+ Welcome once more to Deadly Boring Math! With the tempestuous wight of Quiz 2 rapidly striding towards us (it's tomorrow in the usual studio time), + I'm doing some last-minute studying, and figured I'd post some worked solutions here. These are by no means exhaustive; if you don't do your own studying, + you probably aren't going to pass. +
++ As WS1.2 is trivial and WS2.5 doesn't actually exist, I've skipped over them. The first question in 3.2 is a fairly simple classification task: + we're given a few systems of linear differential equations, and asked to convert them to matrix form and classify their homogeneity and autonomousity. +
++ Interestingly enough, when confronted with some second-order ODEs, we can significantly simplify by rewriting as a system of first-order + linear ODEs. This question asks us to transform the second-order ODE `u'' - 2u' + u = sin(t)` into an SLDE. +
++ The first step is to define some substitutions: `x = u`, `y = u'`, meaning `x' = u'`, and `y' = u''`. Note also that `x' = y`. + Substituting these values into the equation gives us `y' - 2y + x = sin(t)`: because we have the constraint `x' = y`, this is + a system of equations. We do some algebra to get `y' = 2y - x + sin(t), x' = y`, which can be written in matrix form as `X' = \[\[0, 1], \[-1, 2]] X + [0, sin(t)]`. +
++ This entire worksheet is about solving SLDEs from their matrix forms. We're given the equation `X' = \[\[1, 1], \[4, -2]] X`, + and we need to find a general solution. The first step is, of course, to find the eigenvalues and eigenvectors: we can use + the characteristic polynomial method: knowing that the trace is `-1` and the determinant is `-6`, we have `lambda^2 + lambda - 6 = 0`; the roots + are `2` and `-3`. We solve for each eigenvalue by solving the equation `\[\[1 - lambda, 1], \[4, -2 - lambda]] v = 0`: for `lambda = 2`, this is + `v = \[1, 1]`, and for `lambda = -3`, this is `v = \[-1, 4]`. These are linearly independent, so we can immediately construct a general solution: + `X = c_1 e^{-3t} \[-1, 4] + c_2 e^{2t} \[1, 1]`. +
++ Another instance of solving SLDEs from their matrix form! This time, it's `X' = \[\[4, 6], \[2, 5]] X`. We can find the eigenvalues + by finding the roots of the characteristic polynomial `lambda^2 - 9lambda + 8` - `1` and `8`. These have the corresponding eigenvectors + `\[-2, 1]` and `\[3, 2]`. These are linearly independent once again, so we can substitute as usual to get the general solution + `X = c_1 e^{t} \[-2, 1] + c_2 e^{8t} \[3, 2]` +
++ I'm not going to go into phase portraits here, but they will probably be on the test. They are, quite simply, a 2d extension of phase lines: + having solved a system of two linear differential equations, you can plot a bunch of different curves for sensible constant values, and + note the direction of the derivative along them at various intervals. I highly recommend studying these closely! +
++ The quiz is tomorrow in your normal studio room and time. Don't be late! I will, of course, be wearing a hat covered in balloons; + it would be pretty cool to see someone else adopt the trend as well. Good luck! +
+[/] +[=title "Differential Quiz 2"] +[=subject "Calculus"] +[=author "Tyler Clarke"] +[=date "2025-6-2"] \ No newline at end of file diff --git a/site/posts/differential-review-3.html b/site/posts/differential-review-3.html index 7d12b31..9d15734 100644 --- a/site/posts/differential-review-3.html +++ b/site/posts/differential-review-3.html @@ -1,5 +1,8 @@ [!] [=post-] ++ This post is part of a series; you can read the next post here. +
Hello once again, dear readers! It's been yet another big week in summer diffy q (although thankfully a little less big than the preceding - Memorial Day cut out a whole lecture). We've been primarily concerned with solving SLDEs (Systems of Linear Differential Equations), and we've a whole bunch of ways to do that.