第10章 10.3

(1)

$y'' = 2y'$

$y' = P$ おくと $y'' = P\frac{dP}{dy}$ これから

$P\frac{dP}{dy} = 2P$

$\int dP = \int 2dy P = 2y + C_1 \text{元に戻して}$

$\frac{dy}{dx} = 2y + C_1$

$\int \frac{1}{2y+C_1}dy = \int dx \frac{1}{2}\log(2y+C_1) = x + C_2，\log(2y+C_1) = 2x + 2C_2，e^{2x} = C_3 \text{ おくと}$

$2y + C_1 = C_3 e^{2x} y = Ae^{2x} + B \text{(A,B \text{は任意定数})}$

(2)

$xy'' + y' = 0$

$y' = P$ おくと $xP' + P = 0$ $P + \frac{1}{x}P = 0$ $a = \frac{1}{x}，\beta = 0$ おくと

$\int adx = \log x \text{これから}$

$P = \frac{1}{x}\left\{\int 0dx + C_1\right\} = \frac{C_1}{x} \text{元に戻して}$

$\frac{dy}{dx} = \frac{C_1}{x}$

$\int dy = \int\frac{C_1}{x}dx y = C_1 \log x + C_2 C_1 = A，C_2 = B \text{ おくと}$

$y = A \log x + B \text{(A,B \text{は任意定数})}$

(3)

$xy'' + 2y' = 0$

$y' = P$と置くと $xP' + 2P = 0$ $x\frac{dP}{dx} = -2P$

$\int \frac{1}{P}dP = \int \frac{-2}{x}dx$ $\log P = -2\log x + C_1$ $e^{C_1} = C_2$と置くと

$Px^2 = C_2$ 元に戻して

$\frac{dy}{dx}x^2 = C_2$

$\int dy = \int \frac{1}{x^2}C_2 dx$ $y = -\frac{C_2}{x} + C_3$ $C_2 = A、C_3 = B$と置くと

$y = -\frac{A}{x} + B$ （A,B は任意定数）

(4)

$xy'' - y' + 1 = 0$

$y' = P$と置くと $xP' - P + 1 = 0$ $x\frac{dP}{dx} = P - 1$

$\int \frac{1}{P-1}dP = \int \frac{1}{x}dx$ $\log(P-1) = \log x + C_1$ $e^{C_1} = C_2$と置くと

$\frac{P-1}{x} = C_2$ 元に戻して

$\frac{y'-1}{x} = C_2$ $\frac{dy}{dx} = C_2x + 1$

$\int dy = \int (C_2x + 1)dx$ $y = \frac{1}{2}C_2x^2 + x + C_3$ $\frac{1}{2}C_2 = A、C_3 = B$

$y = Ax^2 + x + B$ （A,B は任意定数）

(5)

$(x+1)y'' - 2y' = (x+1)^4$

$y' = P$ とおくと $(x+1)P' - 2P = (x+1)^4$ $\frac{dP}{dx} - \frac{2}{x+1}P = (x+1)^3$

$a = \frac{2}{x+1}$、$\beta = (x+1)^3$ とおくと

$\int \alpha dx = -2\log(x+1)$

$P = (x+1)^2 \{ \int (x+1)dx + C_1 \} = (x+1)^2 (\frac{1}{2}x^2 + x) + C_1\}$ 元に戻して

$\frac{dy}{dx} = (x+1)^2(\frac{1}{2}x^2 + x) + C_1(x+1)^2$

$\int dy = \int (x+1)^2(\frac{1}{2}x^2 + x) + C_1(x+1)^2 dx$ $C_1 - \frac{1}{2} = C_2$ とおくと

$= \int (x+1)^2 [\frac{1}{2}(x+1)^2 + C_2]dx$ $2C_2 = C_3$ とおくと

$= \frac{1}{2}\int (x+1)^4 + C_3(x+1)^2 dx$

$y = \frac{1}{2} \cdot \frac{1}{5}(x+1)^5 + \frac{C_3}{3}(x+1)^3 + C_4\}$ $\frac{C_3}{6} = A$、$\frac{C_4}{2} = B$ とおくと

$= \frac{1}{10}(x+1)^5 + A(x+1)^3 + B$ （A,B は任意定数）

(6)

$y'' = y + 1$

$y' = P$ とおくと $y'' = P\frac{dP}{dy}$ これから

$P\frac{dP}{dy} = y + 1$

$\int PdP = \int y + 1 dy$ $\frac{1}{2}P^2 = \frac{1}{2}y^2 + y + C_1$ $2C_1 = C_2$ とおくと

$P^2 = y^2 + 2y + C_2$ $P = \sqrt{y^2 + 2y + C_2}$ 元に戻して

$\frac{dy}{dx} = \sqrt{y^2 + 2y + C_2} = \sqrt{(y+1)^2 + C_3}$ （$C_2 = C_3 + 1$）

$\int \frac{1}{\sqrt{(y+1)^2 + C_3}} dy = \int dx$

$\log(y+1) + \sqrt{(y+1)^2 + C_3} = x + C_4$

$e^{C_1} = C_5 \text{ とおくと}$

$y + 1 + \sqrt{(y+1)^2 + C_3} = C_5 e^x$

$C_3 = A, C_5 = B \text{ とおくと}$

$y + 1 + \sqrt{(y+1)^2 + A} = Be^x$

$(A, B \text{は任意定数})$

(7)

$yy'' - (y')^2 = 0$

$y' = P \text{ とおくと} \quad y'' = P\frac{dP}{dy}$

$yP\frac{dP}{dy} - P^2 = 0 \quad \alpha = -\frac{1}{y}, \beta = 0 \text{ とおくと}$

$\int \alpha dy = -\log y$

$P = y\left\{ \int 0dy + C_1 \right\} = yC_1$

$\text{元に戻して}$

$\frac{dy}{dx} = yC_1$

$\int \frac{1}{C_1 y} dy = \int dx$

$\frac{1}{C_1} \log y = x + C_2$

$e^{C_1 C_2} = A, C_1 = B \text{ とおくと}$

$y = Ae^{Bx}$

$(A, B \text{は任意定数})$

(8)

$yy'' + (y')^2 = 1$

$y' = P \text{ とおくと} \quad y'' = P\frac{dP}{dy}$

$\text{これから}$

$yP\frac{dP}{dy} + P^2 - 1 = 0$

$y\frac{dP}{dy} = \frac{1-P^2}{P}$

$\int \frac{P}{1-P^2} dP = \int \frac{1}{y} dy$

$-\frac{1}{2}\log(1-P^2) = \log y + C_1$

$-2C_1 = C_2 \text{ とおくと}$

$y^2(1-P^2) = C_2$

$P = \sqrt{1 - \frac{C_2}{y^2}}$

$\text{元に戻して}$

$y' = \sqrt{1 - \frac{C_2}{y^2}}$

$\int \frac{1}{\sqrt{1 - \frac{C_2}{y^2}}} dy = \int dx$

$\int \frac{y}{\sqrt{y^2 - C_2}} dy = \int dx$

$\sqrt{y^2 - C_2} = x + C_3$

$-C_2 = A, C_3 = B \text{ とおくと}$

$(x+B)^2 = y^2 + A$

$(A, B \text{は任意定数})$