week 7

site/posts/differential-quiz-4.html

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 [=post-]
+<p>
+    <i>This post is part of a series. The next post can be found <a href="[^baseurl]/posts/differential-week-7.html">here</a>.</i>
+</p>
 <p>
     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.

site/posts/differential-week-7.html

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+[!]
+[=post-]
+<p>
+    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.
+</p>
+<h2>Briefly: Laplace Features</h2>
+<p>
+    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))`.
+</p>
+<p>
+    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.
+</p>
+<p>
+    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 <i>derivative</i> of the laplace transform of `f(t)`.
+</p>
+<p>
+    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.
+</p>
+<h2>Invertin'!</h2>
+<p>
+    Before we can actually solve, we need to cover one last thing: how do we get the function <i>out</i> 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!
+</p>
+<ul>
+    <li>`L(1) = frac 1 s, L^{-1}(frac 1 s) = 1`</li>
+</ul>
+<p>
+    We'll add to that as we go along. For a full table, consult page 313 in the textbook.
+</p>
+<p>
+    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!
+</p>
+<p>
+    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}`. 
+</p>
+<p>
+    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)`. 
+</p>
+<p>
+    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)`.
+</p>
+<p>
+    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!
+</p>
+<p>
+    These are the inversion substitutions we've covered so far:
+</p>
+<ul>
+    <li>`L^{-1}(frac 1 s) = 1`</li>
+    <li>`L^{-1}(frac s {a^2 + s^2}) = cos(at)`</li>
+    <li>`L^{-1}(frac a {a^2 + s^2}) = sin(at)`</li>
+    <li>`L^{-1}(frac {s - a} {(s - a)^2 + b^2)}) = e^{at} cos(bt)`</li>
+    <li>`L^{-1}(frac {n!} {(s - a)^{n+1}}) = t^n e^{at}`</li>
+</ul>
+<h2>Solving!</h2>
+<p>
+    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.
+</p>
+<p>
+    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`).
+</p>
+<p>
+    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!
+</p>
+<h2>Briefly: Step Functions and Time Shifting</h2>
+<p>
+    The <i>unit step function</i>, 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. 
+</p>
+<p>
+    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.
+</p>
+[/]
+[=author "Tyler Clarke"]
+[=title "Differential Equations Week 7"]
+[=date "2025-6-30"]
+[=subject "Calculus"]
+[#post.html]