Thoughts for Jan. 30 – Feb. 5.

Michael Stone, February 5, 2012, , (src)

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”. Now, let’s try out my definitions on Coddington’s exercise §1.6.1(c).

To begin, let

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: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, \[K = \left\{\lambda x. \frac{-1}{2} (x^2 + 2x + 1) + ce^{2x}\right\}_{c \in \mathbb{C}}\]