Question

If $x{y}^2=1,$ prove that $2\frac{dy}{dx}+y^3=0$

Answer 1

$\mathrm{We}\mathrm{have},x{y}^2=1...\left(\mathrm{i}\right)$
Differentiating with respect to x, we get,

$$\begin{gathered} \frac{d}{dx}\left(x{y}^2\right)=\frac{d}{dx}\left(1\right) \\ \Rightarrow x\frac{d}{dx}\left(y^2\right)+y^2\frac{d}{dx}\left(x\right)=0 \\ \Rightarrow x\left(2y\right)\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}+y^2\left(1\right)=0 \\ \Rightarrow 2xy\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=-y^2 \\ \Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{-y^2}{2xy} \\ \Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{-y}{2x} \\ \mathrm{put}x=\frac{1}{y^2}\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right) \\ \Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{-y}{2\left({\displaystyle \frac{1}{y^2}}\right)} \\ \Rightarrow 2\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=-y^3 \\ \Rightarrow 2\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}+y^3=0 \end{gathered}$$