diff --git a/content/Laplace transform (lec 14-16).md b/content/Laplace transform (lec 14-16).md index 27a11bf..ffed72b 100644 --- a/content/Laplace transform (lec 14-16).md +++ b/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 @@ $$\mathcal{L}^{-1}\left\{ \frac{1}{s^5}+\frac{3}{(2s+5)^2}+\frac{1}{s^2+4s+8}+ \ Isn't that fun! Well, every good thing must come to an end, we are done. -#ex #inv_LT -$\mathcal{L}^{-1}\{\frac{1}{(s-3)(s^2+2s+2)}\}$ notice the numerator is at least one degree lower than the denominator. The limit of the overall term is zero as $s\to \infty$ -It doesn't look like anything in the table, can we factor the denominator? not really, they have complex solutions. So maybe split the terms using partial fractions! +#ex #inv_LT #partial_fractions +Find the inverse LT of: +$$\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 \ No newline at end of file diff --git a/content/_index.md b/content/_index.md index c214206..cf0b285 100644 --- a/content/_index.md +++ b/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.)
[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)
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