Differential Quiz 4

By Tyler Clarke in Calculus on 2025-6-25

Hello and welcome back to Deadly Boring Math! With quiz 4 on the horizon (it's tomorrow), I'm posting some of my worked solutions to worksheets and sample assessments as always.

WS4.4.1: Applications

This is a fun problem, but maybe I'm biased. We're given an inductor-resistor-capacitor (LRC) series circuit, and asked to find the charge on the capacitor at a time `t = 0.01` seconds, given that `L = 0.05`, `R = 2`, `C = 0.01`, `E(t) = 0`, `q(0) = 5`, and `i(0) = 0`.

We need to set up a differential equation. Because we want to find charge, `q` will be our dependent variable. We know by Kirchhoff's law that the voltage drops produced by the resistor, the inductor, and the capacitor will equal `E(t) = 0`. What are those voltage drops? For a resistor, we use Ohm's law `V = IR`; recall that the current over a capacitor is `frac {dq} {dt}`, so we have `V_"resistor" = R frac {dq} {dt}`. For the inductor, there's another simple formula: `V = L frac {dI} {dt}`. Because `I = frac {dq} {dt}`, this is `V = L frac {d^2q} {dt^2}`. Finally, the voltage drop over a capacitor is `frac q C`. Thus, we have the second-order linear homogeneous differential equation `L frac {d^2 q} {dt^2} + R frac {dq} {dt} + frac q C = E(t)`, or `0.05 frac {d^2 q} {dt^2} + 2 frac {dq} {dt} + 100q = 0`.

Step one is to find the roots of the related quadratic `0.05r^2 + 2r + 100 = 0`. I won't bore you with the details; this process is not difficult, and the result is `-20 +- 40i`. This is obviously a complex case, so we use the complex-case solution: `q = e^{-20t}(c_1 cos(40t) + c_2 sin(40t))`. If you graph this, it's a pretty spiral. We have an IV `q(0) = 5, i(0) = q'(0) = 0`, so we need to take the derivative. I hate this part. The result is `q' = -20 c_1 e^{-20t} cos(40t) - 40 c_1 e^{-20t} sin(40t) - 20 c_2 e^{-20t} sin(40t) + 40 c_2 e^{-20t} cos(40t)`, or more suitably `q' = c_1 e^{-20t} (-20 cos(40t) - 40 sin(40t)) + c_2 e^{-20t} (- 20 sin(40t) + 40 cos(40t))`. Substituting our IVP gives us `5 = c_1` and `0 = -20c_1 + 40c_2`. It's very easy to get `c_2 = frac 5 2`.

Substituting what we've found gives us `q(t) = e^{-20t}(5 cos(40t) + frac 5 2 sin(40t))`. `q(0.01) = ~4.57`. Pretty easy, all things considered!

WS4.4.3: Maximization

We're given a problem in much the same vein, but for two differences: `E(t)` is nonzero, and we have to find the maximum charge. Specifically, our constants are `L = frac 5 3, R = 10, C = frac 1 {30}, E(t) = 300`, giving us the equation `frac 5 3 q'' + 10 q' + 30q = 300`. Our IV is `q(0) = 0, i(0) = 0`. How can we find the maximum value of this? I'm actually not sure how we're supposed to do this with just machinery from 4.4, but if we play fast and loose with the rules and just use the Method of Undetermined Coefficients, we'll be fine. Let `q(t) = A`, and we get `30A = 300` or `q_p(t) = A = 10`. The homogeneous part of the solution is not too hard to find to be `q_h(t) = e^{-3t} (c_1 cos(3t) + c_2 sin(3t))`. Thus we have `q(t) = e^{-3t} (c_1 cos(3t) + c_2 sin(3t)) + 10`. I won't bore you with finding the constants; our final result is `q(t) = -10e^{-3t} (cos(3t) + sin(3t)) + 10`. Nice!

Finding the maximum value of this is a problem of solving for `q'(t) = 0`. `q' = 60 e^{-3t} sin(3t)`, so a maxima or minima occurs every time `sin(3t) = 0`. `t = frac {pi} 3 n` for any integer `n` will do it. Because this decays exponentially, it's going to be `n = 0` or `n = 1` - at `t = 0`, we already know `q = 0`, so `t = frac {pi} 3` wins with `q = ~10.43`. Nice!

WS4.5.3: Hooray! Undetermined coefficients!

Regular Deadly Boring Math readers will have no trouble with this UC problem. We're given `y'' + 9y = t^2 e^{3t} + 6`, and asked to simply solve - no IVP. Let's find the complementary solution first. The homogeneous part of this problem is clearly `y'' + 0y' + 9y = 0`, so we have the associated quadratic `r^2 + 9 = 0` - our roots are `+- 3i`. This gives us the fairly nice `y_c = c_1 cos(3t) + c_2 sin(3t)`. Now for the particular solution! Because we're dealing with the form `t^2 e^{3t} + 6`, let's try the form `y = A t^2 e^{3t} + Bt e^{3t} + C e^{3t} + D`: the first part (A through C) is just the normal degree-2 polynomial, multiplied by `e^{3t}`. The trick here is our assumption that this will propagate, so every unwanted term will have a matching term that cancels it out.

Wolfram-Alpha kindly tells us that `y' = e^{3t} (At(3t + 2) + 3Bt + B + 3C)` (what, you didn't think I was going to calculate that by hand, did you?) and the second derivative `y'' = e^{3t}(A(9t^2+12t+2) + B(9t+6) + 9C)`. Terrifying. Substituting into our equation yields `9Ae^{3t}t^2 + 12Ae^{3t}t + 2Ae^{3t} + 9Be^{3t} + 6Be^{3t}+9Ce^{3t} + 9At^2 e^{3t} + 9B t e^{3t} + 9 C e^{3t} + 9D = t^2 e^{3t} + 6`. This is not the worst equation I've written this week.

We want to group for constants and for `t^2 e^{3t}`. Everything else will have to be 0! Doing this yields `18A t^2 e^{3t} + (12A + 9B)t e^{3t} + (2A + 15B + 18C)e^{3t} + 9D = t^2 e^{3t} + 6`. Because the only constant term is `9D`, we know `9D = 6`, or `D = frac 2 3`. We also know that `18A = 1`, `12A + 9B = 0`, and `2A + 15B + 18C = 0`. `A = frac 1 {18}` allows us to easily find `B = - frac 2 {27}` and `C = frac 1 {18}`. This means our particular solution is `y_p = frac 1 {18} t^2 e^{3t} - frac 2 {27} t e^{3t} + frac 1 18 e^{3t} + frac 2 3`. Finally, we substitute to get the general solution: `y = y_c + y_p = c_1 cos(3t) + c_2 sin(3t) + frac 1 {18} t^2 e^{3t} - frac 2 {27} t e^{3t} + frac 1 18 e^{3t} + frac 2 3`.

Awful.
(Editor's Note: my final solution is slightly different from the "official" one. Officially, `B, C = - frac 1 {27}, frac 1 {162}`. I really don't feel like backtracking to figure out what's wrong.)

WS4.6.1: Vibratin'!

We're given our first serious nonhomogeneous vibration problem. An 8lb mass (see: `frac 1 4` slug, a unit that continues to be moronic) stretches a spring by 2ft. A downwards force acts on it, described by `4sin(4t)` lb. It's initially pulled down 6 inches and released. We need to find the position `y` (positive = down) at any time `t`. For 8 pounds of force to stretch the spring by 2ft, the spring constant `k` must be `4`. Thus we have the nice-ish equation `y'' = -16y + 16sin(4t)`, or `y'' + 16y = 16sin(4t)`. Solving the homogeneous part of this is easy, and gives us `y_h = c_1 cos(4t) + c_2 sin(4t)`. Undetermined coefficients with `y_p = A t cos(4t) + B t sin(4t)` (note that without the extra `t` terms, this would be redundant) gives us `y_p = -2t cos(4t)`. Our general solution: `y = c_1 cos(4t) + c_2 sin(4t) - 2t cos(4t)`.

We aren't done. We know the spring starts pulled down by 6 inches, so `y(0) = 0.5`, and it starts at rest, so `y'(0) = 0`. We solve this to get `c_1, c_2 = frac 1 2, frac 1 2`, so we have `y = frac 1 2 cos(4t) + frac 1 2 sin(4t) - 2t cos(4t)`. Yippee!

We're looking for the first four times where `y' = 0`. `y' = -2sin(4t) + 8 t sin(4t)` (after simplifying quite a bit, of course), so for it to be 0 we have either `sin(4t) = 0` or `t = frac 1 4`. `sin(4t) = 0` when `4t` is an integer multiple of `pi`, or when `t` is an integer multiple of `frac pi 4`: hence, our first four zeroes are going to be `t = 0, frac 1 4, frac pi 4, frac pi 2`.

Final Notes

That's everything. It was short this time! Good luck, and see you again soon.