2020/08/02 線形2階非斉次微分方程式の演習

本日のお題

\(1. \quad y'' + y' - 2y = 2x^2 - 4x - 1\)

\(2. \quad y'' - 2y' = 2x - 4\)

\(3. \quad y'' - 3y' + 2y = 3^{-x}\)

\(4. \quad 6y'' - 5y' + y = -13\sin 2x - 33\cos 2x\)

\(5. \quad y'' + 4y = 2\sin 2x + 3\cos 2x\)

はい,かしこまりました
この微分方程式を解けば宜しいのですね

\(1.\quad y'' + y' - 2y = 2x^2 - 4x - 1\)

\(\left(D^2 + D - 2\right)y = 2x^2 - 4x - 1\)

基本解は \(\left(e^{x},\ e^{-2x}\right)\) であり,特殊解を求めると \[y \begin{array}[t]{l} \displaystyle = \frac{1}{(D - 1)(D + 2)}\big[\,2x^2 - 4x - 1\,\big] \\[2px] \displaystyle = \frac{1}{3}\left(\frac{1}{D - 1} - \frac{1}{D + 2}\right)\big[\,2x^2 - 4x -1\,\big] \\[2px] \displaystyle = \frac{1}{3}\left(-\frac{1}{1 - D} -\frac{1}{2}\cdot\frac{1}{1 + \frac{D}{2}}\right)\big[\,2x^2 - 4x -1\,\big] \\[2px] \displaystyle = \frac{1}{3}\left\{-\left(1 + D + D^2\right) - \frac{1}{2}\left(1 - \frac{D}{2} + \frac{D^2}{4}\right)\right\}\big[\,2x^2 - 4x -1\,\big] \\[2px] \displaystyle = \left(-\frac{1}{2} - \frac{1}{4}D - \frac{3}{8}D^2\right)\big[\,2x^2 - 4x -1\,\big] \\[2px] \displaystyle = -\frac{1}{2}\left(2x^2 - 4x - 1\right) - \frac{1}{4}\left(4x - 4\right) - \frac{3}{8}\cdot 4 \\[2px] = -x^2 + x \end{array}\] よって, \(y = C_1\,e^x + C_2\,e^{-2x} - x^2 + x\)

\(2.\quad y'' - 2y' = 2x - 4\)

\(D(D - 2)\,y = 2x - 4\)

基本解は \(\left(e^{2x},\ 1\right)\) であり,特殊解を求めると \[y \begin{array}[t]{l} \displaystyle = \frac{1}{D(D - 2)}\big[\,2x - 4\,\big] \\[2px] \displaystyle = \frac{1}{2}\left(\frac{1}{D - 2} - \frac{1}{D}\right)\big[\,2x - 4\,\big] \\[2px] \displaystyle = \left(-\frac{1}{2}\cdot\frac{1}{1 - \frac{D}{2}} - \frac{1}{D}\right)\big[\,x - 2\,\big] \\[2px] \displaystyle = \left\{-\frac{1}{2}\left(1 + \frac{D}{2}\right) - \frac{1}{D}\right\}\big[\,x - 2\,\big] \\ \displaystyle = -\frac{1}{2}\left\{(x - 2) + \frac{1}{2}\right\} - \left(\frac{1}{2}x^2 - 2x\right) \\[2px] \displaystyle = -\frac{1}{2}x^2 + \frac{3}{2}x + \frac{3}{4} \end{array}\] よって \(\displaystyle y = C_1\,e^{2x} + C_2 - \frac{1}{2}x^2 + \frac{3}{2}x + \frac{4}{3} = C_1\,e^{2x} + C_2 - \frac{1}{2}x + \frac{3}{2}x \)

\(3.\quad y'' - 3y' + 2y = 3e^{-x}\)

\((D - 1)(D - 2)\,y = 3e^{-x}\)

基本解は \(\left(e^x,\ e^{2x}\right)\) であり,特殊解を求めると \[y \begin{array}[t]{l} \displaystyle = \frac{1}{(D - 1)(D - 2)}\big[\,3e^{-x}\,\big] \\[2px] \displaystyle = \frac{3}{(-1 -1)(-1 -2)}\,e^{-x} \\[2px] \displaystyle = \frac{1}{2}\,e^{-x} \end{array}\] よって, \(\displaystyle y = C_1\,e^x + C_2,e^{2x} + \frac{1}{2}\,e^{-x}\)

\(4.\quad 6y'' - 5y' + y = -13\sin 2x - 33\cos 2x\)

\((3D - 1)(2D -1)\,y = -13\sin 2x - 33\cos 2x\)

基本解は \(\left(e^{\frac{x}{3}},\ e^{\frac{x}{2}}\right)\) であり,特殊解を求めるために,まず,\(\displaystyle \frac{1}{6D^2 - 5D + 1}\,\big[\,e^{i2x}\,\big]\) を計算すると \[\begin{array}[t]{l} \displaystyle \frac{1}{6D^2 - 5D + 1}\,\big[\,e^{i2x}\,\big] \displaystyle = \frac{1}{6\cdot(-4) - 10i + 1}\,e^{i2x} \\[2px] \displaystyle = \frac{1}{-23 - 10i}\,e^{i2x} \\[2px] \displaystyle = \frac{-23 + 10i}{629}\,\left(\cos 2x + i\sin 2x\right) \\[2px] \displaystyle = \frac{-23\cos 2x - 10\sin 2x}{629} + i\,\frac{-23\sin 2x + 10\cos 2x}{629} \end{array}\] \[∴\quad\left\{\begin{array}{l} \displaystyle \frac{1}{6D^2 - 5D + 1}\,\big[\,\sin 2x\,\big] = \frac{-23\sin 2x + 10\cos 2x}{629} \\[2px] \displaystyle \frac{1}{6D^2 - 5D + 1}\,\big[\,\cos 2x\,\big] = \frac{-23\cos 2x - 10\sin 2x}{629} \end{array}\right.\] したがって,特殊解は \[y \begin{array}[t]{l} \displaystyle = -13\cdot\frac{-23\sin 2x + 10\cos 2x}{629} - 33\cdot\frac{-23\cos 2x - 10\sin 2x}{629} \\[2px] \displaystyle = \frac{(299 + 330)\sin 2x + (-130 + 759)\cos 2x}{629} \\[2px] = \sin 2x + \cos 2x \end{array}\] 以上から, \(y = C_1\,e^{\frac{x}{3}} + C_2\,e^{\frac{x}{2}} + \sin 2x + \cos 2x\)

\(5.\quad y'' + 4y = 2\sin 2x + 3 \cos 2x\)

基本解は,\(\left(\sin 2x,\ \cos 2x\right)\) であり,特殊解を求めるために,まず,\(\displaystyle \frac{1}{D^2 + 4}\,\big[\,e^{i2x}\,\big]\) を計算すると \[\frac{1}{D^2 + 4}\,\big[\,e^{i2x}\,\big] \begin{array}[t]{l} \displaystyle = \frac{1}{D - 2i}\cdot\frac{1}{D + 2i}\,\big[\,e^{i2x}\,\big] \\[2px] \displaystyle = \frac{1}{D - 2i}\left[\,\frac{1}{4i}\,e^{i2x}\,\right] \\[2px] \displaystyle = -\frac{i}{4}\cdot x\,e^{i2x} \\[2px] \displaystyle = -\frac{i}{4}\,x\,(\cos 2x + i\sin 2x) \\[2px] \displaystyle = \frac{1}{4}\,x\sin 2x - \frac{i}{4}\,x\cos 2x \end{array}\] \[∴\quad\left\{\begin{array}{l} \displaystyle \frac{1}{D^2 + 4}\,\big[\,\sin 2x\,\big] = -\frac{1}{4}\,x\cos 2x \\ \displaystyle \frac{1}{D^2 + 4}\,\big[\,\cos 2x\,\big] = \frac{1}{4}\,x\sin 2x \end{array}\right.\] したがって,特殊解は \[y \begin{array}[t]{l} \displaystyle = \frac{1}{D^2 + 4}\,\big[\,2\sin 2x + 3\cos 2x\,\big] \\[2px] \displaystyle = -\frac{1}{2}\,x\cos 2x + \frac{3}{4}\,x\sin 2x \end{array}\] 以上から, \(\displaystyle y = C_1\,\sin 2x + C_2\,\cos 2x - \frac{1}{2}\,x\cos 2x + \frac{3}{4}\,x\sin 2x\)

Last modified: Friday, 5 March 2021, 5:35 PM