第9章 9.7

(1)

$xy' = x + y$

$y' = 1 + \frac{y}{x} \qquad \frac{y}{x} = u \text{ とおくと} \qquad y' = u + xu' \qquad \text{これから}$

$u + x\frac{du}{dx} = 1 + u \qquad x\frac{du}{dx} = 1$

$\int du = \int \frac{1}{x} dx \qquad u = \log x + C_1 \qquad \text{元に戻して}$

$\frac{y}{x} = \log x + C_1 \qquad C_1 = C \text{ とおくと}$

$y = x(\log x + C) \qquad \text{(C \text{は任意定数})}$

(2)

$yy' + x - 2y = 0$

$y' + \frac{x}{y} - 2 = 0 \qquad \frac{y}{x} = u \text{ とおくと} \qquad y' = u + xu' \qquad \text{これから}$

$u + x\frac{du}{dx} + \frac{1}{u} - 2 = 0 \qquad u^2 + xu\frac{du}{dx} + 1 - 2u = 0 \qquad x\frac{du}{dx} = \frac{-(u^2 - 2u + 1)}{u}$

$\int \frac{u}{u^2 - 2u + 1} du = -\int \frac{1}{x} dx \qquad \int \left[\frac{1}{2} \cdot \frac{2u-2}{u^2-2u+1} + \frac{1}{(u-1)^2}\right] du = -\int \frac{1}{x} dx$

$\frac{1}{2}\log(u-1)^2 - \frac{1}{u-1} = -(\log x + C_1) \qquad \log x(u-1) = \frac{1}{u-1} - C_1 \qquad e^{-C_1} = C \text{ とおくと}$

$x(u-1) = C\exp\left(\frac{1}{u-1}\right) \qquad \text{元に戻して}$

$y = x + C\exp\left(\frac{x}{y-x}\right) \qquad \text{(C \text{は任意定数})}$

(3)

$(x^2+y^2)dx=xydy$

$\left(1+\dfrac{y^2}{x^2}\right)dx=\dfrac{y}{x}dy$ $\dfrac{y}{x}=u$ とおくと $y'=u+xu'$ これから

$1+u^2=u\left(u+x\dfrac{du}{dx}\right)$ $1=ux\dfrac{du}{dx}$

$\displaystyle\int u\,du=\int \dfrac{1}{x}dx$ $\dfrac{1}{2}u^2=\log x+C_1$ 元に戻して

$\dfrac{y^2}{2x^2}=\log x+C_1$ $2C_1=C$ とおくと

$y^2=x^2(2\log x+C)$ （Cは任意定数）

(4)

$(3x-y)=(x+y)y'$

$\left(3-\dfrac{y}{x}\right)=\left(1+\dfrac{y}{x}\right)y'$ $\dfrac{y}{x}=u$ とおくと $y'=u+xu'$ これから

$3-u=(1+u)\left(u+x\dfrac{du}{dx}\right)$ $\dfrac{3-u}{1+u}=u+x\dfrac{du}{dx}$ $\dfrac{3-2u-u^2}{1+u}=x\dfrac{du}{dx}$

$\displaystyle\int \dfrac{1}{x}dx=\int \dfrac{u+1}{-u^2-2u+3}du$ $\log x=-\dfrac{1}{2}\log(-u^2-2u+3)+C_1$

$\log x^2(-u^2-2u+3)=2C_1$ $x^2(-u^2-2u+3)=e^{2C_1}$ 元に戻して

$-y^2-2xy+3x^2=e^{2C_1}$

$-e^{2C_1}=C$ とおくと

$(y+3x)(y-x)=C$ （Cは任意定数）

(5)

$(x^2-y^2)y'=2xy$

$\left(1-\dfrac{y^2}{x^2}\right)y'=\dfrac{2y}{x}$ $\dfrac{y}{x}=u$ とおくと $y'=u+xu'$ これから

$(1-u^2)\left(u+x\dfrac{du}{dx}\right)=2u$ $u+x\dfrac{du}{dx}=\dfrac{2u}{1-u^2}$ $x\dfrac{du}{dx}=\dfrac{2u-u+u^3}{1-u^2}=\dfrac{u^3+u}{1-u^2}$

$\displaystyle\int \dfrac{1}{x}dx=\int \dfrac{1-u^2}{u^3+u}du$ $\displaystyle\int \dfrac{1}{x}dx=\int \dfrac{1}{u}-\dfrac{2u}{u^2+1}\,du$ $\log x+C_1=\log \dfrac{u}{u^2+1}$

$C_1 = \dfrac{u}{x(u^2+1)}$

元に戻して

$C_1 = \dfrac{\dfrac{y}{x}}{x\left(\dfrac{y^2}{x^2}+1\right)} = \dfrac{y}{y^2+x^2}$

$\dfrac{1}{C_1}=C$ とおくと

$Cy = y^2+x^2$

（Cは任意定数）

(6)

$y^2 + (xy+x^2)y' = 0$

$\dfrac{y^2}{x^2} + \left(\dfrac{y}{x}+1\right)y' = 0$

$\dfrac{y}{x}=u$ とおくと $y'=u+xu'$ これから

$u^2 + (u+1)\left(u+x\dfrac{du}{dx}\right) = 0$

$x\dfrac{du}{dx} = -\dfrac{2u^2+u}{u+1}$

$\int -\dfrac{u+1}{2u^2+u}\,du = \int \dfrac{1}{x}\,dx$

$-\int \dfrac{1}{u} + \dfrac{-1}{2u+1}\,du = \int \dfrac{1}{x}\,dx$

$-\log u + \dfrac{1}{2}\log(2u+1) = \log x + C_1$

$-2\log u + \log(2u+1) = 2\log x + 2C_1$

$\log \dfrac{x^2u^2}{2u+1} = -2C_1$

$e^{2C_1}=C$ とおくと

$\dfrac{x^2u^2}{2u+1} = C$

元に戻して

$\dfrac{y^2}{\dfrac{2y+x}{x}} = C$

$xy^2 = C(2y+x)$

（Cは任意定数）