From 1c074faed9a063334653a6ef27c0c0707af0b024 Mon Sep 17 00:00:00 2001 From: Tyler Clarke Date: Mon, 30 Jun 2025 19:50:55 -0400 Subject: [PATCH] week 7 --- site/posts/differential-quiz-4.html | 3 + site/posts/differential-week-7.html | 102 ++++++++++++++++++++++++++++ 2 files changed, 105 insertions(+) create mode 100644 site/posts/differential-week-7.html diff --git a/site/posts/differential-quiz-4.html b/site/posts/differential-quiz-4.html index db7876d..da784b1 100644 --- a/site/posts/differential-quiz-4.html +++ b/site/posts/differential-quiz-4.html @@ -1,5 +1,8 @@ [!] [=post-] +

+ This post is part of a series. The next post can be found here. +

Hello and welcome back to Deadly Boring Math! With quiz 4 on the horizon (it's tomorrow), I'm posting some of my worked solutions to worksheets and sample assessments as always. diff --git a/site/posts/differential-week-7.html b/site/posts/differential-week-7.html new file mode 100644 index 0000000..5f4b65c --- /dev/null +++ b/site/posts/differential-week-7.html @@ -0,0 +1,102 @@ +[!] +[=post-] +

+ Welcome back! It's been one hell of a week, in the best way possible. We are haz Laplace! Last week, we introduced Laplace transforms; + this week, we're actually using them to solve equations. +

+

Briefly: Laplace Features

+

+ The good ol' Laplace transform has a few features we need to cover. First is that `L(e^{ct} f(t)) = F(s - c)`, where `L(f(t)) = F(s)`. This is fairly easy to prove, + but I'm not going to. This feature is incredibly useful for solving some difficult problems: for instance, `L(e^{3t}cos(2t))` is just `F(s - 3)` where `F(s) = L(cos(2t))`. +

+

+ Also of note is that the Laplace transform of a derivative is trivial: if `F(s) = L(f(t))`, then `L(f'(t)) = s F(s) - f(0)`. Easy! This extends nicely to higher derivatives: + consider that, by this formula, `L(f''(t)) = s L(f'(t)) - f''(0)`, which expands nicely to `L(f''(t)) = s (s L(f(t)) - f(0)) - f'(0) = s^2 L(f(t)) - s f(0) - f'(0)`. + This process can be repeated for arbitrarily high derivatives. +

+

+ Finally, given `L(t^n f(t))`, the Laplace transform is `(-1)^n F^(n)(s)`: negative one to the nth power times the nth derivative of the laplace transform of `f(t)`. +

+

+ All of these except the last one are pretty easy to work out a proof yourself. If you're interested in the proofs, I highly recommend reading the textbook on all of them. +

+

Invertin'!

+

+ Before we can actually solve, we need to cover one last thing: how do we get the function out of Laplace-space? This is hard. In fact, it's sufficiently hard + that the easiest way is to use a lookup table of well-known Laplace transforms, pick the one that matches, and use it in reverse. For instance, + the inverse Laplace transform `L^{-1}(frac 1 s)` is easily found to be `= 1`. It's time to build a rolling table! +

+ +

+ We'll add to that as we go along. For a full table, consult page 313 in the textbook. +

+

+ Note that in many cases, there's some algebraic teasing you can do to the Laplace transform to make it invertible. For instance, the linear property of Laplace transforms + means that `L^{-1}(frac {a + s} {s^2 + a^2})` is equivalent to `L^{-1}(frac a {s^2 + a^2}) + L^{-1}(frac s {s^2 + a^2})`, which is incredibly useful: solving the compact + form might be impossible, but in the expanded form, the result is just `f(t) = cos(at) + sin(at)`. Easy-ish! +

+

+ Let's do a serious example. The textbook kindly provides us the problem `L^{-1}(frac 2 {(s + 2)^4} + frac 3 {s^2 + 16} + 5 frac {s + 1} {s^2 + 2s + 5})`. + This looks intimidating, but it's easier than it seems. Let's solve it piece by piece. First up is `L^{-1} (frac 2 {(s + 2)^4})`. The closest match is the + identity `L(t^n e^{at}) = frac {n!} {(s - a)^{n+1}}`, which would give us `n=3` and `a=-2`: `3! = 6`, so we'll have the constant coefficient + `frac 1 3`. Thus: `L^{-1} (frac 2 {(s + 2)^4}) = frac 1 3 t^3 e^{-2t}`. +

+

+ Next problem. `L^{-1}(frac 3 {s^2 + 16})` is best matched by the identity `L(sin(at)) = frac a {s^2 + a^2}`, where `a=4`; this gives us the constant coefficient + `frac 3 4`, to a final result `L^{-1}(frac 3 {s^2 + 16}) = frac 3 4 sin(4t)`. +

+

+ Last but not least: `L^{-1} (5 frac {s + 1} {s^2 + 2s + 5})`. This closest matches the identity `L(e^{at} cos(bt)) = frac {s - a} {(s - a)^2 + b^2)}`: + `a` is obviously `-1`, so we expand this out to get `frac {s + 1} {s^2 + 2s + 1 + b^2}`: for this to match, `b^2 = 4`, so `b = 2`. A hop, skip, and a substitution later, + and we have `L^{-1} (5 frac {s + 1} {s^2 + 2s + 5}) = 5 e^{-t} cos(2t)`. +

+

+ Adding all these together, we get `L^{-1}(frac 2 {(s + 2)^4} + frac 3 {s^2 + 16} + 5 frac {s + 1} {s^2 + 2s + 5}) = frac 1 3 t^3 e^{-2t} + frac 3 4 sin(4t) + 5 e^{-t} cos(2t)`. + Surprisingly easy! +

+

+ These are the inversion substitutions we've covered so far: +

+ +

Solving!

+

+ Solving is actually surprisingly easy. The process, for any differential equation, is to take the Laplace transform of both sides, solve for `L(y)`, and then invert + `L(y)` to get `y`. This makes it possible to represent difficult differential problems as simple algebra problems. +

+

+ Let's start with an example. The textbook kindly gives us `y' + 2y = sin(4t), y(0) = 1`. We'll start by taking the Laplace transform of both sides: we end up with something like + `s L(y) - y(0) + 2L(y) = frac 4 {s^2 + 16}`. Our goal is to isolate `L(y)`, which is pretty easy to do algebraically: we get `L(y) = frac {frac 4 {s^2 + 16} + y(0)} {s + 2}`. + Laplace is easier to do when it's broken up, so we expand to get `L(y) = frac 4 {(s^2 + 16)(s + 2)} + frac 1 {s + 2}` (note that I substituted `y(0) = 1`). +

+

+ This is going to require a partial fraction expansion. We can use `frac 4 {(s^2 + 16)(s + 2)} = frac A {s + 2} + frac {sB + C} {s^2 + 16}`, which solves out to `frac 4 {(s^2 + 16)(s + 2)} = frac 1 5 (frac 1 {s + 2} + frac {2 - s} {s^2 + 16})`. + Recombining and rearranging gives us `L(y) = frac 6 5 frac 1 {s + 2} - frac 1 5 frac {s} {s^2 + 16} + frac 1 10 frac {4} {s^2 + 16}`. This is actually pretty easy + to solve with the identities `L(e^{-2t}) = frac 1 {s + 2}`, `L(cos(4t)) = frac s {s^2 + 16}`, and `L(sin(4t)) = frac 4 {s^2 + 16}`. We get the result + `y = frac 6 5 e^{-2t} - frac 1 5 cos(4t) + frac 1 10 sin(4t)`. Nice! +

+

Briefly: Step Functions and Time Shifting

+

+ The unit step function, or Heaviside function, is a piecewise function defined as `h(t) = 0; t < 0` and `h(t) = 1; t >= 0`: if we're using electric circuits as an analogy, + the Heaviside function throws the switch at precisely time 0. There are a couple other step functions and variants: for instance, the indicator function, which is + `1` between `0` and `d`, and `0` otherwise. +

+

+ Many useful situations involve a shift of `c` units across the `t` axis. The typical way to do this is to multiply by a shifted heaviside function `h_c` (`0` before `c`, `1` after it), + and subtract `c` in the function parameter. That's a mouthful, but it's really quite simple: given `f(t)` and some shift `c`, the shifted version is `h_c(t) f(t - c)`. + The Laplace transform of this is actually very easy: `L(h_c(t) f(t - c)) = e^{-cs} F(s)`. I won't bother proving why, but it's interesting and I highly recommend reading the textbook + about it. +

+[/] +[=author "Tyler Clarke"] +[=title "Differential Equations Week 7"] +[=date "2025-6-30"] +[=subject "Calculus"] +[#post.html] \ No newline at end of file