% 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/