第9章 9.28

(1)

$y' + 5y = 0$ $y_{(0)} = 1$

$\frac{dy}{dx} = -5y$

$\int \frac{1}{y} dy = \int -5dx$ $\log y = -5x + C_1$ $e^{C_1} = C$ と おくと

$y = Ce^{-5x}$

$y_{(0)} = 1$ から $y_{(0)} = 1 = C$ よって

$y = e^{-5x}$

(2)

$y' = \frac{y}{x(x+1)(x+2)}$ $y_{(1)} = 1$

$\frac{dy}{dx} = y\left(\frac{1}{2} \cdot \frac{1}{x} + \frac{-1}{x+1} + \frac{1}{2} \cdot \frac{1}{x+2}\right)$

$\int \frac{1}{y} dy = \int \left(\frac{1}{2} \cdot \frac{1}{x} + \frac{-1}{x+1} + \frac{1}{2} \cdot \frac{1}{x+2}\right) dx$ $\log y = \log \frac{\sqrt{x}\sqrt{x+2}}{x+1} + C_1$ $e^{C_1} = C_2$ と おくと

$y = C_2 \frac{\sqrt{x}\sqrt{x+2}}{x+1}$ $C_2^2 = C$ と おくと

$y^2 = C \frac{x(x+2)}{(x+1)^2}$

$y_{(1)} = \sqrt{C \cdot \frac{3}{4}} = 1$ $C = \frac{4}{3}$ よって

$y^2 = \frac{4x(x+2)}{3(x+1)^2}$

(3)

$y' - \frac{y}{x} = x^2, \quad y_{(1)} = \frac{3}{2}$

$P = -\frac{1}{x}, \quad Q = x^2 \text{ と \text{おくと} } \int Pdx = -\log x \text{ これから}$

$y = x\left\{\int xdx + C\right\} = x\left\{\frac{1}{2}x^2 + C\right\} = \frac{1}{2}x^3 + Cx$

$y_{(1)} = \frac{1}{2} + C = \frac{3}{2}, \quad C = 1 \text{ よって}$

$y = \frac{1}{2}x^3 + x$

(4)

$xy' = 2y - x, \quad y_{(3)} = 0$

$y' - \frac{2}{x}y = -1$

$P = -\frac{2}{x}, \quad Q = -1 \text{ と \text{おくと} } \int Pdx = -2\log x = \log x^{-2} \text{ これから}$

$y = x^2\left\{\int -\frac{1}{x^2}dx + C\right\} = x^2\left\{\frac{1}{x} + C\right\} = x + Cx^2$

$y_{(3)} = 3 + 9C = 0, \quad C = -\frac{1}{3} \text{ よって}$

$y = x - \frac{x^2}{3}$