第9章 9.31

$\int y \, dx = k\int_0 \sqrt{1 + y'^2} \, dx \text{ （\text{曲線の長さ}）}$

$yx = k\sqrt{1 + y'^2} x \quad y = k\sqrt{1 + y'^2} \quad y' = \frac{1}{k}\sqrt{y^2 - k^2}$

$\int \frac{1}{\sqrt{y^2 - k^2}} \, dy = \int \frac{1}{k} \, dx \quad \log(y - \sqrt{y^2 - k^2}) = \frac{x}{k} + C_1 \quad e^{C_1} = C \text{ とおくと}$

$y - \sqrt{y^2 - k^2} = Ce^{x/k} \quad y^2 - k^2 = y^2 - 2Cye^{x/k} + C^2e^{2x/k}$

$y = \frac{1}{2C}(k^2e^{-x/k} + C^2e^{x/k})$

$y(0) = k \text{ から}$

$k = \frac{1}{2C}(k^2 + C^2) \quad 2Ck = k^2 + C^2$

$(C - k)^2 = 0 \quad C = k \text{ これから}$

$y = \frac{1}{2k}(k^2e^{-x/k} + k^2e^{x/k}) = \frac{k}{2}(e^{-x/k} + e^{x/k})$