From e7568ef4482645d3ad20b27f7c2d9537f32a2c16 Mon Sep 17 00:00:00 2001 From: Tyler Clarke Date: Fri, 4 Jul 2025 14:22:03 -0400 Subject: [PATCH] exam 2 review --- site/posts/differential-exam-2.html | 273 ++++++++++++++++++++++++++++ 1 file changed, 273 insertions(+) create mode 100644 site/posts/differential-exam-2.html diff --git a/site/posts/differential-exam-2.html b/site/posts/differential-exam-2.html new file mode 100644 index 0000000..b30a252 --- /dev/null +++ b/site/posts/differential-exam-2.html @@ -0,0 +1,273 @@ +[!] +[=post-] +

+ 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! +

+

+ The exam covers everything from 4.1 to 5.5 inclusive - a total of 12 sections, each with a corresponding worksheet. I'm going to go through a few problems + from each worksheet, and also some of the sample assessments: per Monday's lecture, the material on the exam will be pretty similar. +

+

WS4.1.1: Classification

+

+ This one is a pretty simple classification task. We're given six differential equations, and asked to determine if they're linear and if they're homogeneous. + Remember that "linear" just means it's in the form `a(t)y'' + b(t)y' + c(t)y = d(t)`, extending into arbitrarily high derivatives of `y`, and "homogeneous" + simply means that `d(t) = 0`. +

+
    +
  1. + `y'' - 2y' - 2y = 0`: The form checks out, so it's linear; `d(t) = 0`, so it's homogeneous. +
    2. +
  2. + `2y'' + 2y' + 3y = 0`: Again, this is in the right form and `d(t) = 0`, so it is linear and homogeneous. +
    3. +
  3. + `y''' - y'' = 6`: This is linear in the form `a(t)y''' + b(t)y'' + c(t)y' + d(t)y = e(t)` (`c(t) = d(t) = 0`), but + `e(t) != 0`, so it's linear nonhomogeneous. +
    4. +
  4. + `y'' - 2y' + 2y = e^t tan(t)`: Linear, nonhomogeneous. +
    5. +
  5. + `y'y'' = 4x`: This one actually isn't linear! Multiplying `y^(n)` terms together makes it nonlinear by default. +
    6. +
  6. + `2y'' = 3y^2`: You might think this is linear homogeneous, because it can be rewritten as `2y'' - 3y^2 = 0`, but + in fact the `y^2` term makes this nonlinear. +
    7. +
+

WS4.2.1: Intervals of Existence

+

+ There are a couple problems under 4.2.1, and they're all worth doing, but I'm only going to write down the first one here. + We're given `t(t - 4)y'' + 3ty' + 4y = 2` where `y(3) = 0, y'(3) = -1`, and asked to find the interval on which the IVP + is guaranteed to have a solution. +

+

+ To find the interval, we need an equation in the form `y'' + a(t)y' + b(t)y = c(t)`: to get this, we divide by `t(t - 4)`, to get + `y'' + frac 3 {t - 4} y' + frac 4 {t(t - 4)} y = frac 2 {t(t - 4)}`. It's pretty obvious that `a(t) = frac 3 {t - 4}` and `b(t) = frac 4 {t(t - 4)}`. + The interval of existence is going to be the intersection of the intervals of existence of `a(t)` and `b(t)`. It's pretty obvious that `a(t)` exists + where `t - 4 != 0`: `(-oo, 4) cup (4, oo)`. `b(t)` exists where `t(t - 4) != 0`, so `(-oo, 0) cup (0, 4) cup (4, oo)`. The intersection of these is just + `(-oo, 0) cup (0, 4) cup (4, oo)`. Because the initial `t` value is `t = 3`, we're in the interval `(0, 4)`. +

+

WS4.2.2: A Quick Verification

+

+ We're so far back in time that we haven't yet seen undetermined coefficients, which is kinda weird. This problem is an undetermined coefficients problem. + We're given `y = c_1 e^x + c_2 e^{-x}`, `y'' - y = 0`, `y(0) = 0`, and `y'(0) = 1`, and asked to find `c_1` and `c_2`. +

+

+ The first thing to do is take the derivatives. This is pretty easy: `y' = c_1 e^x - c_2 e^{-x}`, and `y'' = c_1 e^x + c_2 e^{-x}`. Now we substitute! + `y'' - y = 0` works out to `c_1 e^x + c_2 e^{-x} - c_1 e^x - c_2 e^{-x} = 0`, which is obviously true for any `c_1` and `c_2`. `y(0) = 0` works out to + `c_1 + c_2 = 0`, so we know `c_1 = -c_2`. And finally, `y'(0) = 1` works out to `c_1 - c_2 = 1`, which tells us that `c_1 = frac 1 2, c_2 = - frac 1 2`. + Substituting this back gives us `y = frac 1 2 e^x - frac 1 2 e^{-x}`. Easy! +

+

WS4.2.3: Linear Independence (Brief)

+

+ This is a really easy one. We're asked if `y_1 = x, y_2 = x^2, y_3 = 4x - 3x^2` is a linearly independent set of solutions. They clearly aren't; + `y_3 = 4y_1 - 3y_2`. +

+

WS4.3.3: Some Solving

+

+ This is no big trick to anyone who's been paying attention. We're given `4y'' + 4y' + 17y = 0` where `y(0) = -1` and `y'(0) = 2`, and asked to solve the IVP. +

+

+ Remember that we can represent this as `4r^2 + 4r + 17 = 0`, find the roots of that, and then substitute them into one of a few nice general forms. + The roots are `-frac 1 2 +- 2i`, so we're using the imaginary one (derived from Euler's identity): `y = e^{- frac 1 2 t}(c_1 cos(2t) + c_2 sin(2t))`. + Simple enough. +

+

+ Now we just need to solve for `c_1` and `c_2`. Unfortunately, this means taking the derivative: `y'' = e^{- frac 1 2 t}(c_1 (-2 sin(2t) - frac 1 2 cos(2t)) + c_2 (2 cos(2t) - frac 1 2 sin(2t)))`. + Substituting the IVP gives us `-1 = c_1` and `2 = - frac 1 2 c_1 + 2 c_2`. Solve to get `frac 3 4 = c_2`. Thus, our solution is + `y = e^{- frac 1 2 t}(- cos(2t) + frac 3 4 sin(2t))`. +

+

WS4.3.5: A Tall One

+

+ We're asked to solve `y'''' + 2y'' + y = 0`. As it turns out, the polynomial trick actually carries over nicely here: we have the auxiliary polynomial + `r^4 + 2r^2 + 1 = 0`. Finding the roots for this is actually pretty easy: we factor down to `(r^2 + 1)^2 = 0`, so `r = +- i`. Note that this is actually + a repeated roots case! Because the highest power is 4 rather than 2, there are actually 4 solutions: `i, i, -i, -i`. Weird, but true. +

+

+ Applying the normal rule for imaginary along with the normal rule for repeated, we get `c_1 cos(t) + c_2 sin(t) + c_3 t cos(t) + c_4 t sin(t)`. +

+

WS4.3.6: Eulering

+

+ Last one from WS4.3, I promise. We're given `t^2 y'' - 2ty' - 4y = 0`. This is obviously a Cauchy-Euler equation! My favorite way to solve these + is the simplest one: use the formula from Paul's Notes. + We find the roots by solving `r(r - 1) - 2r - 4 = 0`: this simplifies to `r^2 - 3r - 4 = 0`, so `r = -1, 4`. These being real and distinct, + we can plug them into the nice form `y = c_1 x^{lambda_1} + c_2 x^{lambda_2}`: `y = c_1 x^{-1} + c_2 x^4`. +

+

WS4.4.

+

TODO: something from worksheet 4.4 (I hate this worksheet)

+

If you're reading this, I didn't do worksheet 4.4

+

WS4.5.1: Undetermined Coefficients

+

+ All of 4.5 are fairly easy UC problems. I'm just going to do the first one: `y'' - 2y' - 3y = 3e^{2t}` where `y(0) = 0, y'(0) = 1`. + Because the right side is in the form `Ae^{2t}`, we'll use that as our guess: `y = Ae^{2t}`, `y' = 2Ae^{2t}`, `y'' = 4Ae^{2t}`. Inserting + these into the equation gives us `4Ae^{2t} - 4Ae^{2t} - 3Ae^{2t} = 3e^{2t}`. This pretty obviously simplifies down to `A = -1`. + Hence, our particular solution is `y_p = -e^{2t}`. We still need a general solution; this is the solution to the homogeneous part, + `y'' - 2y' - 3y = 0`. That's pretty easy: it's `y_g = c_1 e^{-t} + c_2 e^{3t}`. Our solution is the sum: `y = c_1 e^{-t} + c_2 e^{3t} - e^{2t}`. +

+

+ Solving the IVP requires us to find `y'`. This is, fortunately, easy: `y' = -c_1 e^{-t} + 3 c_2 e^{3t} - 2e^{2t}`. Substituting the IVP + gives us the system `0 = c_1 + c_2 - 1, 1 = -c_1 + 3 c_2 - 2`. `c_1 = frac 1 4`, `c_2 = frac 3 4`. Hence, we have the solution + `y = frac 1 4 e^{-t} + frac 3 4 e^{3t} - e^{2t}`. +

+

+ Note: the worksheet answer is actually different here. I'm not sure what the disconnect is; both solutions substitute correctly, so they appear to be equally valid. +

+

WS4.6.2

+

+ We're given a spring-mass system with spring constant `3 frac N m`, (hooray for metric units), and a mass of `2 kg`. It's suspended in a viscous fluid that resists + motion by `-v` (where `v` is velocity of the mass). The system is being driven downwards by a force `12 cos(3t) - 8 sin(3t)`. We need to find the steady-state response - + the solution that doesn't converge or diverge. +

+

+ Setting this up as a second-order linear differential equation is pretty simple. Operating in terms of `p`, the offset from rest which increases down, we have + `p'' = 6 cos(3t) - 4 sin(3t) - frac 1 2 p' - frac 3 2 p` (note that the force is divided by mass: `p''` is acceleration, not force). This rewrites nicely as + `p'' + frac 1 2 p' + frac 3 2 p = 6 cos(3t) - 4 sin(3t)`. We need to first find the particular solution: this is usually going to be at least part of the steady-state + response. The general solution might be part of it: if the general solution has some `e^{At}` term where `A != 0`, it will converge or diverge, and so we + ignore it. +

+

+ Because the right side is in the form `A cos(3t) + B sin(3t)`, we'll use that for undetermined coefficients. `p_p = A cos(3t) + B sin(3t)`, `p_p' = -3A sin(3t) + 3 B cos(3t)`, + `p_p'' = -9 A cos(3t) - 9 B sin(3t)`. Substituting these into the equation gives us `-9 A cos(3t) - 9 B sin(3t) + -frac 3 2 A sin(3t) + frac 3 2 B cos(3t) + 9A cos(3t) + 9B sin(3t)`. + Yikes. Fortunately, this simplifies pretty far: `-3/2 A sin(3t) + 3/2 B cos(3t) = 6 cos(3t) - 4 sin(3t)`. + It's easy to find `A` and `B` from this; `A = frac 8 3, B = 4`. Hence: `p_p = frac 8 3 cos(3t) + 4 sin(3t)`. +

+

+ We don't have to go past the real part of the solution to the homogeneous equation: it's `-frac 1 4`, meaning the general solution will be a multiple of + `e^{-frac 1 4}`, meaning it will vanish at infinity. Our steady state response is just the particular solution `p = frac 8 3 cos(3t) + 4 sin(3t)`. +

+

WS4.7.1.a: Variation of Parameters: The Force Awakens

+

+ Variation of parameters! Variation of parameters! Variation of parameters! Multiple cheers for variation of parameters! This problem is a pretty classic variation + of parameters case: we're given `x' = \[\[-1, -1], \[0, 1]] x + \[18, 3t]`, and we need to solve for `x`. +

+

+ First, let's find the general solution to the homogeneous case `x' = \[\[-1, -1], \[0, 1]]`. This is done pretty easily with the eigenvector/eigenvalue method: + `x_g = c_1 \[e^{-t}, 0] + c_2 \[-e^{t}, 2e^{t}]`. This means the fundamental set of solutions is `\[\[e^{-t}, -e^t], \[0, 2e^t]]`. The Wronskian of that fundamental set + is important: calculating it is, fortunately, not very hard; `W = 2`. +

+

+ The particular solution for a situation like this is the fundamental matrix times the integral of the inverse of the fundamental matrix times `f`, where `f` is the nonhomogeneous term (`\[18, 3t]` in our case). + Finding the inverse matrix to use the form `x = A int A^{-1} f dt` is pretty hard, but (linear) algebra saves the day: if we let `u = A^{-1}x`, and then if we take the derivative of both sides, we get + `u' = A^{-1} f`, and multiply the left side of both terms by `A`, we get `A u' = f`. This is easy to solve with Gaussian elimination! Once we have `u`, we can get + `x` with the formula `x = A u`. +

+

+ Let's do it. The augmented matrix we're solving works out to `\[\[e^{-t}, -e^t | 18], \[0, 2e^t | 3t]]`, so our result is `u' = \[18e^t + frac 3 2 t e^t, frac 3 2 t e^{-t}]`. + Integrating yields `u = \[frac 33 2 e^t + frac 3 2 t e^t, - frac 3 2 e^{-t} - frac 3 2 t e^{-t}]` (assuming `C = 0`, which works). To get the final result, we need to multiply this by the + fundamental matrix: `x_p = \[18 + 3t, -3 - 3t]`. +

+

+ And we're done! Adding them all together yields `x = x_g + x_p = c_1 \[e^{-t}, 0] + c_2 \[-e^{t}, 2e^{t}] + \[18 + 3t, -3 - 3t]`. Easy! Ish! +

+

WS4.7.2.b: Variation of Parameters: Generalizin', always generalizin'

+

+ Hooray for even more variation of parameters! This technique is one hell of a drug. This time, we're given a second-order linear differential equation + `y'' + 2y' + y = 3e^{-t}`. We could solve this by converting it into a system of two first-order ODEs and using the method above, but there's + a quicker and easier substitution: given fundamental solutions `y_1` and `y_2`, and the Wronskian of the fundamental solution set `W`, and the nonhomogeneous + term `g`, the particular solution is `y_p = y_2 int frac {y_1 g} W dt - y_1 int frac {y_2 g} W dt`. +

+

+ We find the fundamental solution set pretty easily: the normal tricks give us `y_g = c_1 e^{-t} + c_2 t e^{-t}`, so our solutions to `\[y_g, y_g']` are + `y_1 = \[e^{-t}, -e^{-t}]` and `y_2 = \[t e^{-t}, - t e^{-t} - e^{-t}]`. The Wronskian of `\[y_1, y_2]` is `- e^{-2t}`. `g(t) = 3 e^{-t}`. + Substituting into the general form gives us `y_p = t e^{-t} int frac {3e^{-2t}} {e^{-2t}} dt - e^{-t} int frac {3 t e^{-2t}} {e^{-2t}} dt`. + Simplify to get `y_p = t e^{-t} int 3 dt - e^{-t} int 3 t dt`, and integrate to `y_p = 3 t^2 e^{-t} - frac 3 2 t^2 e^{-t} = frac 3 2 t^2 e^{-t}`. +

+

+ Adding it all together, we get `y = c_1 e^{-t} + c_2 t e^{-t} + frac 3 2 t^2 e^{-t}`. Pretty straightforward. +

+

WS5.1.1: A Little Laplace (Briefly)

+

+ Ahhh, Laplace, my favorite transform... In this problem, we're given the piecewise function + `f(t)` defined with `f(t) = 0, 0 <= t <= 1; f(t) = 1, 1 < t <= 2; f(t) = 0, t > 2`, and asked to find the Laplace transform. This is pretty easy: + for a piecewise function, the Laplace transform breaks up into several simple integrals, like so: `L(f(t)) = int_0^1 e^{-st} * 0 dt + int_1^2 e^{-st} * 1 dt + int_2^{oo} e^{-st} * 0 dt`. + The first and third terms both evaluate to 0, so the only interesting part is the middle: `L(f(t)) = int_1^2 e^{-st} dt`. This is very easy to evaluate: + we end up with `1/s (e^{-s} - e^{-2s})`. +

+

WS5.2.1: A Little More Laplace (Briefly)

+

+ Even more Laplace! This time, `f(t) = e^{-2t} sin(3t)`. This is pretty easy if you remember that, fundamentally, `L(e^{at} f(t)) = F(s - c)`. The identity + `L(sin(at)) = frac a {a^2 + s^2}` works nicely for the `sin` part, and we combine these ideas to get `F(s) = frac a {a^2 + (s + 2)^2}` +

+

WS5.2.4: Transforming Equations with Laplace

+

+ A pretty important trick in our arsenal, now that we've got the Laplace transform, is using it to turn differential equations into algebra problems. + This question asks us to do exactly that: given a differential equation, we use the Laplace transform to turn it into an algebraic problem and solve for `L(y)` + - we don't have to do the tedious final step, yet, which would be inverting it to get `y`. +

+

+ The equation is `9y'' + 12y' + 4y = 0, y(0) = 2, y'(0) = -1`. Taking the Laplace transform is as easy as applying it individually to each term: + `9 L(y'') + 12 L(y') + 4 L(y) = L(0)`. `L(f')` for any `f` is always `s L(f) - f(0)`, so we have `L(y') = s L(y) - y(0)` and + `L(y'') = s L(y') - y'(0)`. We can expand the `L(y')` term to get `L(y'') = s (s L(y) - y(0)) - y'(0) = s^2 L(y) - s y(0) - y'(0)`. +

+

+ Substituting these knowns into the equation gives us `9 s^2 L(y) - 9 s y(0) - 9 y'(0) + 12 s L(y) - 12 y(0) + 4 L(y) = 0`. + We want something in the form `L(y) = f(s)`, so we group a bit: `L(y)(9 s^2 + 12 s + 4) = 9 s y(0) + 9 y'(0) + 12 y(0)`. Substituting the IVs gives us + `L(y)(9 s^2 + 12 s + 4) = 18 s + 15`, and rearranging yields `L(y) = frac {18s + 15} {9 s^2 + 12 s + 4}`. +

+

WS5.3.1: Everybody was Laplace Invertin', doo doo doo doo doo...

+

+ As it turns out, it's somewhere around the 200th line of mathjax-augmented HTML that my naming conventions start getting really stupid. +

+

+ We're given a pretty simple problem: we need to find the inverse Laplace transform of `frac 2 {s^2 + 3s - 4}`. Why do I call this simple? Because + finding inverse Laplace transforms organically is really hard - sufficiently hard, in fact, that the best way to do it is just to pattern match against + known transforms, which is really pretty easy. In this case, we need to first factorize the denominator, then apply partial fraction decomposition; + we'll end up with a linear combination of a few well-known forms that can be inverted easily. +

+

+ Factorizing the denominator is simple: `s^2 + 3s - 4 = (s + 4)(s - 1)`. This is an easy case of PFD: our substitution will be `frac A {s + 4} + frac B {s - 1}`. + Doing a bit of algebra gives us `frac 2 {(s + 4)(s - 1)} = frac {As - A + Bs + 4B} {(s + 4)(s - 1)}`, which groups and simplifies to + `(A + B)s + (4B - A) = 2`. Hence, `A + B = 0` and `4B - A = 2`, so `A = - frac 2 5, B = frac 2 5`. +

+

+ Now we're doing the inverse Laplace of a much nicer function: `L(f) = - frac 2 5 frac 1 {s + 4} + frac 2 5 frac 1 {s - 1}`. This is nice because + `L(e^{at}) = frac 1 {s - a}`: if we apply that in reverse, we can pretty easily find `f = - frac 2 5 e^{-4t} + frac 2 5 e^{t}`. +

+

WS5.4.1: Scooby Dooby Laplace!

+

+ Because we're (finally!) solving a problem. Get it? +

+

+ We're given `y'' - 4y' - 12y = 0, y(0) = 8, y'(0) = 0`. Using the formulae from WS5.2.4, we take the Laplace transform of both sides: + `s^2 L(y) - s y(0) - y'(0) - 4s L(y) + 4 y(0) - 12L(y) = 0`. Group for `L(y)` and substitute the IVs to get `L(y)(s^2 - 4s - 12) = 8s - 32`, and flip + to `L(y) = frac {8s - 32} {s^2 - 4s - 12}`. +

+

+ Now we need to solve. Step one is, as always, factorize and simplify: `L(y) = 8 frac {s - 4} {(s + 2)(s - 6)}`. Next, we need to PFD: given the PFD substitution + `8 frac {s - 4} {(s + 2)(s - 6)} = frac A {s + 2} + frac B {s - 6}`, we get `8s - 32 = As - 6A + Bs + 2B`. Grouping once more: `8s - 32 = (A + B)s + (2B - 6A)`. Thus, + `A + B = 8` and `2B - 6A = -32`. These solve handily to `A = 6` and `B = 2` (yes, I did use Gaussian elimination for that). This turns our + problem into `L(y) = frac 6 {s + 2} + frac 2 {s - 6}`. +

+

+ We already know that `L(e^{at}) = frac 1 {s - a}`. Both of these terms are in the perfect form! A quick legerdemain, and we have + `y = 6e^{-2t} + 2e^{6t}`. Yay! +

+

WS5.5.6: Periodic!

+

+ WS5.5 mostly rehashes stuff we've already done, but the problems are harder. I highly recommend going through them. This problem in particular is the only one + I considered surprising enough to merit inclusion. We're given a periodic function `f(t)`, defined as `f(t) = 1, 0 <= t < 1; f(t) = -1, 1 <= t < 2` with period + 2. This simply means that it repeats: `f(t + 2) = f(t)` for any given `t`. +

+

+ The best way to deal with the Laplace transform of this is with the window function `f_T`, which is defined to be equal to `f` over a single period `0 <= t <= 2`, + and `0` otherwise. The Laplace transform of our periodic function `f(t)` can then be found with the formula `L(f(t)) = frac L(f_T(t)) {1 - e^{-2s}}`. The Laplace + transform of the window function will work out to simply `int_0^2 e^{-st} f(t) dt`. That works out to `int_0^1 e^{-st} dt - int_1^2 e^{-st} dt`. Evaluating + this gives us `L(f(t)) = frac 1 s frac {-2e^{-s} + 1 + e^{-2s}} {1 - e^{-2s}}`. Unpleasant, but not too bad! +

+

Final notes

+

+ This ran long. It ran really long. I didn't even include any of the promised supplemental-assessment problems and it still ran really long. + The midterm is on June 9th during the normal lecture time. Don't be late! I'll be wearing the long-missing balloon hat (marching around in 90f weather wearing + that menace got unreasonable, fast), and if this helped you, you should wear one too! It would be pretty funny. +

+

+ Sayonara, and good luck! +

+[/] +[=title "Differential Equations Exam 2 Review"] +[=author "Tyler Clarke"] +[=date "2025-7-4"] +[=subject "Calculus"] +[#post.html] \ No newline at end of file