear combination Heaviside)

Answer 22

\[f(t)=5\mathcal{U}(t-1)\]

Answer 23

\[\mathcal{L}(y')+\mathcal{L}(y)=\mathcal{L}(5\mathcal{U}(t-1))\]\[\mathcal{L}(y')+\mathcal{L}(y)=5\mathcal{L}(\mathcal{U}(t-1))\]
\[[sY(s)-y(0)]+Y(s)=\frac{5e^{-s}}{s^2}\]

Answer 24

\[Y(s)=\frac{5e^{-s}}{s^3}\]
(Forgot to mention an initial condition, y'(0)=0).

Answer 25

\[\mathcal{L^{-1}}(Y(s))=\mathcal{L^{-1}}\left\{\vphantom{}\frac{5e^{-s}}{s^3}\right\}\]

Answer 26

Making the identification that\[a=-1, \ \ \ f(t)=\mathcal{L^{-1}}\left\{\vphantom{}\frac{1}{s^3}\right\},\]We know that our answer will be in the form\[f(t-1)\mathcal{U}(t-1).\]

Answer 27

\[\mathcal{L^{-1}}\left\{\vphantom{}\frac{1}{s^3}\right\}=\frac{1}{2!}\mathcal{L^{-1}}\left\{\vphantom{}\frac{2!}{s^3}\right\}=\frac{1}{2!}t^2=\frac{t^2}{2}\]

Answer 28

\[y(t)=\frac{5}{2}(t-1)^2\mathcal{U}(t-1)\]