数学 - Python - JavaScript - 面積、体積、長さ - 積分法の応用 – 簡単な微分方程式 - 2階微分方程式(定数係数の2階線形同次微分方程式の一般解、場合分け)

学習環境

Surface 3 (4G LTE)、Surface 3 タイプカバー、Surface ペン(端末)
Windows 10 Pro (OS)
数式入力ソフト(TeX, MathML): MathType
MathML対応ブラウザ: Firefox、Safari
MathML非対応ブラウザ(Internet Explorer, Microsoft Edge, Google Chrome...)用JavaScript Library: MathJax
参考書籍

数学読本〈5〉微分法の応用/積分法/積分法の応用/行列と行列式(松坂和夫(著)、岩波書店)の第20章(面積、体積、長さ - 積分法の応用)、20.4(簡単な微分方程式)、2階微分方程式、問48.を取り組んでみる。

1. 2次方程式が2つの異なる実数解α、βを持つ場合。
  $\begin{array}{l} y' = C_{1} α e^{α x} + C_{2} β e^{β x} \\ y'' = C_{1} α^{2} e^{α x} + C_{2} β^{2} e^{β x} \\ y'' + p y' + q y \\ = C_{1} α^{2} e^{α x} + C_{2} β^{2} e^{β x} + p (C_{1} α e^{α x} + C_{2} β e^{β x}) + q (C_{1} e^{α x} + C_{2} e^{β x}) \\ = C_{1} e^{α x} (α^{2} + p α + q) + C_{2} e^{β x} (β^{2} + p β + q) \\ = C_{1} e^{α x} 0 + C_{2} e^{β x} 0 \\ = 0 \end{array}$
2. 重解αをもつ場合。
  $\begin{array}{l} y' = C_{1} (e^{α x} + x e^{α x} α) + C_{2} e^{α x} α \\ = e^{α x} (C_{1} (1 + x α) + C_{2} α) \\ y'' = C_{1} (e^{α x} α + e^{α x} α + x e^{α x} α^{2}) + C_{2} e^{α x} α^{2} \\ = C_{1} (2 e^{α x} α + x e^{α x} α^{2}) + C_{2} e^{α x} α^{2} \\ = e^{α x} (C_{1} (2 α + x α^{2}) + C_{2} α^{2}) \\ y'' + p y' + q y \\ = e^{α x} (C_{1} (2 α + x α^{2}) + C_{2} α^{2}) + p e^{α x} (C_{1} (1 + x α) + C_{2} α) + q (C_{1} x e^{α x} + C_{2} e^{α x}) \\ = e^{α x} (C_{1} (2 α + x α^{2} + p + p x α + q x) + C_{2} (α^{2} + p α + q)) \\ = e^{x} (C_{1} (2 α + p + x (α^{2} + p α + q)) + C_{2} \cdot 0) \\ = e^{x} C_{1} (2 α + p) \\ α = \frac{- p}{2} \\ e^{x} C_{1} (2 α + p) \\ = e^{x} C_{1} (2 \cdot \frac{- p}{2} + p) \\ = e^{x} C_{1} (- p + p) \\ = 0 \\ y'' + p y' + q y = 0 \end{array}$
3. - $\begin{array}{l} {(a + b i)}^{2} + p (a + b i) + q = 0 \\ (a^{2} - b^{2} + p a + q) + (2 a b + p b) i = 0 \\ a^{2} - b^{2} + p a + q = 0 \\ 2 a b + p b = 0 \\ b (2 a + p) = 0 \\ b \neq 0 \\ 2 a + p = 0 \\ y' = A (e^{a x} a \sin (b x + β) + e^{a x} b \cos (b x + β)) \\ = A e^{a x} (a \sin (b x + β) + b \cos (b x + β)) \\ y'' = A (e^{a x} a (a \sin (b x + β) + b \cos (b x + β)) + e^{a x} (a b \cos (b x + β) - b^{2} \sin (b x + β))) \\ = A e^{a x} (a^{2} \sin (b x + β) + a b \cos (b x + β) + a b \cos (b x + β) - b^{2} \sin (b x + β)) \\ = A e^{a x} ((a^{2} - b^{2}) \sin (b x + β) + 2 a b \cos (b x + β)) \\ y'' + p y' + q \\ = A e^{a x} ((a^{2} - b^{2}) \sin (b x + β) + 2 a b \cos (b x + β)) + p A e^{a x} (a \sin (b x + β) + b \cos (b x + β)) + q A e^{a x} \sin (b x + β) \\ = A e^{a x} ((a^{2} - b^{2}) \sin (b x + β) + 2 a b \cos (b x + β) + p a \sin (b x + β) + p b \cos (b x + β) + q \sin (b x + β)) \\ = A e^{x} ((a^{2} - b^{2} + p a + q) \sin (b x + β) + b (2 a + p) \cos (b x + β)) \\ = A e^{x} (0 \sin (b x + β) + b 0 \cos (b x + β)) \\ = 0 \end{array}$
  - $\begin{array}{l} y' = A (e^{a x} a \cos (b x + γ) - e^{a x} b \sin (b x + γ)) \\ = A e^{a x} (a \cos (b x + γ) - b \sin (b x + γ)) \\ y'' = A (e^{a x} a (a \cos (b x + γ) - b \sin (b x + γ)) + e^{a x} (- a b \sin (b x + γ) - b^{2} \cos (b x + γ))) \\ = A e^{a x} (a^{2} \cos (b x + γ) - a b \sin (b x + γ) - a b \sin (b x + γ) - b^{2} \cos (b x + γ)) \\ = A e^{a x} ((a^{2} - b^{2}) \cos (b x + γ) - 2 a b \sin (b x + γ)) \\ y'' + p y' + q \\ = A e^{a x} ((a^{2} - b^{2}) \cos (b x + γ) - 2 a b \sin (b x + γ)) + p A e^{a x} (a \cos (b x + γ) - b \sin (b x + γ)) + q (A e^{a x} \cos (b x + γ)) \\ = A e^{a x} ((a^{2} - b^{2}) \cos (b x + γ) - 2 a b \sin (b x + γ) + p (a \cos (b x + γ) - b \sin (b x + γ)) + q \cos (b x + γ)) \\ = A e^{a x} ((a^{2} - b^{2} + p a + q) \cos (b x + γ) - b (2 a + p) \sin (b x + γ)) \\ = A e^{a x} (0 \cos (b x + γ) - b 0 \sin (b x + γ)) \\ = 0 \end{array}$

コード(Emacs)

Python 3

#!/usr/bin/env python3
from sympy import pprint, symbols, solve, exp, Derivative

p, q, X, a, b, C1, C2, x = symbols('p, q, X, a, b, C1, C2, x')
eq = X ** 2 + p * X + q
a0, b0 = solve(eq, X)

y = C1 * exp(a * x) + C2 * exp(b * x)
D1 = Derivative(y, x, 1)
D2 = Derivative(y, x, 2)
y1 = D1.doit()
y2 = D2.doit()
eq = y2 + p * y1 + q * y

for t in [D1, y1, D2, y2, eq, eq.subs({a: a0, b: b0}).expand() == 0]:
    pprint(t)
    print()

入出力結果(Terminal, Jupyter(IPython))

=$ ./sample48.py
∂ ⎛    a⋅x       b⋅x⎞
──⎝C₁⋅ℯ    + C₂⋅ℯ   ⎠
∂x                   

      a⋅x         b⋅x
C₁⋅a⋅ℯ    + C₂⋅b⋅ℯ   

  2                   
 ∂ ⎛    a⋅x       b⋅x⎞
───⎝C₁⋅ℯ    + C₂⋅ℯ   ⎠
  2                   
∂x                    

    2  a⋅x       2  b⋅x
C₁⋅a ⋅ℯ    + C₂⋅b ⋅ℯ   

    2  a⋅x       2  b⋅x     ⎛      a⋅x         b⋅x⎞     ⎛    a⋅x       b⋅x⎞
C₁⋅a ⋅ℯ    + C₂⋅b ⋅ℯ    + p⋅⎝C₁⋅a⋅ℯ    + C₂⋅b⋅ℯ   ⎠ + q⋅⎝C₁⋅ℯ    + C₂⋅ℯ   ⎠

True

$

HTML5

<div id="graph0"></div>
<pre id="output0"></pre>
<label for="r0">r = </label>
<input id="r0" type="number" min="0" value="0.5">
<label for="dx">dx = </label>
<input id="dx" type="number" min="0" step="0.0001" value="0.005">
<br>
<label for="x1">x1 = </label>
<input id="x1" type="number" value="-10">
<label for="x2">x2 = </label>
<input id="x2" type="number" value="10">
<br>
<label for="y1">y1 = </label>
<input id="y1" type="number" value="-10">
<label for="y2">y2 = </label>
<input id="y2" type="number" value="10">
<br>
<label for="n0">n = </label>
<input id="n0" type="number" min="0" value="1">
<label for="ε0">ε = </label>
<input id="ε0" type="number" min="0" value="0.1">
<label for="a0">A = </label>
<input id="a0" type="number" value="1">
<label for="α0">α = </label>
<input id="α0" type="number" value="2">

<button id="draw0">draw</button>
<button id="clear0">clear</button>

<script type="text/javascript" src="https://cdnjs.cloudflare.com/ajax/libs/d3/4.2.6/d3.min.js" integrity="sha256-5idA201uSwHAROtCops7codXJ0vja+6wbBrZdQ6ETQc=" crossorigin="anonymous"></script>

<script src="sample48.js"></script>

JavaScript

=$ ./sample48.py
∂ ⎛    a⋅x       b⋅x⎞
──⎝C₁⋅ℯ    + C₂⋅ℯ   ⎠
∂x                   

      a⋅x         b⋅x
C₁⋅a⋅ℯ    + C₂⋅b⋅ℯ   

  2                   
 ∂ ⎛    a⋅x       b⋅x⎞
───⎝C₁⋅ℯ    + C₂⋅ℯ   ⎠
  2                   
∂x                    

    2  a⋅x       2  b⋅x
C₁⋅a ⋅ℯ    + C₂⋅b ⋅ℯ   

    2  a⋅x       2  b⋅x     ⎛      a⋅x         b⋅x⎞     ⎛    a⋅x       b⋅x⎞
C₁⋅a ⋅ℯ    + C₂⋅b ⋅ℯ    + p⋅⎝C₁⋅a⋅ℯ    + C₂⋅b⋅ℯ   ⎠ + q⋅⎝C₁⋅ℯ    + C₂⋅ℯ   ⎠

True

$

r = dx =
x1 = x2 =
y1 = y2 =
n = ε = A = α =

Kamimura's blog

2017年11月5日日曜日

数学 - Python - JavaScript - 面積、体積、長さ - 積分法の応用 – 簡単な微分方程式 - 2階微分方程式(定数係数の2階線形同次微分方程式の一般解、場合分け)

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