[!]
[=post-]
<p>
    Hello, everyone! I'm a bit late to this one; yesterday was pretty hectic. Fortunately, this quiz is only on 4.1 to 4.3 - at the risk of taunting Murphy, it should be a milk run.
</p>
<h2>WS4.1.1: Classification</h2>
<p>
    The only question in worksheet 4.1 is just a classification task. We're given a bunch of equations and asked if they're linear or nonlinear, and if linear, if they're
    homogeneous or nonhomogeneous.
</p>
<ol>
    <li>`y'' - 2y' - 2y = 0`: This is clearly linear; it's in the form `l(t)y'' + p(t)y' + q(t)y = r(t)`. It's also homogeneous, because `r(t)` is 0.</li>
    <li>`2y'' + 2y' + 3y = 0`: This is in the same form, and `r(t)` is still `0`, so this is a linear homogeneous differential equation.</li>
    <li>`y''' - y'' = 6`: This is still linear! The `y'''` term might be a bit disconcerting, but the equation is still in the third-order linear form.
        It's nonhomogeneous because `6` is not equal to `0`.
    </li>
    <li>`y'' - 2y' + 2y = e^x tan(x)`: Linear nonhomogeneous. Note that if the right-hand side had any mention of `y` it might not be linear; in this case, though
        everything works out.
    </li>
    <li>
        `y' y'' = 4x`: This isn't linear; it require either the `y'` or the `y''` coefficient to be in terms of the other, which isn't possible.
    </li>
    <li>
        `2y'' = 3y^2`: This also isn't linear. Even if we rewrite like `2y'' - 3y^2 = 0`, we have a pesky `y^2` term that can't exist in a linear differential equation. 
    </li>
</ol>
<h2>WS4.2.1.a: Intervals of Existence</h2>
<p>
    Finally we're doing existence in second-order linear DEs! The idea is pretty much the same as first-order: having an equation in the form `y'' + p(t)y' + q(t)y = r(t)`,
    the interval of existence is the intersection of the intervals of existence of `p(t)`, `q(t)`, and `r(t)`. We handle IVPs much the same way, as well: the interval around
    our initial point `t_0` is the interval of existence for that specific solution.
</p>
<p>
    We're given the IVP `t(t - 4)y'' + 3ty' + 4y = 2, y(3) = 0, y'(3) = 1`. This is clearly a nonhomogeneous 2OLDE; to find our intervals of existence, we need to get the coefficient
    of `y''` to be just `1`. We do this by dividing the entire equation by `t(t - 4)`: `y'' + frac 3 {t - 4} y' + frac 4 {t(t - 4)} y = frac 2 {t(t - 4)}`. Clearly,
    `p(t) = frac 3 {t - 4}`, `q(t) = frac 4 {t(t - 4)}`, and `r(t) = frac 2 {t(t - 4)}`. `q(t)` and `r(t)` have the same interval of existence; they're defined whenever
    `t(t - 4)` is nonzero, so `(-oo, 0) cup (0, 4) cup (4, oo)`. `p(t)` is defined for `(-oo, 4) cup (4, oo)`. Since our starting point is `t=3`, it falls inside the
    `(0, 4)` range - meaning the interval of existence of our solution is `0 < t < 4`.
</p>
<h2>WS4.2.2: Solving Some Coefficients</h2>
<p>
    This question is fairly cryptic; as best as I can tell, it's asking us to find some variant of `y = c_1 e^{x} + c_2 e^{-x}` that matches the IVP
    `y'' - y = 0, y(0) = 0, y'(0) = 1`. Because we already know `y` in terms of some arbitrary constants, we can immediately skip to solving the IVP:
    differentiating gives us `y' = c_1 e^x - c_2 e^{-x}`, and substituting into our equations for `y` and `y'` respectively gives us
    `0 = c_1 + c_2, 1 = c_1 - c_2`. These easily solve to `c_1 = frac 1 2, c_2 = - frac 1 2`, so our final result is `y = frac 1 2 e^x - frac 1 2 e^{-x}`.
</p>
<h2>WS4.3.1: Solving A 2OLDE</h2>
<p>
    Finally, some solving! We're given the pleasant linear homogeneous equation `2y'' - 5y' - 3y = 0`, and asked to solve and graph. I'm not going to post the graphs here,
    but you should do them for practice!
</p>
<p>
    Because this is homogeneous and has constant coefficients, we can instantly turn it into an <i>auxiliary</i> polynomial, find the roots `lambda_1` and `lambda_2`, and
    substitute them into the general solution `y = c_1 e^{lambda_1 t} + c_2 e^{lambda_2 t}`. For an equation `a y'' + by' + cy = 0`, the auxiliary polynomial is
    `a lambda^2 + b lambda + c = 0`, so we have `2 lambda^2 - 5 lambda - 3 = 0`. This solves easily to `lambda_1 = - frac 1 2`, `lambda_2 = 3`. Substituting these in
    gives us `y = c_1 e^{- frac 1 2 t} + c_2 e^{3t}`.
</p>
<p>
    Remember that phase portraits are parametric graphs in terms of `y` and `y'`, and solution graphs are normal solution graphs in terms of `y` and `t`.
    You can find `y'` by differentiating `y` if you didn't use a technique that provides `y'` as a side effect.
</p>
<h2>SA36: Verifying Fundamental Sets</h2>
<p>
    This isn't hard at all, but it's important to understand. We're given an equation `x^2y'' - 6xy' + 12y = 0`, and asked to verify that `y_1(x) = x^3, y_2(x) = x^4` is a fundamental
    solution set. The idea is that two solutions are a fundamental set if the corresponding SLDE matrix they produce is not singular; it's probably easier to
    just do the question than to explain the process. Our first step is to verify that these are actually solutions: the substitution is fairly simple, and I'll leave it as an exercise
    to the reader.
    We now take the derivatives `y_1' = 3x^2` and `y_2' = 4x^3` and populate a matrix:
    `\[\[x^3, x^4], \[3x^2 + 4x^3]]`. Note that this is actually an SLDE matrix. As long as the determinant of this is nonzero, we have a linearly independent solution.
    The determinant is `x^6`, which is not always 0, so the solution is linearly independent. Easy-peasy!
</p>
<h2></h2>
[/]
[=author "Tyler Clarke"]
[=date "2025-6-17"]
[=subject "Calculus"]
[=title "Differential Quiz 3"]
[=unpub]
[#post.html]