Differential Equations Final Exam

By Tyler Clarke in Calculus on 2025-7-27

It's finally here! The final exam! The big one! This exam is going to be cumulative, sadly, and so we have a lot to review. I'm going to do the normal thing, going through one or two questions from each worksheet; I highly recommend reading my previous materials for quizzes and tests as well, since the questions will probably be similar.

As there are 36 individual worksheets and I do need to preserve my sanity, I'm largely going to skip the "less-interesting" ones.

Lastly, this being the grand finale, it's your last chance this semester to wear a silly hat to a calculus exam! Anyone who wears a balloon hat gets a handful of rubber ducks! We might not have a chance at being the best class Prof. Mamillapalle has ever taught, but we can definitely be the most surreal.

WS2.2.1: Some Simple Separatin'

Nothing before this point is really worth inclusion. This problem is simple enough: we're given `ty' + 2y = sin(t)`, and asked to solve for `y`. This is obviously a linear system: dividing the whole thing by `t` gives us `y' + frac 2 t y = frac {sin(t)} t`, so `p = frac 2 t` and `q = frac {sin(t)} t`. Remember how to solve this using the integrating factor method: let `mu = e^{int p(t) dt}`, and `frac {d} {dt} (mu y) = mu q`. Solving for `y`, then: `y = frac { int mu q dt } { mu }`. In this case, `mu = e^{int frac 2 t dt} = e^{2ln(t) + C} = Ct^2` and so `y = frac {int t sin(t) dt} {t^2}`. Integrating by parts yields the not-entirely-horrible `y = frac {sin(t) - t cos(t)} {t^2} + C`.

WS2.7.1

Substitutions! We're given `frac {dy} {dx} = frac {x + 3y} {3x + y}`, and asked to solve. The best way to do this is with a substitution. If we let `y = vx`, and thus `frac {dy} {dx} = v + frac {dv} {dx} x`, we can rewrite as `v + frac {dv} {dx} x = frac {x + 3vx} {3x + vx}`. Factoring out the `x`s in the right hand side and doing some rearranging yields `frac {3 + v} {(1 + v)(1 - v)} dv = frac 1 x dx`. Decomposing to partial fractions yields `frac 1 {1 + v} + frac 2 {1 - v} dv = frac 1 x dx`; integrating gives us `ln|1 + v| - 2 ln|1 - v| = ln|x| + C`. Finally raise `e` to both sides to get `frac {1 + v} {(1 - v)^2} = Cx`.

Resubstituting `v = frac y x` and doing some algebra, we end up with `frac {x + y} {(1 - y)(x - y)} = C`. This can't really be reduced to an explicit solution, but implicit is Good Enough ™!

WS2.7.2: Bernoulli

We're given `frac {dy} {dt} - frac y t = - frac {y^2} {t^2}`. This is clearly an `n=2` Bernoulli differential equation, so we use the substitution `u = y^{1 - n} = frac 1 y`. `frac {du} {dt} = - frac 1 {y^2} frac {dy} {dt}`. We rearrange the original equation by dividing `- y^2`, to get `- frac 1 {y^2} frac {dy} {dt} + frac 1 { t y } = frac 1 {t^2}`. Perhaps not entirely surprisingly, we substitute in `u` to get a much nicer equation: `frac {du} {dt} + frac 1 t u = frac 1 { t^2 }`. This is a first-order linear equation! We solve with the integrating factor method to get `u = frac {ln(t)} {t} + frac c t`, and resubstitute to get `y = frac t {c + ln(t)}`.

WS3.3.1: Matrix Methods

Given `x' = [[1, 1], [4, -2]] x`, solve for `x`. This one isn't too hard: we just use the old eigenvalue/eigenvector method! Our eigenvalues are `lambda_1 = -3, lambda_2 = 2`, so our eigenvectors are `v_1 = [1, -4], v_2 = [1, 1]`. Thus we just plug in: `x = C_1 e^{-3t} [1, -4] + C_2 e^{2t} [1, 1]`.

WS3.4.1: Makin' It Complex

In the case of `x' = [[1, 2], [-5, 1]]x`, we end up with complex eigenvalues `1 +- sqrt(10) i` and complex eigenvectors `[+- sqrt(10) i, -5]`. Plug into the general form: `e^t (C_1 cos(sqrt(10)t) [sqrt(10) i, -5] + C_2 sin(sqrt(10)t) [-sqrt(10) i, -5])`.

WS3.5.1: Constant Repetition!

`x' = [[3, -4], [1, -1]] x`. This is a repeated eigenvalue case with `lambda = 1, v = [2, 1]`. General form: `x = C_1 e^t [2, 1] + C_2 t e^t [2, 1]`. Note that this is not quite complete - we also need a generalized eigenvector, found by solving `[[3 - lambda, -4], [1, -1 - lambda]] x = [2, 1]`. One option is `w = [1, 0]`. We insert this into the second part of the solution: `x = C_1 e^t [2, 1] + C_2 e^t ([2t, t] + [1, 0])`

WS4.3.1: Second Order

Solve the second order linear differential equation `2y'' - 5y' - 3y = 0`. And scramble around for some acronym to make these names shorter. I can't remember if there was a "right" way to do this back at 4.3, but the associated polynomial method works nicely: `2r^2 - 5r - 3 = 0`. We get real and distinct roots `1, frac 3 2`, so th simplest general form works: `y = C_1 e^{t} + C_2 e^{frac 3 2 t}`.

WS4.5.1: Undetermined Coefficients

We're given `y'' - 2y' - 3y = 3e^{2t}` and asked to solve for `y`. This is clearly a UC problem. We start by finding the complementary solution: `y_c'' - 2y_c' - 3y_c = 0`, so via the associated polynomial method, `y_c = C_1 e^{-t} + C_2 e^{3t}`. Because the RHS is `3e^{2t}`, we can guess `y_p = Ae^{2t}`, which substitutes to give us `4Ae^{2t} - 4Ae^{2t} - 3Ae^{2t} = 3e^{2t}`. Obviously `A = -1`. Hence: our final solution is `y = y_c + y_p = C_1 e^{-t} + C_2 e^{3t} - e^{2t}`.

WS4.7.1.a: Variation of Parameters

We're given `x' = [[-1, -1], [0, 1]] x + [18, 3t]`. Recall that `x = A u` and `A u' = g` where `g` is the nonhomogeneous term and `A` is the fundamental solution matrix. We need to find the fundamental solution matrix. This is actually not very hard: we just use the eigenvalue method, getting the complementary solution `x = C_1 e^t [1, -2] + C_2 e^{-t} [1, 0]`, so `A = [[e^t, e^{-t}], [-2 e^t, 0]]`. To find `u'`, we solve with Gaussian elimination: the augmented matrix is `[[e^t, e^{-t} | 18], [-2e^t, 0 | 3t]]`, so `u' = [-frac 3 2 t e^{-t}, frac 3 2 t e^t + 18 e^t]`. Integrating (and assuming 0 constants) gives us `u = [frac 3 2 e^{-t} (t + 1), frac 3 2 e^t (t - 1) + 18 e^t]`. Finally: because `x = A u`, we matrix multiply to get `x = [3t + 18, -3t - 3]`. Nice!

WS6.1.1

I skipped all of Laplace. There will be a Laplace formula sheet on the exam.

This problem asks us to turn `t(t - 1)y'''' + e^t y'' + 4t^2 y = 0` into a first-order system. This actually isn't too hard. The trick is to pick a reasonable `x`: `x = [y, y', y'', y''', y'''']`. We also need to find `y''''`: we do this by rearranging, to get `y'''' = - frac {e^t} {t(t - 1)} y'' - frac {4t^2} {t(t - 1)} y`. Finally a matrix can be filled, giving us an equation in the form `x' = A x` (in this case, `x' = [y', y'', y''', y'''', y''''']`): `x' = [[0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1], [ - frac {4t^2} {t(t - 1)}, 0, - frac {e^t} {t(t - 1)}, 0]] x`. This is actually the whole thing!

WS6.3.1

We're given the matrix `X' = [[1, -2, 2], [-2, 1, -2], [2, -2, 1]] X` and asked to solve for the general solution. This being a first-order linear homogeneous system, we just use the eigenvalue/eigenvector method: we end up with `lambda_1 = 5, v_1 = [1, -1, 1]`, `lambda_2 = -1, v_2 = [-1, 0, 1]`, and `lambda_3 = -1, v_3 = [1, 1, 0]`. The solution, thus, is `X = C_1 e^{5t} [1, -1, 1] + C_2 e^{-t} [-1, 0, 1] + C_3 e^{-t} [1, 1, 0]`. This is also pretty easy.

WS7.1.1.a

Find the critical points of `x' = 1 + 5y, y' = 1 - 6x^2`. This is relatively easy: `5y = -1, 6x^2 = 1` trivially gives us `y = - frac 1 5, x = +- sqrt(frac 1 6)`. These are in fact the only two possible critical points.

WS7.2.1

Find the critical points of `x' = x - y^2, y' = x - 2y + x^2`. This gives us obviously `x = y^2, 2y = x(1 + x)`. There's obviously a critical point at `(0, 0)`. At `x = -1`, there's no real-valued solution (`y = i`); if we take `y = frac {x(1 + x)} 2` and substitute, we get `x = frac {x^2(1 + x)^2} 4`, or `x^3 + 2x^2 + x - 4 = 0`: this has a real solution at `x = 1`, for which `y = 1`: thus, we have two critical points, `(0, 0)` and `(1, 1)`. Let's also find the linear system at those points. Recall that the linear system is written in terms of the partial derivatives at the point: if `x' = F(x, y)` and `y' = G(x, y)`, at some point `a`, `x' = [[F_x(a), F_y(a)], [G_x(a), G_y(a)]] x`. The Jacobian matrix in this case is `[[1, -2y], [1 + 2x, -2]]`, and so for `(0, 0)` we have `x' = [[1, 0], [1, -2]]x` and at `(1, 1)` we have `x' = [[1, -2], [3, -2]]`.

WS8.3.1

Hooray. Improved Euler.

I'm only going to do the first step of this because, frankly, I really hate using Euler. As we've talked about before, the Improved Euler Formula gives us `y_{n + 1} = y_n + h frac { f(t_n, y_n) + f(t_{n + 1}, y_n + h f(t_n, y_n))} 2`. Our step size is `h = 0.05`, and `y' = f(t, y) = 2y - 3t, y(0) = 1`.

We already know that `y_1 = 1, t_1 = 0`: plugging those in gives us `f(t_1, y_1) = 2`, so we can substitute known values into the equation to get `y_{2} = 1 + 0.05 frac { 2 + f(t_2, 1.1)} 2`: `t_2 = 0.05`, so `y_2 = 1 + 0.05 frac { 4.05 } 2 = 1.10125`. Tedious, but not difficult.

Final Notes

Well, folks, this is the end! There will be no more Deadly Boring Math posts on differential equations. It's been quite a wild ride.

Stay tuned over break for more Math Gun and some fancy calculations regarding the speed of a robot my (non-GT-affiliated) team's been building.

If anyone really liked Deadly Boring Math, consider becoming an author! As I'm leaving physics for computer engineering, I won't be writing as much; shoot me an email at plupy44@gmail.com if you wanna help keep DBMUS alive.

Godspeed, and good luck!