deadlyboringmath.us/site/posts/differential-final-exam.html

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[=post-]
<p>
    It's finally here! The final exam! The big one! This exam is going to be cumulative, sadly, and so we have a <i>lot</i> 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.
</p>
<p>
    As there are 36 individual worksheets and I do need to preserve my sanity, I'm largely going to skip the "less-interesting" ones.
</p>
<p>
    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 <i>best</i> class Prof. Mamillapalle has ever taught, but we can <i>definitely</i> be the most surreal.
</p>
<h2>WS2.2.1: Some Simple Separatin'</h2>
<p>
    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`.
</p>
<h2>WS2.7.1</h2>
<p>
    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`.
</p>
<p>
    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 ™!
</p>
<h2>WS2.7.2: Bernoulli</h2>
<p>
    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)}`.
</p>
<h2>WS3.3.1: Matrix Methods</h2>
<p>
    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]`. 
</p>
<h2>WS3.4.1: Makin' It Complex</h2>
<p>
    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])`.
</p>
<h2>WS3.5.1: Constant Repetition!</h2>
<p>
    `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])`
</p>
<h2>WS4.3.1: Second Order</h2>
<p>
    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}`.
</p>
<h2>WS4.5.1: Undetermined Coefficients</h2>
<p>
    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}`.
</p>
<h2>WS4.7.1.a: Variation of Parameters</h2>
<p>
    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!
</p>
<h2>WS6.1.1</h2>
<p>
    I skipped all of Laplace. There will be a Laplace formula sheet on the exam.
</p>
<p>
    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!
</p>
<h2>WS6.3.1</h2>
<p>
    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.
</p>
<h2>WS7.1.1.a</h2>
<p>
    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.
</p>
<h2>WS7.2.1</h2>
<p>
    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]]`.
</p>
<h2>WS8.3.1</h2>
<p>
    Hooray. Improved Euler.
</p>
<p>
    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`.
</p>
<p>
    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.
</p>
<h2>Final Notes</h2>
<p>
    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.
</p>
<p>
    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.
</p>
<p>
    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 <a href="mailto:plupy44@gmail.com">plupy44@gmail.com</a> if you wanna help keep DBMUS alive.
</p>
<p>
    Godspeed, and good luck!
</p>
[/]
[=author "Tyler Clarke"]
[=date "2025-7-27"]
[=subject "Calculus"]
[=title "Differential Equations Final Exam"]
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