revised Laplace and added LT tables

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Sasserisop 2023-10-30 22:20:43 -06:00
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@ -87,7 +87,7 @@ $\mathcal{L}\{t^n\}= \frac{n!}{s^{n+1}}$
Today covers all midterm material. Yay!
#end of lec 15 #start of lec 16
He advises us to learn the table of common LT's, however a sheet will be provided for the exam.
You can find the table of common LT's in the course textbook. There's a big table and a small table, the small table is the one he recommends we learn.
You can find the table of common LT's in the course textbook (I saved it and put it on the index page). There's a big table and a small table, the small table is the one he recommends we learn.
## Examples
#ex #LT
@ -116,7 +116,7 @@ $\mathcal{L}^{-1}\{\alpha F(s)+\beta G(s)\}=\alpha \mathcal{L}^{-1}\{F\}+\beta \
This can be proven rather easily due to the linearity of the forward transform (wasn't done in class unfortunately).
## Examples
#ex
#ex #inv_LT I
Compute this inverse LT:
$$\mathcal{L}^{-1}\left\{ \frac{1}{s^5}+\frac{3}{(2s+5)^2}+\frac{1}{s^2+4s+8}+ \frac{{s+1}}{s^2+2s+10} \right\}$$
Notice that all these terms approach 0 as s approaches inf.
@ -137,9 +137,11 @@ $$\mathcal{L}^{-1}\left\{ \frac{1}{s^5}+\frac{3}{(2s+5)^2}+\frac{1}{s^2+4s+8}+ \
Isn't that fun!
Well, every good thing must come to an end, we are done.
#ex #inv_LT
$\mathcal{L}^{-1}\{\frac{1}{(s-3)(s^2+2s+2)}\}$ notice the numerator is at least one degree lower than the denominator. The limit of the overall term is zero as $s\to \infty$
It doesn't look like anything in the table, can we factor the denominator? not really, they have complex solutions. So maybe split the terms using partial fractions!
#ex #inv_LT #partial_fractions
Find the inverse LT of:
$$\mathcal{L}^{-1}\{\frac{1}{(s-3)(s^2+2s+2)}\}$$
notice the numerator is at least one degree lower than the denominator. The limit of the overall term is zero as $s\to \infty$
It doesn't look like anything in the table, can we factor the denominator? not really, $s^2+2s+2$ has complex solutions. We also can't do any completing the square here. What we can do is split the terms using partial fractions!
$=\mathcal{L}^{-1}\left\{ \frac{A}{{s-3}}+\frac{{Bs+C}}{s^2+2s+2} \right\}$
$=\mathcal{L}^{-1}\left\{ \frac{{As^2+2As+2A+Bs^2-3Bs+Cs-3C}}{(s-3)(s^2+2s+2)} \right\}$
we get a linear system of equations:
@ -155,34 +157,61 @@ $=\frac{1}{17}\mathcal{L}^{-1}\left\{ \frac{1}{(s-3)} \right\}-\frac{1}{17}\math
Final answer:
$$\frac{1}{17}e^{3t}-\frac{1}{17}e^{-t}\cos(t)-\frac{4}{17}\sin(t)e^{-t}$$
partial fractions, hopefully you remember from math 101:
If you need a refresher on #partial_fractions , hopefully you remember from math 101:
Partial fractions:
For each term $(s+a)^k$
we include $\frac{A_{1}}{s+a}+\frac{A_{2}}{(s+a)^2}+\dots + \frac{A_{k}}{(s+a)^k}$
for each term $(s^2+as+b)^k$
we include: $\frac{B_{1}s+c_{1}}{(s^2+as+b)}+\frac{{B_{2}s+c_{2}}}{(s^2+as+b)^2}+\dots+\frac{{B_{k}s+c_{k}}}{(s^2+as+b)^k}$
#ex #inv_LT
$\mathcal{L}^{-1}\left\{ \frac{{3s^2+5s+3}}{s^4+s^3} \right\}$
factor out s^3 (s+1)s^3
Find the inverse LT of:
$$\mathcal{L}^{-1}\left\{ \frac{{3s^2+5s+3}}{s^4+s^3} \right\}$$
We try to manipulate it to match something in the LT table.
factor out $s^3$ :
$\mathcal{L}^{-1}\left\{ \frac{{3s^2+5s+3}}{s^3(s+1)} \right\}$
Partial fraction time!
$\mathcal{L}^{-1}\left\{ \frac{A}{s}+\frac{B}{s^2}+\frac{C}{s^3}+\frac{D}{s+1} \right\}$
$=\mathcal{L}^{-1}\left\{ \frac{{As^2(s+1)+Bs(s+1)+C(s+1)+Ds^3}}{(s+1)s^3} \right\}$
we get a linear system:
$\begin{matrix}A+D=0 \\A+B=3 \\B+C=5 \\C=3\end{matrix}$
solving the linear system yeilds:
solving the linear system yields:
$A=1,\ B=2,\ C=3,\ D=-1$
so:
$\mathcal{L}^{-1}\left\{ \frac{1}{s}+\frac{2}{s^2}+\frac{3}{s^3}+\frac{-1}{s+1} \right\}$
$$=1+2t+\frac{3}{2}t^2-e^{-t}$$
is the final answer.
#ex #inv_LT
$\mathcal{L}^{-1}\left\{ \ln \frac{{s^2+9}}{s^2+1} \right\}$
how do we get rid of the ln? work with the derivative.
property: $\mathcal{L}\{tf(t)\}=-\frac{dF}{ds}$
#ex #inv_LT
Find the inverse LT of:
$$\mathcal{L}^{-1}\left\{ \ln \frac{{s^2+9}}{s^2+1} \right\}$$
notice again, as $s\to \infty$, the inside approaches 0.
What can we match it against in the table? There are no $\ln$'s on the table!
How do we get rid of that $ln$? Work with the derivative.
Recall property 7: $\mathcal{L}\{t^nf(t)\}=(-1)^{n} \frac{d^nF}{ds^n}$
In this case we use $n=1$ :
$\mathcal{L}\{tf(t)\}=-\frac{dF}{ds}$
Take the inverse LT of both sides:
$\mathcal{L}^{-1}\left\{ \frac{d}{ds}\ln \frac{{s^2+9}}{s^2+1} \right\}=-tf(t)$
partial fractions:
$\mathcal{L}^{-1}\left\{ -\frac{16s}{(s^2+1)(s^2+9)} \right\}=2\mathcal{L}^{-1}\left\{ \frac{s}{{s^2+9}}-\frac{s}{{s^2+1}} \right\}=2(\cos 3t-\cos t)$
$\mathcal{L}^{-1}\left\{ \frac{s^2+1}{{s^2+9}} \cdot \frac{(s^2+1)2s-(s^2+9)2s}{(s^2+1)^2} \right\}=-tf(t)$
>Aha! This is looking a lot more familiar now. Solving it from here will be pretty similar to the previous examples.
$f(t)=-\frac{2}{t}(\cos 3t-\cos t)$
Simplifying gives:
$\mathcal{L}^{-1}\left\{ \frac{1}{{s^2+9}} \cdot \frac{(s^2+1)2s-(s^2+9)2s}{s^2+1} \right\}=-tf(t)$
>psst. you don't even need to use partial fractions for this question, it's already telling you that your coefficients are $2$ and $-2$ above^
$\mathcal{L}^{-1}\left\{ \frac{\cancel{ 2s^3 }+2s\cancel{ -2s^3 }-18s}{{(s^2+9)(s^2+1)}} \right\}=-tf(t)$
partial fractions:
$\mathcal{L}^{-1}\left\{ -\frac{16s}{(s^2+1)(s^2+9)} \right\}=\mathcal{L}^{-1}\left\{ \frac{A+Bs}{s^2+1}+\frac{{C+Ds}}{s^2+9} \right\}$
$=\mathcal{L}^{-1}\left\{ \frac{As^2+9A+Bs^3+9Bs+Cs^2+C+Ds^3+Ds}{(s^2+9)(s^2+1)} \right\}$
We get a system of linear equations:
$\begin{matrix}B+D&=0 \\A+C&=0 \\9B+D&=-16 \\9A+C&=0 \end{matrix}$
solving gives:
$\implies B=-2,\quad D=2,\quad A=0,\quad C=0$
$=2\mathcal{L}^{-1}\left\{ \frac{s}{{s^2+9}}-\frac{s}{{s^2+1}} \right\}=2(\cos 3t-\cos t)=-tf(t)$
divide both sides by $-t$ :
$$f(t)=-\frac{2}{t}(\cos 3t-\cos t)$$
We are done.
#end of lec 16

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@ -16,7 +16,7 @@ Good luck on midterms! <3 -Oct 18 2023
[Reduction of order (lec 11)](reduction-of-order-lec-11.html)
[Free vibrations (lec 11-12)](free-vibrations-lec-11-12.html)
[Resonance & AM (lec 13-14)](resonance-am-lec-13-14.html)
[Laplace transform (lec 14-16)](laplace-transform-lec-14-16.html) (raw notes, not reviewed or revised yet.)
[Laplace transform (lec 14-16)](laplace-transform-lec-14-16.html)
[Solving IVP's using Laplace transform (lec 17-18)](solving-ivps-using-laplace-transform-lec-17-18.html) (raw notes, not reviewed or revised yet.)
[(Heaviside) Unit step function (lec 18)](heaviside-unit-step-function-lec-18.html) (raw notes, not reviewed or revised yet.)
[Periodic functions (lec 19)](periodic-functions-lec-19.html) (raw notes, not reviewed or revised yet.)
@ -25,4 +25,6 @@ Good luck on midterms! <3 -Oct 18 2023
[Power series (lec 22)](power-series-lec-22.html) (raw notes, not reviewed or revised yet.)
</br>
[How to solve any DE, a flow chart](Solve-any-DE.png) (Last updated Oct 1st, needs revision. But it gives a nice overview.)
[Big LT table (.png)](drawings/bigLTtable.png)
[Small LT table (.png)](drawings/smallLTtable.png)
</br>

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