第10章 10.8

(1)

$xy''-(1+x)y'+y = 0$

$m^2x - m(1+x)+1 = 0 \quad (m-1)(mx-1) = 0 \quad m=1 \quad \text{よって}y = e^x\text{を特殊解に持つ}$

$y = e^x\text{とおくと} \quad y' = e^xv + e^xv' \quad y'' = e^xv + 2e^xv' + e^xv'' \quad \text{これを与式に代入して}$

$x(e^xv + 2e^xv' + e^xv'') - (1-x)(e^xv + e^xv') + e^xv = 0 \quad xy'' + xy' - y' = 0$

$xy'' + y'(x+1) = 0 \quad y' = P\text{とおくと}$

$xP' + (x-1)P = 0 \quad \frac{1}{P}\frac{dP}{dx} = -\frac{x-1}{x} = -1 + \frac{1}{x}$

$\int\frac{1}{P}dP = \int\left(-1+\frac{1}{x}\right)dx \quad \log P = -x + \log x + C_1 \quad e^{C_1} = C_2\text{とおくと}$

$P = C_2xe^{-x} \quad \text{元に戻して}$

$y' = C_2xe^{-x}$

$y = C_2\{-xe^{-x} + \int e^{-x}dx + C_3\} = C_2\{-xe^{-x} - e^{-x} + C_3\} \quad \text{元に戻して}$

$ye^{-x} = C_2\{-(x+1)e^{-x} + C_3\} \quad -C_2 = A、C_2C_3 = B\text{とおくと}$

$y = A(x+1) + Be^x \quad \text{（A,B \text{は任意定数}）}$

(2)

$xy''-(2x+1)y'+(x+1)y = 0$

$m^2x - m(2x+1)+x+1 = 0 \quad \{mx-(x+1)\}(m-1) = 0 \quad m=1 \quad \text{よって}y = e^x\text{を特殊解に持つ}$

$y = e^xv\text{とおくと} \quad y' = e^xv + e^xv' \quad y'' = e^xv + 2e^xv' + e^xv'' \quad \text{これを与式に代入して}$

$x(e^xv + 2e^xv' + e^xv'') - (2x+1)(e^xv + e^xv') + (x+1)e^xv = 0 \quad xy'' - y' = 0 \quad y' = P\text{とおくと}$

$xP' - P = 0$

$\int\frac{1}{P}dP = \int\frac{1}{x}dx \quad \log P = \log x + C_1 \quad e^{C_1} = C_2\text{とおくと}$

$P = C_2x \quad \text{元に戻して}$

$y' = C_2x$

$v = \frac{1}{2}C_2 x^2 + C_3$ 元に戻して

$ye^{-x} = \frac{1}{2}C_2 x^2 + C_3$ $\frac{1}{2}C_2 = A, C_3 = B$ とおくと

$y = e^x(Ax^2 + B)$ (A,B は任意定数)

(3)

$(x + 3)y'' - (2x + 7)y' + 2y = 0$

$m^2(x + 3) - m(2x + 7) + 2 = 0$ $\{m(x+3) - 1\}(m-2) = 0$ $m = 2$ よって$y = e^{2x}$を特殊解に持つ

$y = e^{2x}v$ とおくと $y' = 2e^{2x}v + e^{2x}v'$ $y'' = 4e^{2x}v + 4e^{2x}v' + e^{2x}v''$ これを与式に代入して

$(x+3)(4e^{2x} + 4e^{2x}v' + e^{2x}v'') - (2x+7)(2e^{2x}v + e^{2x}v') + 2e^{2x}v = 0$

$(x+3)v'' + (2x + 5)v = 0$ $v' = P$ とおくと

$(x+3)P' + (2x + 5)P = 0$

$\int \frac{1}{P}dP = \int \frac{2x+5}{x+3}dx = \int\left(-2 + \frac{1}{x+3}\right)dx$

$\log P = -2x + \log(x+3) + C_1$ $e^{C_1} = C_2$ とおくと

$P = (x+3)e^{-2x}C_2$ 元に戻して

$v' = C_2(x+3)e^{-2x}$

$v = C_2\left\{-\frac{1}{2}(x+3)e^{-2x} + \int \frac{1}{2}e^{-2x}dx + C_3\right\} = C_2\left\{-\frac{1}{2}(x+3)e^{-2x} - \frac{1}{4}e^{-2x} + C_3\right\}$

元に戻して $-\frac{1}{4}C_2 = A, -\frac{1}{4}C_2C_3 = B$ とおくと

$y = A(2x + 7) + Be^{2x}$ (A,B は任意定数)