2.5 解答

(a) 補助方程式 $y^{\prime\prime} + 2y^{\prime} + y = 0$ の特性方程式は $m^2 + 2m + 1 = (m + 1)^2 = 0$

より特性根

を得る．よって余関数は

$\displaystyle y_{c} = (c_{1} + c_{2}x)e^{-x}$

$\displaystyle y_{p} = u_{1}e^{-x} + u_{2}xe^{-x}$

$\displaystyle \left\{\begin{array}{ll} u_{1}^{\prime}e^{-x} + u_{2}^{\prime}xe^... ...x}) + u_{2}^{\prime}(e^{-x} - xe^{-x}) & = \frac{e^{-x}}{x} \end{array}\right.$

$\displaystyle u_{1}^{\prime}$	$\displaystyle =$	$\displaystyle \frac{\left\vert\begin{array}{cc} 0 & xe^{-x}\\ \frac{e^{-x}}{x}... ...-x} & e^{-x} - xe^{-x} \end{array}\right\vert } = \frac{-e^{-2x}}{e^{-2x}} = -1$
$\displaystyle u_{2}^{\prime}$	$\displaystyle =$	$\displaystyle \frac{\left\vert\begin{array}{cc} e^{-x} & 0\\ -e^{-x} & \frac{e... ...{x} \end{array}\right\vert }{e^{-2x}} = \frac{e^{-2x}/x}{e^{-2x}} = \frac{1}{x}$

$\displaystyle u_{1} = -x, \ u_{2} = \log{\vert x\vert} \$ (定数無視) $\displaystyle$

$\displaystyle y_{p} = -xe^{-x} + x\log{\vert x\vert}e^{-x}.$

$\displaystyle y = y_{c} + y_{p} = (c_{1} + c_{2}x)e^{-x} -xe^{-x} + x\log{\vert x\vert}e^{-x}$

(b) 補助方程式 $y^{\prime\prime} + 4y = 0$ の特性方程式は $m^2 + 4 = 0$

より特性根 $m = \pm 2i$ を得る．よって余関数は

$\displaystyle y_{c} = c_{1}\cos{2x} + c_{2}\sin{2x}$

$\displaystyle y_{p} = u_{1}\cos{2x} + u_{2}\sin{2x}$

$\displaystyle \left\{\begin{array}{ll} u_{1}^{\prime}\cos{2x} + u_{2}^{\prime}\... ...prime}(-2\sin{2x}) + u_{2}^{\prime}(2\cos{2x}) & = \tan{2x} \end{array}\right.$

$\displaystyle u_{1}^{\prime}$	$\displaystyle =$	$\displaystyle \frac{\left\vert\begin{array}{cc} 0 & \sin{2x}\\ \tan{2x} & 2\co... ...tan{2x}\sin{2x}}{2(\cos^{2}{2x} + \sin^{2}{2x})} = - \frac{\tan{2x}\sin{2x}}{2}$
$\displaystyle u_{2}^{\prime}$	$\displaystyle =$	$\displaystyle \frac{\left\vert\begin{array}{cc} \cos{2x} & 0\\ -2\sin{2x} & \tan{2x} \end{array}\right\vert }{2} = \frac{\sin{2x}}{2}$

$\displaystyle u_{1}$	$\displaystyle =$	$\displaystyle -\frac{1}{2}\int \frac{\sin^{2}{2x}}{\cos{2x}} = \frac{1}{2}\int \frac{\cos^{2}{2x} - 1}{\cos{2x}} dx$
	$\displaystyle =$	$\displaystyle \frac{1}{2}\int (\cos{2x} - \sec{2x})dx$
	$\displaystyle =$	$\displaystyle -\frac{1}{4}(\sin{2x} + \log{\vert\sec{2x} + \tan{2x}\vert})$
$\displaystyle u_{2}$	$\displaystyle =$	$\displaystyle \int \frac{\sin{2x}}{2}dx = - \frac{\cos{2x}}{4}$

$\displaystyle y_{p} = -\frac{1}{4}\cos{2x}( \sin{2x} + \log{\vert\sec{2x} + \tan{2x}\vert}) - \frac{1}{4}\cos{2x} \sin{2x} .$

$\displaystyle y = y_{c} + y_{p}$	$\displaystyle =$	$\displaystyle c_{1}\cos{2x} + c_{2}\sin{2x}$
	$\displaystyle -$	$\displaystyle \frac{1}{4}\cos{2x} \log{\vert\sec{2x} + \tan{2x}\vert} - \frac{1}{2}\sin{2x}\cos{2x}$

より特性根

を得る．よって余関数は

$\displaystyle y_{c} = (c_{1} + c_{2}x)e^{2x}$

$\displaystyle y_{p} = u_{1}e^{2x} + u_{2}xe^{2x}$

$\displaystyle \left\{\begin{array}{ll} u_{1}^{\prime}e^{2x} + u_{2}^{\prime}xe^... ...}) + u_{2}^{\prime}(e^{2x} + 2xe^{2x}) & = \frac{e^{2x}}{x} \end{array}\right.$

$\displaystyle u_{1}^{\prime}$	$\displaystyle =$	$\displaystyle \frac{\left\vert\begin{array}{cc} 0 & xe^{2x}\\ \frac{e^{2x}}{x}... ...x} & e^{2x} + 2xe^{2x} \end{array}\right\vert } = - \frac{e^{4x}}{e^{4x}} = - 1$
$\displaystyle u_{2}^{\prime}$	$\displaystyle =$	$\displaystyle \frac{\left\vert\begin{array}{cc} e^{2x} & 0\\ 2e^{2x} & \frac{e... ...{2x} + 2xe^{2x} \end{array}\right\vert} = \frac{e^{4x}/x}{e^{4x}} = \frac{1}{x}$

$\displaystyle y_{p} = - xe^{2x} + xe^{2x}\log{\vert x\vert}.$

$\displaystyle y = y_{c} + y_{p} = (c_{1} + c_{2}x)e^{2x} - xe^{2x} + xe^{2x}\log{\vert x\vert}$

(d) 補助方程式 $y^{\prime\prime} - 3y^{\prime} + 2y = 0$ の特性方程式は $m^2 - 3m + 2 = (m -1)(m-2) = 0$

より特性根 $m = 1, \ 2$ を得る．よって余関数は

$\displaystyle y_{c} = c_{1}e^{x} + c_{2}e^{2x}$

$\displaystyle y_{p} = u_{1}e^{x} + u_{2}e^{2x}$

$\displaystyle \left\{\begin{array}{ll} u_{1}^{\prime}e^{x} + u_{2}^{\prime}e^{2... ...}(e^{x}) + u_{2}^{\prime}(2e^{2x}) & = \frac{1}{1 + e^{-x}} \end{array}\right.$

$\displaystyle u_{1}^{\prime}$	$\displaystyle =$	$\displaystyle \frac{\left\vert\begin{array}{cc} 0 & e^{2x}\\ \frac{1}{1 + e^{-... ...ght\vert } = - \frac{e^{2x}/(1 + e^{-x})}{e^{3x}} = - \frac{e^{-x}}{1 + e^{-x}}$
$\displaystyle u_{2}^{\prime}$	$\displaystyle =$	$\displaystyle \frac{\left\vert\begin{array}{cc} e^{x} & 0\\ e^{x} & \frac{1}{1... ...y}\right\vert} = \frac{e^{x}/(1 + e^{-x})}{e^{3x}} = \frac{e^{-2x}}{1 + e^{-x}}$

$\displaystyle u_{1} = -\int \frac{e^{-x}}{1 + e^{-x}} dx = \log{(1 + e^{-x})}$

$\displaystyle u_{2}$	$\displaystyle =$	$\displaystyle \int \frac{e^{-2x}}{1 + e^{-x}} dx = \int (e^{-x} - \frac{e^{-x}}{1 + e^{-x}})dx$
	$\displaystyle =$	$\displaystyle - e^{-x} + \log{(1 + e^{-x})}$

$\displaystyle y_{p} = e^{x} \log{(1 + e^{-x})} + e^{2x}(- e^{-x} + \log{(1 + e^{-x}})).$

$\displaystyle y = y_{c} + y_{p} = c_{1}e^{x} + c_{2}e^{2x} + (e^{x} + e^{2x})\log{(1 + e^{-x})} - e^{x}$

(e) 補助方程式 $y^{\prime\prime\prime} + y^{\prime} = 0$ の特性方程式は $m^3 + m = m(m^2 + 1) = 0$

より特性根 $m = 0, \pm i$ を得る．よって余関数は

$\displaystyle y_{c} = c_{1} + c_{2}\cos{x} + c_{3}\sin{x}$

$\displaystyle y_{p} = u_{1} + u_{2}\cos{x} + u_{3}\sin{x}$

$\displaystyle \left\{\begin{array}{ll} u_{1}^{\prime} + u_{2}^{\prime}\cos{x} +... ...}^{\prime}(-\cos{x}) + u_{3}^{\prime}(-\sin{x}) & = \tan{x} \end{array}\right.$

$\displaystyle u_{1}^{\prime}$	$\displaystyle =$	$\displaystyle \frac{\left\vert\begin{array}{ccc} 0 & \cos{x} & \sin{x}\\ 0 & -... ...in{x} & \cos{x}\\ 0 & - \cos{x} & - \sin{x} \end{array}\right\vert } = \tan{x}$
$\displaystyle u_{2}^{\prime}$	$\displaystyle =$	$\displaystyle \frac{\left\vert\begin{array}{ccc} 1 & 0 & \sin{x}\\ 0 & 0 & \co... ...& - \cos{x} & - \sin{x} \end{array}\right\vert } = - \tan{x}\cos{x} = - \sin{x}$
$\displaystyle u_{3}^{\prime}$	$\displaystyle =$	$\displaystyle \frac{\left\vert\begin{array}{ccc} 1 & \cos{x} & 0\\ 0 & - \sin{... ...cos{x}\\ 0 & - \cos{x} & - \sin{x} \end{array}\right\vert } = - \sin{x}\tan{x}$

$\displaystyle u_{1} = \int \tan{x} dx = \int \frac{\sin{x}}{\cos{x}}dx = - \log{\vert\cos{x}\vert} = \log{\vert\sec{x}\vert}$

$\displaystyle u_{2} = - \int \sin{x}dx = \cos{x}$

$\displaystyle u_{3}$	$\displaystyle =$	$\displaystyle - \int \sin{x}\tan{x}dx = - \int \frac{\sin^{2}{x}}{\cos{x}} dx = \int \frac{\cos^{2}{x} - 1}{\cos{x}}dx$
	$\displaystyle =$	$\displaystyle \int (\cos{x} - \sec{x})dx = \sin{x} - \log{\vert\sec{x} + \tan{x}\vert}$

$\displaystyle y_{p}$	$\displaystyle =$	$\displaystyle \log{\vert\sec{x}\vert} + \cos^{2}{x} + \sin{x}(\sin{x} - \log{\vert\sec{x} + \tan{x}\vert})$
	$\displaystyle =$	$\displaystyle \log{\vert\sec{x}\vert} - \sin{x} \log{\vert\sec{x} + \tan{x}\vert} + 1$

$\displaystyle y = y_{c} + y_{p} = c_{1} + c_{2}\cos{x} + c_{3}\sin{x} + \log{\vert\sec{x}\vert} - \sin{x} \log{\vert\sec{x} + \tan{x}\vert}$

補助方程式 $y^{\prime\prime} + y = 0$ の特性方程式は $m^2 + 1 = 0$

より特性根 $m = \pm i$ を得る．よって余関数は

$\displaystyle y_{c} = c_{1}\cos{x} + c_{2}\sin{x}$

$\displaystyle y_{p} = u_{1}\cos{x} + u_{2}\sin{x}$

$\displaystyle \left\{\begin{array}{ll} u_{1}^{\prime}\cos{x} + u_{2}^{\prime}\s... ...u_{1}^{\prime}(-\sin{x}) + u_{2}^{\prime}(\cos{x}) & = f(x) \end{array}\right.$

$\displaystyle u_{1}^{\prime}$	$\displaystyle =$	$\displaystyle \frac{\left\vert\begin{array}{cc} 0 & \sin{x}\\ f(x) & \cos{x} \... ...os{x} & \sin{x}\\ - \sin{x} & \cos{x} \end{array}\right\vert } = - f(x)\sin{x}$
$\displaystyle u_{2}^{\prime}$	$\displaystyle =$	$\displaystyle \frac{\left\vert\begin{array}{cc} \cos{x} & 0\\ - \sin{x} & f(x)... ...\cos{x} & \sin{x}\\ - \sin{x} & \cos{x} \end{array}\right\vert } = f(x)\cos{x}$

$\displaystyle u_{1} = - \int f(x)\sin{x} dx = - \int_{a}^{x}f(t)\sin{t}dt$

$\displaystyle u_{2} = \int f(x)\cos{x}dx = \int_{a}^{x}f(t)\cos(t)dt$

$\displaystyle y_{p}$	$\displaystyle =$	$\displaystyle - \int_{a}^{x}f(t)\sin{t}dt \cos{x} + \int_{a}^{x}f(t)\cos(t)dt \sin{x}$
	$\displaystyle =$	$\displaystyle - \int_{a}^{x}f(t)\sin{t} \cos{x} dt + \int_{a}^{x}f(t)\cos(t) \sin{x} dt$
	$\displaystyle =$	$\displaystyle \int_{a}^{x} f(t)(\sin{x}\cos{t} - \cos{x}\sin{t})dt = \int_{a}^{x} f(t)(\sin{x}\cos{t} - \cos{x}\sin{t})dt$
	$\displaystyle =$	$\displaystyle \int_{a}^{x} f(t)\sin{(x - t)}dt$