diff --git a/site/posts/differential-exam-2.html b/site/posts/differential-exam-2.html index b30a252..b23106b 100644 --- a/site/posts/differential-exam-2.html +++ b/site/posts/differential-exam-2.html @@ -1,5 +1,8 @@ [!] [=post-] +
+ This post is part of a series. You can read the next post here. +
Our second and final midterm exam is in less than a week, and it's gonna be a big one! The light at the end of the tunnel, it do approacheth. After this, we've only one quiz and then the final! diff --git a/site/posts/differential-week-7.html b/site/posts/differential-week-7.html index 5f4b65c..4b1c72f 100644 --- a/site/posts/differential-week-7.html +++ b/site/posts/differential-week-7.html @@ -1,5 +1,8 @@ [!] [=post-] +
+ This post is part of a series. You can read the next post here. +
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. diff --git a/site/posts/differential-week-8.html b/site/posts/differential-week-8.html new file mode 100644 index 0000000..eeeed84 --- /dev/null +++ b/site/posts/differential-week-8.html @@ -0,0 +1,53 @@ +[!] +[=post-] +
+ Hello again! Our second midterm is getting close (you can read the review bit on it here), but we aren't talking about + that now: the sujet du jour is week 8, and we're getting deep + into Laplace transforms. +
++ Last week was pretty reduced compared to our normal lecture load; about 66% less material than we'd typically cover. Hence, this will be a short post. +
++ Often in physics we'll have a situation where a very quick impulse is applied to a system. This involves a piecewise function: at some time `t`, the function + will go to a high value, and then shortly after it will go low. The integral of that function is the total momentum produced, in physics terms. +
++ Impulses are so useful there's actually a function specifically for representing them: the Dirac delta function, represented as `delta (t)`. + This function is not explicitly defined: instead, we define it `delta(t) = 0, t != 0` with the constraint that `int_{-oo}^{oo} delta (t) dt = 1`. + There aren't any actual functions we can define that satisfy this: if you aren't convinced, take a minute to think about a function whose integral + is `1` despite only being nonzero over an infinitely small interval. `delta(0) = oo` doesn't satisfy! Integrating this would yield `oo`, not + `1`. +
++ This property makes the delta function incredibly useful for modelling things that happen instantly. However, it also makes it hard to think about! + Consider that we can actually take the Laplacian of `delta(t)`: `L(delta(t)) = 1`. This is actually fairly hard to prove; sufficiently so + that I'm just going to accept it as a fact (yikes!). +
++ Sometimes things don't happen at `t = 0` \[citation needed], so we also sometimes want to translate the delta function; that is, we want it to + do its thing at `t = t_0` rather than `t = 0`. We do this the normal way: `delta(t - t_0)`. The Laplace transform is just `L(delta(t - t_0)) = e^{-s t_0}`. +
++ Let's do an example, straight from the textbook. We're given `2y'' + y' + 2y = delta (t - 5), y(0) = 0, y'(0) = 0`: + to solve this, we take the Laplace transform of both sides. I won't bore you with the details; we end up with `L(y) (2s^2 + s + 2) + y(0) (-1 - 2s) - 2y'(0) = e^{-5st}`, + which substitutes to `L(y) (2s^2 + s + 2) = e^{-5st}` because both IVs are `0`. `L(y) = frac {e^{-5s}} {2s^2 + s + 2}`. Consider that we can use some + square-completion magic to rewrite this as `L(y) = frac {e^{-5s}} 2 frac 1 {(s + frac 1 4)^2 + frac 15 16}`. Why did we do this? Because + of the theorem `L(u_c(t) f(t - c)) = e^{-cs} F(s)`. In this case, `c` is clearly + `5`, so we end up with `u_5(t) L^{-1} (frac 1 {(s + frac 1 4)^2 + frac 15 16})`. +
++ Now we apply the second theorem: `L(e^{at} sin(bt)) = frac b {(s - a)^2 + b^2}`. `b = frac {sqrt(15)} 4`, and `a = frac 1 4`. We also have a constant multiplier + `frac 4 { sqrt(15) }`. Using both of these in concert + gives us our final result: `f(t) = u_5(t) frac 2 {sqrt(15)} e^{-frac (t - 5) 4} sin(frac {sqrt(15)} 4 (t - 5))`. Easy enough! +
++ This was pretty quick. More of a blurb than anything, really. See you next week and good luck on the exam! +
+[/] +[=title "Differential Equations Week 8"] +[=date "2025-7-7"] +[=author "Tyler Clarke"] +[=subject "Calculus"] \ No newline at end of file