+ Hello again! With the spector of quiz 1 looming on the horizon (it's tomorrow, May 22nd, in the usual studio), I figured I'd post some last-minute + review materials. If you haven't already read it, I'd recommend checking out differential week 1 review, as + it covers the concepts behind many of these questions in much greater detail. +
++ Note: while these are mostly faithful to the originals, in some cases I've added my own modifications to the questions to make them more difficult or interesting. + All of my modifications are noted. +
++ Spoiler alert: This document includes full worked solutions for a number of worksheet problems. If you haven't already taken a crack at them on your own, you should! + The best way to use this review is to solve the problems yourself, and treat the explanations as guardrails and hints if you get stuck. +
++ The first question on worksheet 1.3 isn't terribly hard, but it deserves mention here anyways. We're asked to find the order and linearity of a list of ODEs - + we don't have to solve them, thankfully. +
++ I skipped over the previous two as they're fairly simple but tedious; this fourth one is a challenge. We're asked not to solve the differential equation + `frac {dy} {dt} = e^{3t + 2y}`, but to verify that it is in fact solved by `3e^{-2y} + 2e^{3t} = k`. +
++ My first instinct is to work backwards. If we can solve `3e^{-2y} + 2e^{3t} = k` for some reasonably-easily-differentiable `y(t)`, we can just substitute! + This is easier said than done. I won't include the algebraic steps here (work through 'em yourself if you aren't convinced!), but the final result is + something nasty like `y = - frac 1 2 ln(frac {k - 2e^{3t}} 3)`. +
++ Can we even differentiate this? It would seem luck is indeed on our side - we can use chain rule. Taking `g(t) = frac {k - 2e^{3t}} 3`, we've `- frac 1 2 ln(g(t))`, + which differentiates to `- frac 1 2 frac { g'(t) } { g(t) }` for `g(t) > 0`. Now we just need to find `g'(t)` - this is just a very simple exponential differentiation, + which yields `g'(t) = -2e^(3t)`. No explicit chain rule needed. Substituting back in gives us `y'(t) = 3 frac { e^(3t) } { k - 2e^{3t} }`. +
++ Substituting this into the differential equation gives us `3 frac { e^(3t) } { k - 2e^{3t} } = e^{3t} e^{-ln(frac {k - 2e^{3t}} 3)}`. This looks bad, but + we can simplify considerably: +
++ Nice! It's useful to note that the differential equation is variable-separable, so we could have found the solution by integrating `e^{-2y} dy = e^{3t} dt`, + which yields `-frac 1 2 e^{-2y} = frac 1 3 e^{3t} + C`, which does indeed simplify to `3 e^{-2y} + 2e^{3t} = C`. Which method you use depends largely + on the properties of the equation, and indeed your own intuition. +
++ Most of 2.1 is very simple variable-separable equations; this one is no exception, but I'm doing it anyways as it's a good example of an initial value problem. + We're asked to solve the equation `frac {dy} {dt} = e^{y + t}` where `y(0) = 0`. Solving this equation is easy: we can separate it to `e^{-y} dy = e^t dt`, + integrate to `-e^{-y} = e^t + C`, and solve to get `y = -ln(-e^t + C)`. To find `C`, we substitute: `0 = -ln(-e^0 + C)` becomes `0 = ln(C - 1)` becomes `e^0 = C - 1` + becomes `C = 2`. That means our solution is simply `y = -ln(2 - e^t)`. +
++ This is actually a continuation of the previous question, and it's quite interesting. `e^t` is defined for all `t`, so the only restrictive part is the `ln` function: + `ln(t)` is defined for all `t > 0` (because `e^t` can never be negative or zero), so this equation is defined when `2 - e^t > 0`. Solving gives us `e^t < 2`, or + `t < ln(2)`. +
++ This one is a pretty good example of a linear ODE. We're asked to solve `t frac {dy} {dx} + 2y = sin(t)`, for `t > 0`. This can be rewritten as + `frac {dy} {dx} + frac {2} {t} y = frac {sin(t)} {t}`, meaning it's a first-order linear equation with `p(t) = frac {2} {t}` and `q(t) = frac {sin(t)} {t}`. + The integrating factor is `mu = e^{int p(t) dt}`, which is in this case `mu = e^{2ln|t| + C} = Ct^2` (where did the absolute value sign go? Because `t > 0` is a constraint, we don't need it). + Substituting this into the standard equation `frac {d} {dx} (y mu) = q(t) mu` gives us `frac {d} {dx} (Cyt^2) = Ctsin(t)`. We can integrate to get `yt^2 = int tsin(t) dt` (canceled out C), + which is solvable with integration by parts to get `yt^2 = sin(t) - tcos(t) + C`. This is an implicit solution - we do one final step to get + `y = frac {sin(t)} {t^2} - frac {cos(t)} t + frac C {t^2}`, the explicit solution. + The rest of worksheet 2.2 is more of the same - you should definitely go through it, especially the IVP problems, if you haven't fully memorized the general solution to + a first-order linear ODE. +
++ Note: I haven't yet gone through the proof for the first-order linear ODE formula as it's lengthy and not entirely germane. If anyone's interested in a Deadly Boring + explanation of that, email me and I'll put up a quick post on it. +
++ I'm getting physics 2 flashbacks. We're asked to find the current in an LR circuit with respect to time: we have an Emf of 30 volts pushing through a 0.1-Henry inductor + and a 50-Ohm resistor. The initial current is 0, of course, which gives us an initial condition: `I(0) = 0`. The voltage produced by + an inductor is represented by the differential equation `V = L frac {dI} {dt}`, and is always opposing the voltage already running through the circuit. Ohm's law states that + `V = IR`, and we know by the Kirchhoff rule for voltage that `V_"inductor" + V_"resistor" = Emf`, meaning we can build the model `0.1 frac {dI} {dt} + 50I = 30`. +
++ The first question we should ask when solving a strange new differential equation is "can it be separated?" Unfortunately, the answer is "definitely not": this equation is + just 'bout as linear as they come. We can rephrase it as `frac {dI} {dt} + 500I = 300`, so `p(t) = 500` and `q(t) = 300`. This is actually fairly simple! +
++ (Aside: if you've read Inductance Hell, you understand just how insane it is for me to say that + an LR problem is "fairly simple" - maybe I should have taken differential before physics 2). +
++ We know the integrating factor `mu = e^{int p(t) dt} = e^{int 500 dt} = Ce^{500t}`, so we can substitute in to get `Ce^{500t}I = C int 300e^{500t} dt`. This + solves to `e^{500t}I = 1.5e+5 e^{500t} + C`, or more explicitly, `I = 1.5e+5 + Ce^{-500t}`. +
++ We still aren't done: this is an IVP! Above we determined the initial condition `I(0) = 0`, and we can substitute that in to get `0 = 1.5e+5 + C`, or `C = -1.5e+5`. + Substitute this in to get `I = 1.5e+5(1 - e^{-500t})`. +
++ Editor's note: I changed this problem to make it a little harder. The original one didn't involve actually solving the IVP, but I felt it would be interesting to do so. +
+
+ BIG spoiler alert: the next few problems are all pulled from the Sample Assessments. It's pretty critical that you do these on your own before reading
+ my explanations!
+ Further warning: for some inexplicable reason, the answers to all the SA questions are included inline. This means they are essentially spoilers for themselves.
+ Thus, it's quite important that you be able to explain why each answer is what it is and find the answers on your own.
+
+ The first question in the Sample Assessments document is extremely simple, but worth mentioning for completeness. We're given a set of equations, and asked + to determine if they're autonomous or non-autonomous. +
++ This one is a bit more interesting: we have to solve the differential equation `frac {y} {1 + y^2} frac {dy} {dt} = te^{t^2}`. + You'll note that this is not only variable-separable but actually already separated for us. Integrating it is a hop, skip, and u-substitution: + `frac 1 2 ln|1 + y^2| = frac 1 2 e^{t^2} + C`; this solves "nicely" to `y = sqrt(Ce^{e^{t^2}} - 1)`. Yuck. +
++ I've omitted quite a bit: none of the homework problems are on here, nor are most of the SA questions included on the quiz. There are also quite a few ignored worksheet problems: + I tried to pick one or two of the most hard/interesting problems from each worksheet, rather than doing them exhaustively. I'm fairly confident, however, that this review covers + everything integral to the quiz - at the very least, it should give you a good idea of what you need to study. +
++ If you've already gone through everything else, make sure you have these memorized (they probably won't be provided): +
++ I think that's it. Good luck on the quiz, and don't forget to wear a balloon hat! It's not mandatory, but it's fun. +
+[/] +[=title "Differential Quiz 1"] +[=author "Tyler Clarke"] +[=date "2025-5-21"] +[=subject "Calculus"] +[#post.html] \ No newline at end of file diff --git a/site/posts/differential-review-1.html b/site/posts/differential-review-1.html index 35246b9..1ed3c27 100644 --- a/site/posts/differential-review-1.html +++ b/site/posts/differential-review-1.html @@ -1,5 +1,8 @@ [!] [=post-] ++ Note: the next post in this series can be found here. +
Hello, everyone! Break is over (for me, at least... the lucky buggers at KSU get another two weeks), and the first week of summer classes is almost over. As you probably can tell by the title, one of the classes I'm taking is differential equations - which means, of course, the entire Internet gets to hear about it.