Question

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

Answer 1

We have,

`xy^2 = 1`            .............(1)

Differentiating with respect to x, we get,

\[\frac{d}{dx}\left( x y^2 \right) = \frac{d}{dx}\left( 1 \right)\]

\[ \Rightarrow x\frac{d}{dx}\left( y^2 \right) + y^2 \frac{d}{dx}\left( x \right) = 0 \]

\[ \Rightarrow x\left( 2y \right)\frac{d y}{d x} + y^2 \left( 1 \right) = 0\]

\[ \Rightarrow 2xy\frac{d y}{d x} = - y^2 \]

\[ \Rightarrow \frac{d y}{d x} = \frac{- y^2}{2xy}\]

\[ \Rightarrow \frac{d y}{d x} = \frac{- y}{2x}\]

\[\text{ put x } = \frac{1}{y^2} \text{ from equation } \left( 1 \right)\]

\[ \Rightarrow \frac{d y}{d x} = \frac{- y}{2\left( \frac{1}{y^2} \right)}\]

\[ \Rightarrow 2\frac{d y}{d x} = - y^3 \]

\[ \Rightarrow 2\frac{d y}{d x} + y^3 = 0\]