operatornameuv^2x=vx-u^2 Denklemini Çözme

$$\operatorname { uv } ^ { 2 } x = v x - u ^ { 2 }$$

u için çözün (complex solution)

$$u=-\left(xv^{2}+1\right)^{-\frac{1}{2}}\sqrt{v}\sqrt{x}$$

$$u=\left(xv^{2}+1\right)^{-\frac{1}{2}}\sqrt{v}\sqrt{x}\text{, }x=0\text{ or }\left(v\neq -ix^{-\frac{1}{2}}\text{ and }v\neq ix^{-\frac{1}{2}}\right)$$

v için çözün (complex solution)

$$\left\{\begin{matrix}v=\frac{\sqrt{x\left(x-4u^{4}\right)}+x}{2xu^{2}}\text{; }v=\frac{-\sqrt{x\left(x-4u^{4}\right)}+x}{2xu^{2}}\text{, }&x\neq 0\text{ and }u\neq 0\\v=0\text{, }&u=0\text{ and }x\neq 0\\v\in \mathrm{C}\text{, }&x=0\text{ and }u=0\end{matrix}\right,$$

u için çözün

$$u=\sqrt{\frac{vx}{xv^{2}+1}}$$

$$u=-\sqrt{\frac{vx}{xv^{2}+1}}\text{, }v>\sqrt{-\frac{1}{x}}\text{ or }v=0\text{ or }\left(v\leq 0\text{ and }v\neq \sqrt{-\frac{1}{x}}\text{ and }v>-\sqrt{-\frac{1}{x}}\text{ and }x<0\right)\text{ or }x\geq 0$$

v için çözün

$$\left\{\begin{matrix}v=\frac{\sqrt{x\left(x-4u^{4}\right)}+x}{2xu^{2}}\text{; }v=\frac{-\sqrt{x\left(x-4u^{4}\right)}+x}{2xu^{2}}\text{, }&x\neq 0\text{ and }u\neq 0\text{ and }\left(|u|\leq \frac{\sqrt{2}\sqrt[4]{x}}{2}\text{ or }x<0\right)\\v=0\text{, }&u=0\text{ and }x\neq 0\\v\in \mathrm{R}\text{, }&x=0\text{ and }u=0\end{matrix}\right,$$