Differential Equations Week 6

By Tyler Clarke in Calculus on 2025-6-23

This post is part of a series. You can read the next post here.

Welcome back for yet another week! We're officially in the latter half of the semester now, and things are heating up. We have yet another quiz on Thursday, June 26th; the second midterm is in just two and a half weeks on Wednesday, July 9th. The semester ends with our very long final exam on July 31st, so the light at the end of the tunnel is getting close!

The quiz was entirely Cauchy-Euler equations. I do believe this is foreshadowing; either way, it's probably a war crime. If you had trouble with allat Cauchy-Euler, I highly recommend reading the Paul's Notes on the subject - he has a trick to solve most Cauchy-Euler problems in two or three simple steps.

Some More Oscillations

In week 5, we covered a bunch of oscillation material - specifically, damped ones. While we can easily solve the basic problems, some more complicated problems with a forcing term require a more serious consideration of the structure. The general equation for an oscillating system is `m y'' + b y' + k y = F(t)`, where `m` is mass, `b` is the damping coefficient, `k` is the spring constant (from Hooke's law!), and `F(t)` is some external force. This is also the general formulation for an SLDE. More useful because of how it simplifies some math we'll get to in a second is `y'' + 2 delta y' + omega_0^2 y = f(t)`, where `delta = frac b {2m}`, `omega_0^2 = frac k m`, and `f(t) = frac {F(t)} m`. Generally, we expect `f(t)` to be a linear combination of `A cos(omega t)` and `A sin(omega t)` where `omega` is frequency and `A` is amplitude. The most useful variant is `f(t) = Ae^{i omega t} = Acos(omega t) + i A sin(omega t)` because we can represent both trigonometric functions with a single, easily-mathed-upon exponential.

Generally speaking, the easiest way to solve vibration systems is with the method of undetermined coefficients. Finding the general solution is done as normal; for a general vibration system in the form `y'' + 2 delta y' + omega_0^2 y = Ae^{i omega t}`, the particular solution is found by plugging into the handy formula `y_p = frac {Ae^{i omega t}} {-omega^2 + 2 delta i omega + omega_0^2}`.

Let's do an example. The textbook gives us the pleasant equation `y'' + frac 1 8 y' + y = 3 cos(omega t)`. I won't bore you with the steps to find the general solution; it's `e^{- frac t {16}} (2 cos(frac {sqrt(255)} {16} t) + frac 2 {sqrt(255)} sin(frac {sqrt(255)} {16} t))`. To solve the particular solution, we need to first determine what `delta`, `omega_0`, and `A` are. In this case, we can just read off: `2 delta = frac 1 8`, `omega_0^2 = 1`, `A = 3` so `delta = frac 1 {16}` and `omega_0 = 1`. Plugging into the formula above gives us `frac {3e^{i omega t}} {-omega^2 + frac {i omega} 8 + 1}`.

Now we do an unpleasant little trick: because we know that `cos(omega t) = Re(e^{i omega t})`, we take the real part of the above formula to get `y_p`. To do that, we first need to get `i` out of the denominator: the complex conjugate of the denominator is `-omega^2 + 1 - frac {i omega} 8`, so multiplying by `frac {-omega^2 + 1 - frac {i omega} 8} {-omega^2 + 1 - frac {i omega} 8}` yields a formula with a completely real denominator: `frac {(-omega^2 + 1 - frac {i omega} 8) (3cos(omega t) + 3i sin(omega t))} {(1 - omega^2)^2 + frac {o^2} {64}}`.

Isolating the real part is pretty easy: we get `frac {-3 omega^2 cos(omega t) + 3cos(omega t) + frac {3 omega sin(omega t)} 8} {(1 - omega^2)^2 + frac {o^2} {64}}`. Hence, our final solution is `y = e^{- frac t {16}} (2 cos(frac {sqrt(255)} {16} t) + frac 2 {sqrt(255)} sin(frac {sqrt(255)} {16} t)) + frac {-3 omega^2 cos(omega t) + 3cos(omega t) + frac {3 omega sin(omega t)} 8} {(1 - omega^2)^2 + frac {o^2} {64}}`. Yikes.

Variation of Parameters

Note: I skipped over resonance as it doesn't seem that we're doing much with it in this course. If anyone wants to see a deeper exploration of oscillations, shoot me an email! (Or just read the textbook.)

Finally we're on variation of parameters! This is a very nice way to solve a wide variety of 2OLDEs, without relying on the messy algebra of undetermined coefficients. The drawback is that we have to do integration. Given an equation in the form `y'' + q(t) y' + r(t) y = g(t)`, we first find the solution to the homogeneous case `y'' + q(t) y' + r(t) y = 0` in the form `y_g = p(t) vec y_1 + q(t) vec y_2`, then find `W`, the determinant of the matrix `[vec y_1, vec y_2]`, then substitute into the formula `y_p(t) = y_2 int frac {y_1 g(t)} {W} dt - y_1 int frac {y_2 g(t)} { W } dt`. Finally, use superposition just like in undetermined coefficients to get `y = y_g + y_p`. The derivation of this is fascinating, but I'm not going to include it here; I highly recommend reading the Paul's notes.

Let's do an example. Given `y'' - 2y' + y = frac {e^t} {t^2 + 1}`, find a general solution. Our first step is going to be to find the `y_h` homogeneous solution to `y'' - 2y' + y = 0`: this can be found the Normal Way to be `y_h = c_1 e^t + c_2 t e^t` (note the repeated roots). There's a twist: to use variation of parameters, we need to have a vector solution - we need to know `Y = [y_h, y_h']`, not just `y_h`. Fortunately, this is very easy to find: `y_h' = c_1e^t + c_2 e^t + c_2 t e^t`. Thus, we have `Y = c_1 [e^t, e^t] + c_2 [t e^t, e^t + t e^t]`. Note that `y_1 = e^t, y_2 = t e^t`, not the full vectors - that actually tripped me up while writing this post.

To find `W`, we need to take the determinant of `[[e^t, t e^t], [e^t, e^t + t e^t]]`. This is pretty easy: it's `W = e^{2t}`. We know also that `g(t) = frac {e^t} {t^2 + 1}`, so we can substitute into our general form to get `y_p = t e^t int frac { e^{2t} } { e^2t (t^2 + 1) } dt - e^t int frac { t e^{2t} } { e^{2t}(t^2 + 1) } dt`. Integrating this isn't too hard, but it is somewhat tedious: I won't bore you with the details; the final result is `y_p = t e^t atan(t) - frac 1 2 e^t ln|t^2 + 1|`. (note that I ignored the constants of integration: this is because they would multiply out to be constant multiples of `y_1` and `y_2`, which would be redundant in our final solution).

Finally, we put the pieces together: `y = c_1 e^t + c_2 t e^t + t e^t atan(t) - frac 1 2 e^t ln|t^2 + 1|`. Nice! This required a lot less algebra than UC would, and is pretty elegant. I would argue that, if you can memorize the formula, Variation of Parameters is much quicker and easier than UC, but it depends on the person and the problem.

Enter Laplace

A new chapter! This one is gonna be fun. We're finally introducing Laplace transforms! The core idea of Laplace is the same as every other substitution or rewriting method in calculus: that we can solve a hard problem by hoisting the problem to a system where it's simpler, then solve the simple problem, then drop the solution back to our old system. It's really very elegant. Moving a difficult differential problem from `t`-space, being our normal representation, to `s`-space, being the Laplace transform's space, allows us to work out a solution as simple algebra. This eliminates many of the problems with solving complex differential equations.

Given a function `f(t)` that exists for every `t` where `0 <= t < oo`, the Laplace transform is `F(s) = lim_{A -> oo} int_0^{A} e^{-st} f(t) dt`. This is bulky, so there's a shorthand: `F(s) = L{f(t)} = int_0^{oo} e^{-st} f(t) dt` (note: the L is supposed to be curly, but I'm using ASCIIMath and there's no support for that symbol). There are a bunch of well-known Laplacian functions that are worthwhile to memorize, such as `L(1) = frac 1 s`; I'm not going to go through all of them, but I recommend consulting the textbook for this. It's useful to have them memorized so you don't have to calculate them on your own every time.

Note that `L(a + b) = L(a) + L(b)`. This is immediately obvious if you substitute into the expanded formula: `lim_{A -> oo} int_0^{A} e^{-st} (a + b) dt = lim_{A -> oo} int_0^{A} a e^{-st} + b e^{-st} dt = lim_{A -> oo} int_0^{A} a e^{-st} dt + int_0^A b e^{-st} dt = lim_{A -> oo} int_0^{A} a e^{-st} dt + lim_{B -> oo} int_0^B b e^{-st} dt` Note also that constants propagate out of the Laplace transform. These two facts mean that the Laplace transform is a linear operator.

Laplace transforms can also be applied to piecewise continuous functions, meaning they can be used to solve problems with essentially any nonsmooth function that is defined on `[0, oo)`. This is done by breaking up the integral. For instance, for a piecewise function defined on `[0, 4)` and `[4, oo)` separately, your Laplace transform will look like `F(s) = lim_{A -> oo} int_0^4 e^{-st} f(t) dt + int_4^A e^{-st} f(t) dt`.

Note that Laplace transforms are not guaranteed to exist. The function `f(t)` has to be convergent and defined on all `[0, oo)`, and has to be of exponential order - it has to be bounded by an exponential function, meaning `f(t) <= Ke^{at}` for some finite real `K` and `a`. Note that this doesn't have to be true over the entire set - as long as it eventually fits the bound, we're fine. I won't cover how to check or prove this here.

Final Notes

It's been an exciting week! We covered a lot of material that I was very, very excited for, and there's more around the corner (solving differential equations with Laplace transforms is gonna be fun). As previously mentioned, we have a quiz coming up; I'll post some review materials here tomorrow or Wednesday.

For unpleasant personal reasons, I wasn't able to post any review materials for quiz 3. My draft isn't complete enough to post: if anyone's interested in posthumous study resources, shoot me an email and I'll finish 'em up.

That's everything for now. See ya later, and good luck!