% Thoughts for Jan. 30 -- Feb. 5.
% Michael Stone
% February 5, 2012
As I mentioned [last week], I've recently become interested in trying to fill
in some gaps in my mathematical education, beginning with the theory of
differential equations. To that end, I began by recalling a (very mildly
careful) definition of the word ["equation"](../equations-vs-definitions).
Now, let's try out my definitions on [Coddington's][diffeq] exercise §1.6.1(c).
To begin, let
* $\mathbb{N}$ be the non-negative integers,
* $\mathbb{C}$ be the complex numbers,
* $I = [p, q] \subset \mathbb{R}$ be a closed real interval,
* $(A, B)$ be the space of all functions from $A$ to $B$,
* $C^1\!\!(A, B)$ be the space of continuously differentiable functions from $A$ to $B$,
* $L(A, B)$ be the space of linear transformations from $A$ to $B$,
* $D : C^1\!\!(A, B) \to (A, L(A,B))$ be the total derivative operator.
Exercise §1.6.1(c) asks us to find all the solutions of the equation
$$y' - 2y = x^2 + x$$
This means that we're looking for a parameterization of the subset $K \subset
C^1\!\!(I, \mathbb{C})$ satisfying the identity, for all $k \in K$ and $x \in I$:
$$(Dk)(x) - 2k(x) - x^2 - x = 0$$
We are also given, via Theorem §1.6.2 that for all $a \in \mathbb{C}$ and $b
\in C^0\!\!(I, \mathbb{C})$ continuous on $I$, the sets of solutions $V$ of
differential equations of the form:
$$y' + ay = b(x)$$
all have, for some $x_0 \in I$ and for all $c \in \mathbb{C}$, the form
$$V = \left\{\phi : \phi(x) = e^{-ax} \int_{x_0}^{x} e^{at} b(t) \mathrm{d}t + ce^{-ax}\right\}$$
Unifying the signature of Theorem §1.6.2 with the equation of Exercise
§1.6.1(c), we find that $a = -2$ and $b(x) = x^2 + x$ and that
$$K = \left\{\lambda x. e^{2x} \int_{x_0}^{x} e^{-2t}(t^2 + t) \mathrm{d}t + ce^{2x}\right\}_{c \in \mathbb{C}}$$
Our remaining task is to simplify $K$ by simplifying
$$\int e^{-2t}(t^2 + t) \mathrm{d}t$$
To that end, we start off knowing that, for all $a, b, c \in \mathbb{C}$ and
for all $n \in N$,
$$\int e^{c t} = \frac{1}{c} e^{c t}, \int t^n = \frac{1}{n+1} t^{n+1}, \mbox{ and } \int \left(af + bg\right) = a \int f + b \int g$$
Next, via the product rule for differentiation, we know that
$$D(e^{ct} t^n) = D(e^{ct}) t^n + e^{ct} D(t^n) = c e^{ct} t^{n} + n e^{ct} t^{n-1}$$
Next, while we don't yet know how to compute $\int e^{-2t}(t^2 + t)$, we might
hope to find an appropriate antiderivative by solving
$$e^{-2t}(t^2 + t) = D\left(aF + bG + cH\right)$$
for some $a, b, c \in \mathbb{C}$ and for $F = e^{-2t} t^2$, $G = e^{-2t} t^1$, and $H = e^{-2t} t^0$.
Fortunately indeed, this question reduces to solving the following system of
three linear equations in the Euclidean vector space generated by the basis
$\left\{x^a e^{-2bx}\right\}_{a, b \in \mathbb{N}, a < 3}$ with coefficients in
$\mathbb{C}$:
$$
\begin{align*}
\left[F + G\right] &= \left[\begin{array}{c}DF \\ DG \\ DH\end{array}\right]^T \left[\begin{array}{c}a \\ b \\ c\end{array}\right] \\ \\
\left[F + G\right] &= \left[\begin{array}{c}-2F + 2G \\ -2G + H \\ -2H \end{array}\right]^T \left[\begin{array}{c}a \\ b \\ c\end{array}\right] \\ \\
\left[\begin{array}{c}F \\ G \\ H\end{array}\right]^T \left[\begin{array}{c}1 \\ 1 \\ 0\end{array}\right] &= \left[\begin{array}{c}F \\ G \\ H\end{array}\right]^T \left[\begin{array}{ccc}-2 & 0 & 0 \\ 2 & -2 & 0 \\ 0 & 1 & -2 \end{array}\right] \left[\begin{array}{c}a \\ b \\ c\end{array}\right] \\ \\
\left[\begin{array}{c}1 \\ 1 \\ 0\end{array}\right] &= \left[\begin{array}{ccc}-2 & 0 & 0 \\ 2 & -2 & 0 \\ 0 & 1 & -2 \end{array}\right] \left[\begin{array}{c}a \\ b \\ c\end{array}\right]
\end{align*}
$$
From here, a quick trip through [Octave] yields:
~~~~ { .octave }
octave:1> A = [-2, 0, 0; 2, -2, 0; 0, 1, -2]
A =
-2 0 0
2 -2 0
0 1 -2
octave:2> inv(A)
ans =
-0.50000 0.00000 0.00000
-0.50000 -0.50000 0.00000
-0.25000 -0.25000 -0.50000
octave:3> inv(A) * [1; 1; 0]
ans =
-0.50000
-1.00000
-0.50000
~~~~
Thus we have
$$\left[\begin{array}{c}a \\ b \\ c\end{array}\right] = \frac{-1}{2} \left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right]$$
and
$$\int e^{-2t}(t^2 + t) \mathrm{d}t = \int D\left(aF + bG + cH\right) = aF + bG + cH$$
and, finally,
$$K = \left\{\lambda x. \frac{-1}{2} (x^2 + 2x + 1) + ce^{2x}\right\}_{c \in \mathbb{C}}$$
[diffeq]: http://www.amazon.com/Introduction-Ordinary-Differential-Equations-Mathematics/dp/0486659429
[last week]: http://mstone.info/posts/thoughts-20120129/
[Octave]: http://www.gnu.org/software/octave/