微分方程式 2階微分方程式定係数2階微分方程式非同次方程式の解法、未定係数法による解法累乗、指数関数、三角関数、正弦と余弦、積

学習環境

力学・電磁気学・熱力学のための基礎数学 (松下貢(著)、裳華房)の第3章(微分方程式)、3.3(2階微分方程式)、3.3.2(定係数2階微分方程式)、(3)非同次方程式の解法、未定係数法による解法、問題11の解答を求めてみる。

特解の1つを

y_{p} = a_{0} + a_{1} x + a_{2} x^{2}

と推測。

\frac{{dy}_{p}}{dx} = a_{1} + 2 a_{2} x

\frac{d^{2} y_{p}}{{dx}^{2}} = 2 a_{2}

2 a_{2} - 5 (a_{1} + 2 a_{2} x) + 4 (a_{0} + a_{1} x + a_{2} x^{2}) = x^{2}

これがすべてのx に対して成り立つには、

a_{2} = \frac{1}{4}

\begin{array}{l} 4 a_{1} - \frac{5}{2} = 0 \\ a_{1} = \frac{5}{8} \end{array}

\begin{array}{l} 4 a_{0} - \frac{25}{8} + \frac{1}{2} = 0 \\ a_{0} = \frac{21}{32} \end{array}

よって、

y_{p} = \frac{21}{32} + \frac{5}{8} x + \frac{1}{4} x^{2}

同次微分方程式

y'' - 5 y' + 4 y = 0

の特性方程式の解は、

\begin{array}{l} λ^{2} - 5 λ + 4 = 0 \\ (λ - 1) (λ - 4) = 0 \\ λ = 1, 4 \end{array}

よって同次微分方程式の一般解は、

y_{g} = c_{1} e^{x} + c_{2} e^{4 x}

よって、求める問題の非同次方程式の一般解は、

y = y_{p} + y_{q} = \frac{21}{32} + \frac{5}{8} x + \frac{1}{4} x^{2} + c_{1} e^{x} + c_{2} e^{4 x}

実際に確認。

\begin{array}{l} \frac{1}{2} + c_{1} e^{x} + 16 c_{2} e^{4 x} \\ - 5 (\frac{5}{8} + \frac{1}{2} x + c_{1} e^{x} + 4 c_{2} e^{4 x}) \\ + \frac{1}{4} (\frac{21}{32} + \frac{5}{8} x + \frac{1}{4} x^{2} + c_{1} e^{x} + c_{2} e) \\ = x^{2} \end{array}

y_{p} = a e^{2 x}

4 a e^{2 x} - 4 a e^{2 x} + 2 a e^{2 x} = e^{2 x}

\begin{array}{l} 2 a = 1 \\ a = \frac{1}{2} \end{array}

λ^{2} - 2 λ + 2 = 0

λ = 1 \pm \sqrt{1 - 2} = 1 \pm i

y_{g} = e^{x} (C_{1} \cos x + C_{2} \sin x)

y = \frac{1}{2} e^{2 t} + e^{x} (C_{1} \cos x + C_{2} \sin x)

y_{p} = A \cos x + B \sin x

- A \cos x - B \sin x - 6 (- A \sin x + B \cos x) + 8 (A \cos x + B \sin x) = \cos x

\begin{array}{l} - A - 6 B + 8 A = 1 \\ - B + 6 A + 8 B = 0 \end{array}

\begin{array}{l} 7 A - 6 B = 1 \\ 6 A + 7 B = 0 \end{array}

\begin{array}{l} 42 A - 36 B = 6 \\ 42 A + 49 B = 0 \end{array}

\begin{array}{l} 13 B = 6 \\ B = - \frac{6}{85} \\ A = \frac{7}{6} \cdot \frac{6}{85} = \frac{7}{85} \end{array}

y_{p} = \frac{7}{85} \cos x - \frac{6}{85} \sin x

\begin{array}{l} λ^{2} - 6 λ + 8 = 0 \\ (λ - 2) (λ - 4) = 0 \\ x = 2, 4 \end{array}

y_{g} = c_{1} e^{2 x} + c_{2} e^{4 x}

y = \frac{7}{85} \cos x - \frac{6}{85} \sin x + c_{1} e^{2 x} + c_{2} e^{4 x}

y_{p} = (A \cos x + B \sin x) e^{x}

\frac{{dy}_{p}}{dx} = (- A \sin x + B \cos x) e^{x} + (A \cos x + B \sin x) e^{x}

= ((A + B) \cos x + (B - A) \sin x) e^{x}

\frac{d^{2} y_{p}}{{dx}^{2}} = ((- A - B) \sin x + (B - A) \cos x) e^{x} + ((A + B) \cos x + (B - A) \sin x) e^{x}

= (2 B \cos x - 2 A \sin x) e^{x}

\begin{array}{l} 2 B + 5 (A + B) + 6 A = 0 \\ - 2 A + 5 (B - A) + 6 B = 1 \end{array}

\begin{array}{l} 11 A + 7 B = 0 \\ - 7 A + 11 B = 1 \end{array}

B = - \frac{11}{7} A

- 7 A - \frac{121}{7} A = 1

- \frac{170}{7} A = 1

A = - \frac{7}{170}

B = \frac{11}{170}

y_{p} = (- \frac{7}{170} \cos x + \frac{11}{170} \sin x) e^{x}

\begin{array}{l} λ^{2} + 5 λ + 6 = 0 \\ λ = \frac{- 5 \pm \sqrt{25 - 24}}{2} = - 3, - 2 \end{array}

y_{g} = c_{1} e^{- 3 x} + c_{2} e^{- 2 x}

y = (- \frac{7}{170} \cos x + \frac{11}{170} \sin x) e^{x} + c_{1} e^{- 3 x} + c_{2} e^{- 2 x}

コード(Wolfram Language)

y[x_] := 1/2 Exp[2x]+Exp[x](c1 Cos[x] + c2 Sin[x])

y''[x] -2y'[x]+2y[x] // Simplify

y[x_] := 7/85 Cos[x]-6/85 Sin[x]+c1 Exp[2x] + c2 Exp[4x]

y''[x]-6y'[x]+8y[x] // Simplify

y[x_] := (-7/170 Cos[x] + 11/170 Sin[x]) Exp[x] + c1 Exp[-3x] + c2 Exp[-2x]

y''[x] + 5y'[x] + 6y[x] // Simplify