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Tyler Clarke
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<p>
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 <a href="[^baseurl]/posts/differential-review-1.html">differential week 1 review</a>, as
it covers the concepts behind many of these questions in much greater detail.
</p>
<p>
Note: while these are <i>mostly</i> 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.
</p>
<p>
<b>Spoiler alert:</b> 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.
</p>
<h2>WS1.3.1: Classification Tasks</h2>
<p>
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.
</p>
<ol>
<li>
`(1 + y^2) frac {d^2y} {dt^2} + t frac {dy} {dt} + y = e^t`: The highest derivative of `y` is `frac {d^2y} {dy^2}`, meaning this is a <b>second-order</b> equation.
The definition of a second-order linear differential equation is the general form `p(t) y'' + q(t) y' + r(t)y = s(t)` - this situation has `p(t) = 1 + y^2`, which is
impossible (`p`, `q`, `r`, and `s` must always be functions of solely the independent variable). Hence, this is <b>nonlinear</b>.
</li>
<li>
`frac {dy} {dt} + t y^2 = 0`. This is clearly <b>first-order</b>, because the highest derivative of `y` is `frac {dy} {dt}`. It is just as clearly <b>nonlinear</b>,
because the `ty^2` term cannot be rewritten as a function of `t` times `y` - it would require the function to be something like `q(t) = ty`, which isn't valid.
</li>
<li>
`frac {d^3y} {dt^3} + t frac {dy} {dt} + cos^2(t)y = t^3`. The order of this one is also obvious - it's <b>third-order</b>. Surprisingly enough, it's linear! It can be rewritten as
`p(t) frac {d^3 y} {dt^3} + q(t) frac {d^2y} {dt^2} + r(t) frac {dy} {dt} + s(t) y = u(t)` - `p(t) = 1`, `q(t) = 0`, `r(t) = t`, `s(t) = cos^2(t)`, `u(t) = t^3`. This
is fully correct for a third-order linear ODE.<br><br>
Though it's not terribly important, note that the <i>degree</i> of this equation is 3, because 3 is the highest power of `t` (nonlinear differential equations do not have a degree).
You shouldn't have to worry about degree; make sure not to confuse it with order, though!
</li>
</ol>
<h2>WS1.3.4: Verifying A Solution</h2>
<p>
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`.
</p>
<p>
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)`.
</p>
<p>
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} }`.
</p>
<p>
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:
</p>
<ol>
<li>
`9 frac { 1 } { k - 2e^{3t} } = e^{-ln(frac {k - 2e^{3t}} 3)}`
</li>
<li>
`frac { 3 } { k - 2e^{3t} } = (frac {k - 2e^{3t}} 3)^{-1}`
</li>
<li>
`frac { 3 } { k - 2e^{3t} } = frac 3 {k - 2e^{3t}}`
</li>
</ol>
<p>
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.
</p>
<h2>WS2.1.4: An Initial Value Problem</h2>
<p>
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)`.
</p>
<h2>WS2.1.5: A Simple Solution Interval</h2>
<p>
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)`.
</p>
<h2>WS2.2.1: Solving a Linear ODE</h2>
<p>
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 <i>implicit solution</i> - we do one final step to get
`y = frac {sin(t)} {t^2} - frac {cos(t)} t + frac C {t^2}`, the <i>explicit solution</i>.
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.
</p>
<p>
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.
</p>
<h2>WS2.3.2: Electrical Problem</h2>
<p>
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`.
</p>
<p>
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!
</p>
<p>
(Aside: if you've read <a href="[^baseurl]/posts/inductance-hell.html">Inductance Hell</a>, you understand just how insane it is for <i>me</i> to say that
an LR problem is "fairly simple" - maybe I should have taken differential before physics 2).
</p>
<p>
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}`.
</p>
<p>
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})`.
</p>
<p>
<i>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.</i>
</p>
<p>
<b>BIG spoiler alert:</b> 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!<br>
<b>Further warning:</b> 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 <i>why</i> each answer is what it is and find the answers on your own.
</p>
<h2>SA1</h2>
<p>
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 <i>autonomous</i> or <i>non-autonomous</i>.
</p>
<ul>
<li>`y' = y^2 - y^3`. Because this equation can be written as `frac {dy} {dt} = f(y)`, it is autonomous - there is no `t` except in the differential.</li>
<li>`y' = e^{y + t}`. Try writing this without `t` anywhere but the differential. I'll wait. It's non-autonomous!</li>
<li>`y' + frac 1 t y = ln(t)`. Can <i>you</i> write this as `frac {dy} {dt} = f(y)`? Try it- you can't. Non-autonomous.</li>
<li>`t y' = 13t`. If you divide both sides by `t`, you get `frac {dy} {dt} = 13`, which is in the correct form for an autonomous equation. This is autonomous! <i>(Editors note: this one
isn't actually in the SA; I added it to demonstrate a situation where an autonomous equation can include `t`)
</i></li>
</ul>
<h2>SA13</h2>
<p>
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.
</p>
<h2>Final Notes</h2>
<p>
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.
</p>
<p>
If you've already gone through everything else, make sure you have these memorized (they probably won't be provided):
</p>
<ul>
<li>Newton's law of cooling</li>
<li>Basic circuit equations - Kirchhoff, Ohm, and inductors</li>
<li>IVP/U-Sub/etc</li>
<li>The general first-order linear ODE formula</li>
<li>The logarithm laws</li>
<li>Definitions of order/autonomousity/linearity</li>
</ul>
<p>
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.
</p>
[/]
[=title "Differential Quiz 1"]
[=author "Tyler Clarke"]
[=date "2025-5-21"]
[=subject "Calculus"]
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<p>
<i>Note: the next post in this series can be found <a href="[^baseurl]/posts/differential-quiz-1.html">here</a>.</i>
</p>
<p>
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.