admin/deadlyboringmath.us
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity

week 4 and 5
All checks were successful
Build / Build-Docker-Image (push) Successful in 44s
Details

Browse Source
This commit is contained in:
Tyler Clarke
2025-06-15 12:09:04 -04:00
parent 2802ec90a5
commit 0fae262081
2 changed files with 174 additions and 0 deletions
Download Patch File Download Diff File Expand all files Collapse all files

3
site/posts/differential-exam-1.html
View File

@@ -1,5 +1,8 @@
[!]
[=post-]
<p>
<i>This post is part of a series. You can read the next post <a href="[^baseurl]/posts/differential-review-4-and-5.html">here</a>.</i>
</p>
<p>
Hello once more, Internet! Our little Diffy Q adventuring party has passed the first two random encounters, and Rivendell is close in sight - but a ferocious
band of orcs lies in the way; our first midterm rapidly approaches. It's going to be on June 9th in the normal lecture room and normal time; don't be late!

171
site/posts/differential-review-4-and-5.html Normal file
View File

@@ -0,0 +1,171 @@
[!]
[=post-]
<p>
Hello, dear readers! This post is a bit strange, even by Deadly Boring standards; because of an exam on June 9th I was unable to write the week 4 review,
so this post covers all of week 4 and week 5 together.
</p>
<p>
<b>Editor's note:</b> It has come to my attention that the math in this post is unreadably painful, even by DBM standards. Blame the course.
</p>
<h2>Shifted Systems</h2>
<p>
The first topic we need to cover is <i>shifted systems</i>. When given a nonhomogeneous SLDE, you have a trivial solution for situations where the
`c` term is constant. The trick lies in a substitution: you find the equilibrium solution `q` of `X' = AX + C`, define `Y = X - q`, and solve
`Y' = AY`. The solution to this can then be substituted back to get `X`. This actually isn't too hard; let's do an example. Given the equation
`X' = \[\[1, 2], \[3, 4]] X + \[5, 6]`, we clearly can't solve this the normal way; to start on a solution, we need to find an equilibrium point.
This is as simple as solving the equation `\[\[1, 2], \[3, 4]]Q + \[5, 6] = \[0, 0]`. This can be algebraically rewritten as `\[\[1, 2], \[3, 4]]Q = \[-5, -6]`,
which can be solved with Gaussian elimination to get `Q = \[4, - frac 9 2]`.
</p>
<p>
This means we have `Y = X - Q = \[x, y] - \[4, - frac 9 2] = \[x - 4, y + frac 9 2]`. We set up an equation `Y' = \[\[1, 2], \[3, 4]]Y`. We know that
`Y' = \[x', y']` (don't believe me? calculate it yourself!), so this is homogeneous and can be solved the normal way. If this confuses you,
note that this is equivalent to saying `X' = A (X - Q) = AX - AQ`, and that `AQ = \[-5, -6] = -C` as we solved above: this equation is, by the magic
of matrices, exactly equivalent to our equation in terms of `X`.
</p>
<p>
I won't bore you with the details of finding eigenvalues and eigenvectors. They're `lambda_1 = frac {5 + sqrt(33)} 2, v_1 = \[-3 + sqrt(33), 6], lambda_2 = frac {5 - sqrt(33)} 2,
v_2 = \[-3 - sqrt(33), 6]`. I probably should have picked a nicer matrix from the start. These awful eigenvalues and eigenvectors give us the nice solution
`Y = c_1 e^{t frac {5 + sqrt(33)} 2} \[-3 + sqrt(33), 6] + c_2 e^{ frac {5 - sqrt(33)} 2 } \[-3 - sqrt(33), 6]`. Finally, we can backsubstitute `Y = X - Q`, to get
`X = c_1 e^{t frac {5 + sqrt(33)} 2} \[-3 + sqrt(33), 6] + c_2 e^{ frac {5 - sqrt(33)} 2 } \[-3 - sqrt(33), 6] + \[4, - frac 9 2]`. Not too hard!
</p>
<h2>2ODEs</h2>
<p>
Second order differential equations! We've talked a bit about these already, but only in the context of transforming them into simpler first order differential equations.
If you don't remember the definition from way back in chapter 1, it's pretty simple: a second order differential equation is a differential equation in terms of the
second derivative of a function. You can also have the first derivative as a term, which complicates things immensely. There is a form for linear equations
in a second-order ODE: `y'' + P(t)y' + Q(t)y = R(t)`.
</p>
<p>
Solving a second-order linear differential equation is actually pretty easy. All you have to do, in fact, is turn it into an SLDE! The substitution process
we've covered several times works like a charm. Once you have your SLDE, you can solve <i>that</i> the normal way, and resubstitute. We go now to your
regularly scheduled example. Given the IVP `u'' = 4u' - 3u, u(0) = 2, u'(0) = 5`, we can rewrite with `vec X = \[u, u']` as `X' = \[\[0, 1], \[4, -3]] X`.
The eigenvalues and eigenvectors of this are `lambda_1 = -4, v_1 = \[-1, 4]` and `lambda_2 = 1, v_2 = \[1, 1]`. These are conveniently real and different,
so we have an easy solution: `X = c_1 e^{-4t} \[-1, 4] + c_2 e^{t} \[1, 1]`.
</p>
<p>
How can we use this to solve the second-order equation? That part is actually pretty easy.
Remember that we're looking for `u` - and `X` is conveniently defined as `\[u, u']`. Taking the dot product of `X` and `\[1, 0]` yields `u = -c_1 e^{-4t} + c_2 e^{t}`. Don't
try to merge the constants yet; we need to first solve: we have `u(0) = 2, u'(0) = 5`, so we have the equations `u(0) = 2 = -c_1 + c_2` and `u'(0) = 5 = 4c_1 + c_2` (note
how the `e^{ lambda t }` terms reduce to `1`, because `t = 0`). We can construct a matrix equation: `\[\[-1, 1], \[4, 1]] \[c_1, c_2] = \[2, 5]`.
This is very easy to solve with Gaussian elimination; the result is `\[c_1, c_2] = \[frac 3 5, frac {13} 5]`.
</p>
<p>
Finally, we resubstitute to get `u = -frac 3 5 e^{-4t} + frac {13} 5 e^t`. Nice and easy! The exact same method applies for a wide variety of homogeneous
2ODEs that can be transformed into a homogeneous SLDE. Any method to solve the SLDE for complex and/or repeated eigenvalues will work the same way, as long
as you can get a vector equation for `X`. Note that if the 2ODE is nonautonomous, you won't be able to apply this trick.
</p>
<h2>A Hard Problem</h2>
<p>
To get acquainted with 2ODEs, let's solve <b>HW7.3</b>.
<b>At least take a crack at it before reading this</b>. This will actually be the first homework problem solved on Deadly Boring Math this semester;
I usually only do a few of the hardest ones, to avoid making DBM a general cheating tool.
</p>
<p>
The problem is a fairly simple physical one: we have a 4lb ball on a spring, which stretches the string by half a foot. It starts from equilibrium with a downwards force of
9 feet per second. In our coordinate plane, positive is "down" and negative is "up". Gravitational acceleration is a constant 32 feet per second squared in the "down" direction.
We also have air resistance equal to negative four times the velocity. Note that, because imperial units are horrendously stupid, `4lb` really means four pounds of <i>weight</i>,
not mass; because we're assuming `32` feet per second squared Earth gravity, the mass is `frac 1 8`... <i>slug</i>. That's really what it's called. You can't make this stuff up.
</p>
<p>
The first thing we need to do is analyze the spring. If you remember physics 1, you'll remember that Hooke's law says `F = -xk`, where `x` is displacement and `k` is the <i>spring constant</i>.
If you don't remember physics 1, that's fine, neither do I. Google to the rescue. Spring force is, assuming `y=0` is equilibrium, `F = -yk`; we need to find `k`.
Because we know the spring was already displaced by `frac 1 2` feet, and the force acting on the ball is `frac 1 8 * 32 = 4` slug-feet per second squared ("pounds") opposing the displacement,
then assuming we have strong enough stomachs to slog through imperial units, we have an equation `-4 = -frac 1 2 k`. This is very easy to solve to find
`k = 8`.
</p>
<p>
Let's formalize our initial values. We know the ball is starting at `y=0` with a velocity of `9` feet per second downwards, so we have `y(0) = 0, v(0) = 9`
where `v(t) = y'(t)`. So far so good. Also note that `a = v' = y''`.
</p>
<p>
We have three forces acting on the system: gravity, air resistance, and the spring. We want to find the acceleration from each, so we can
write a differential equation for net acceleration. Air resistance is stated in the question as being a value in pounds,
so we need to divide by mass to get acceleration: `a_"air" = frac {-4v} {frac 1 8} = -32v`. Note the negative; this force is always opposing our current velocity. Finally, we already know
the spring force thanks to Hooke's law: dividing by mass gives us `a_"spring" = frac {-8y} {frac 1 8} = -64y`. Because the spring is counteracting gravity,
we simply ignore it; adding these together gives us
`a = a_"air" + a_"spring" = - 32v - 64y`.
</p>
<p>
We <i>could</i> rewrite as a system, but that would be pointless; we already know that the answer will be in the form `y = C_1 e^{lambda_1 t} + C_2 e^{lambda_2 t}` assuming
the eigenvalues are real and different. If you aren't sure why we can say this, look closely at the solution for a 2ODE: you can always write each eigenvector as
`\[1, v]`, and we only care about the first value in each vector. Furthermore, because of the way this will be rewritten, we can immediately turn `ay'' + by' + cy = 0` into
a characteristic equation `a lambda^2 + b lambda + c = 0`. Convenient!
</p>
<p>
Rewriting gives us `y'' + 32y' + 64y = 0`, so our characteristic equation is `lambda^2 + 32lambda + 64 = 0`. The roots are easily found to be `lambda_1 = -16 + 8 sqrt(3)` and `lambda_2 = -16 - 8 sqrt(3)`.
Substituting these into our general form yields `y = C_1 e^{-16t + 8sqrt(3)t} + C_2e^{-16t + -8sqrt(3)t}`. We can differentiate to find
`y' = C_1 (-16 + 8sqrt(3)) e^{-16t + 8sqrt(3)t} + C_2 (-16 + -8sqrt(3)) e^{-16t + -8sqrt(3)t}`.
</p>
<p>
To find `C_1` and `C_2`, we'll just substitute the initial values into these equations: `0 = C_1 + C_2`, `9 = C_1 (-16 + 8sqrt(3)) + C_2 (-16 + -8sqrt(3))`. These solve easily
to `C_1 = frac {3sqrt(3)} 16, C_2 = -frac {3sqrt(3)} 16`, meaning we have a final solution `y = frac {3sqrt(3)} 16 (e^{-16t + 8sqrt(3)t} - e^{-16t + -8sqrt(3)t})`.
Nice!
</p>
<h2>Undetermined Coefficients</h2>
<p>
Note: I skipped over oscillation as it's mostly covered by our discussion of a spring-mass system. It may be worthwhile to go through it if you're hazy on the mechanics of oscillation.
Watch the lectures!
</p>
<p>
The method of undetermined coefficients is yet another fascinating substitution that can be used to very quickly solve a second-order linear nonhomogeneous differential equation.
The core idea is that we can occasionally guess at the <i>form</i> of a solution, leaving the coefficients as unknowns; substituting into the equation will yield
an equation in terms of the coefficients, which can be solved and back-substituted. This finds a <i>particular</i> solution, and we can complete the solution by
solving the homogeneous part to get a <i>complementary solution</i>. The final result is the particular solution plus the complementary solution. It sounds confusing, but it isn't too bad.
</p>
<p>
Let's do an example. Straight from the textbook, we have `y'' - 3y' - 4y = 3e^{2t}`. Let's first find the complementary solution: this is done by simply ignoring
the right hand side and solving via the characteristic equation method. We have the characteristic equation `lambda^2 - 3lambda - 4 = 0`, the roots of which are
`lambda_1 = -1` and `lambda_2 = 4`, so our complementary solution is trivially `y = C_1e^{-1t} + C_2e^{4t}`.
</p>
<p>
Now for the fun part. Because `3e^{2t}` is exponential, our solution form will probably be exponential: a decent guess is `y = Ae^{2t}`. Given this, we have
`y' = 2Ae^{2t}` and `y'' = 4Ae^{2t}`, which we substitute back into the original equation to get `4Ae^{2t} - 6Ae^{2t} - 4Ae^{2t} = 3e^{2t}`. This is promising!
A bit of simplification reduces this whole thing down to just `-6Ae^{2t} = 3e^{2t}`, meaning `A` must be `- frac 1 2`; this means our <i>particular solution</i>
is going to be `y = - frac 1 2 e^{2t}`. Summing the particular and complementary solutions gives us `y = C_1e^{-1t} + C_2e^{4t} - frac 1 2 e^{2t}`.
</p>
<p>
This is pretty quick and easy! The same technique works for most "simple" functions that propagate through derivatives in a predictable way; for instance,
`y'' - 3y' - 4y = 2sin(t)` will probably have a particular solution in the form `y = A sin(t) + B cos(t)`. Why not `y = A sin(t)`? The reason is easiest to explain
by trying it. Let's say `y = A sin(t)`, so `y' = A cos(t)` and `y'' = -A sin(t)`. This substitutes back to `-Asin(t) + 3Acos(t) - 4Asin(t) = 2sin(t)`, which simplifies to
`3Acos(t) = (6 + 5A)sin(t)`, which can't be solved for general `t` (note that `t = frac {pi} 2, A = - frac 6 5` is a solution, but we can't extend that to
any `t`). Doing it with `y = Asin(t) + Bcos(t)` gives us `y' = A cos(t) - B sin(t)` and `y'' = -A sin(t) - B cos(t)`, which substitutes to get
`-A sin(t) - B cos(t) - 3A cos(t) + 3B sin(t) - 4A sin(t) - 4B cos(t) = 2sin(t)`. What a mess! Fortunately, most of these terms cancel or combine, and we get
a much nicer `(-5A + 3B) sin(t) (-5B - 3A)cos(t) = 2sin(t)`. This implies that `-5A + 3B = 2` and `-5B - 3A = 0`; these solve to `A = - frac 1 5` and
`B = frac 1 3`.
</p>
<p>
Sometimes we have to deal with the product of two simple functions (for instance, `e^t` and `sin(t)` multiplied to get `e^t sin(t)`). Individually these are easy to solve, but how do we
compose an undetermined coefficient equation that will propagate correctly? In these cases, we can actually just multiply together the forms. For instance, because
for `e^t` we'd use `Ae^t` and for `sin(t)` we'd use `A sin(t) + B cos(t)`, for `e^t sin(t)` we'd use `A e^t sin(t) + B e^t cos(t)` (the undetermined coefficients merge; we only need one per term).
</p>
<p>
In cases where you have to deal with the <i>sum</i> of two simple functions, use the superposition; that is, the sum of partial solutions for each term is
a full particular solution. For instance, given an equation `y'' + p(t) y' + q(t) y = g(t) + h(t)`, you solve `y'' + p(t) y' + q(t) y = g(t)` and
`y'' + p(t) y' + q(t) y = h(t)` individually to find solutions `y_g` and `y_h`, and get the particular solution by adding them: `y = y_g + y_h`.
</p>
<p>
There are several situations where you'll end up, for some reason or another, with cancelled coefficients, leaving you with a specific solution
but no solution for general `t`. For instance, `y'' + 4y = sin(2t)`: the obvious guess, `Asin(2t)` (adding the usual `B cos(2t)` term won't change the result),
yields `-4Asin(2t) + 4Asin(2t) = sin(2t)`, which cancels to `0 = sin(2t)`.
In these cases, your best bet is to multiply by some power of `t` and hope it improves the situation. For instance, if we let `y = At sin(2t) + Bt cos(2t)`,
we end up with `y'' = 4A cos(2t) - 4Bt cos(2t) - 4At sin(2t) - 4B sin(2t)`. `y'' + 4y = sin(2t)` now has a much better expansion:
`4A cos(2t) - 4Bt cos(2t) - 4At sin(2t) - 4B sin(2t) + 4At sin(2t) + 4Bt cos(2t) = sin(2t)`, simplifying to `4A cos(2t) - 4B sin(2t) = sin(2t)`.
Obviously, `A = 0`, and `B = -frac 1 4`, so our final answer is `y = - frac t 4 cos(2t)`.
</p>
<p>
This technique is obviously very powerful. I won't go through all of the permutations here, but it's useful to memorize the common substitutions, and
get familiar enough to rederive some less common ones.
</p>
<h2>Final Notes</h2>
<p>
There's a quiz coming up soon (Tuesday). Don't forget! I'll hopefully get some review up here tomorrow.
</p>
<p>
I think that's everything. Until tomorrow, sayonara, do svidanya, and au revoir!
</p>
[/]
[=author "Tyler Clarke"]
[=date "2025-6-15"]
[=subject "Calculus"]
[=title "Differential Equations Week 4 & 5"]
[#post.html]