udydy/udy+ydu=y+sqrt{u^2y^2-y^2} を解きます| Microsoft 数学ソルバー

$$u d y \frac { d y } { u d y + y d u } = y + \sqrt { u ^ { 2 } y ^ { 2 } - y ^ { 2 } }$$

d を解く (複素数の解)

$$d=\frac{2\left(\sqrt{y^{2}\left(u^{2}-1\right)}+y\right)}{y}$$

$$\left(u\neq \sqrt{2}\text{ or }arg(y)<\pi \right)\text{ and }\left(u\neq -\sqrt{2}\text{ or }arg(y)<\pi \right)\text{ and }y\neq 0\text{ and }u\neq 0$$

u を解く (複素数の解)

$$\left\{\begin{matrix}u=1\text{; }u=-1\text{, }&d=2\text{ and }y\neq 0\\u=-\frac{i\sqrt{-d^{2}+4d-8}}{2}\text{; }u=\frac{i\sqrt{-d^{2}+4d-8}}{2}\text{, }&d\neq 0\text{ and }y\neq 0\text{ and }d\neq 2-2i\text{ and }d\neq 2+2i\text{ and }arg(\frac{dy}{2}-y)<\pi \end{matrix}\right.$$

d を解く

$$d=\frac{2\left(|y|\sqrt{u^{2}-1}+y\right)}{y}$$

$$\left(y\neq 0\text{ and }u\neq -\sqrt{2}\text{ and }u\leq -1\right)\text{ or }\left(y\neq 0\text{ and }u\neq \sqrt{2}\text{ and }u\geq 1\right)\text{ or }\left(y>0\text{ and }|u|\geq 1\right)$$

u を解く

$$u=\frac{\sqrt{d^{2}-4d+8}}{2}$$

$$u=-\frac{\sqrt{d^{2}-4d+8}}{2}\text{, }\left(d\neq 0\text{ and }d\leq 2\text{ and }y<0\right)\text{ or }\left(d\geq 2\text{ and }y>0\right)$$