Differential Quiz 2
By Tyler Clarke in Calculus on 2025-6-2
This post is part of a series; you can view the next post here.
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.
WS3.2.1: Rewriting as Matrix Equations
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.
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`x' = -x + ty, y' = tx - y`. If we let `X = [x, y]`, and thus `X' = [x', y']`, we can write this as a matrix multiplication
`X' = [[-1, t], [t, -1]] X`. Because there isn't a constant vector, this is homogeneous; because the right side of the equation
has `t` in it, this is non-autonomous.
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`x' = 3x - z, y' = x + y - z, z' = z + t`. This one is a bit different in that it's 3-dimensional, but the process is still essentially the same.
Let `X = [x, y, z]` and `X' = [x', y', z']`, and this can be written as `X' = [[3, 0, -1], [1, 1, -1], [0, 0, 1]] X + [0, 0, t]`.
This is non-homogeneous, because the constant term is nonzero; it is non-autonomous because `t` appears on the right side.
WS3.2.2: Second-Order ODEs to SLDEs
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)]`.
WS3.3.1
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]`.
SA23
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]`
Phase Portraits
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!
Final Notes
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!
