revised Laplace and added LT tables

content/Laplace transform (lec 14-16).md

@@ -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
+#ex #inv_LT #partial_fractions
-$\mathcal{L}^{-1}\{\frac{1}{(s-3)(s^2+2s+2)}\}$ notice the numerator is at least one degree lower than the denominator. The limit of the overall term is zero as $s\to \infty$
+Find the inverse LT of:
-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!
+$$\mathcal{L}^{-1}\{\frac{1}{(s-3)(s^2+2s+2)}\}$$
+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:
@@ -155,21 +157,25 @@ $=\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
-$\mathcal{L}^{-1}\left\{ \frac{{3s^2+5s+3}}{s^4+s^3} \right\}$
+Find the inverse LT of:
-factor out s^3 (s+1)s^3
+$$\mathcal{L}^{-1}\left\{ \frac{{3s^2+5s+3}}{s^4+s^3} \right\}$$
+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:
 $\mathcal{L}^{-1}\left\{ \frac{1}{s}+\frac{2}{s^2}+\frac{3}{s^3}+\frac{-1}{s+1} \right\}$
@@ -177,12 +183,35 @@ $$=1+2t+\frac{3}{2}t^2-e^{-t}$$
 is the final answer.
 
 #ex #inv_LT 
-$\mathcal{L}^{-1}\left\{ \ln \frac{{s^2+9}}{s^2+1} \right\}$
+Find the inverse LT of:
-how do we get rid of the ln? work with the derivative.
+$$\mathcal{L}^{-1}\left\{ \ln \frac{{s^2+9}}{s^2+1} \right\}$$
-property: $\mathcal{L}\{tf(t)\}=-\frac{dF}{ds}$
+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:
+$\mathcal{L}^{-1}\left\{ \frac{s^2+1}{{s^2+9}} \cdot \frac{(s^2+1)2s-(s^2+9)2s}{(s^2+1)^2} \right\}=-tf(t)$
-$\mathcal{L}^{-1}\left\{ -\frac{16s}{(s^2+1)(s^2+9)} \right\}=2\mathcal{L}^{-1}\left\{ \frac{s}{{s^2+9}}-\frac{s}{{s^2+1}} \right\}=2(\cos 3t-\cos t)$
+>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

content/_index.md

@@ -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>