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Question
If xy = 4, prove that \[x\left( \frac{dy}{dx} + y^2 \right) = 3 y\] ?
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Solution
\[\text{We have, xy } = 4\]
\[ \Rightarrow y = \frac{4}{x}\]
Differentiate it with respect to x,
\[\frac{d y}{d x} = \frac{d}{dx}\left( \frac{4}{x} \right)\]
\[ \Rightarrow \frac{d y}{d x} = 4\frac{d}{dx}\left( x^{- 1} \right)\]
\[ \Rightarrow \frac{d y}{d x} = 4\left( - 1 \times x^{- 1 - 1} \right)\]
\[ \Rightarrow \frac{d y}{d x} = 4\left( - \frac{1}{x^2} \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{- 4}{x^2}\]
\[ \Rightarrow \frac{d y}{d x} = - \frac{4}{\left( \frac{4}{y} \right)^2} \left[ \because x = \frac{4}{y} \right]\]
\[ \Rightarrow \frac{d y}{d x} = - \frac{4 y^2}{16}\]
\[ \Rightarrow \frac{d y}{d x} = - \frac{y^2}{4}\]
\[ \Rightarrow 4\frac{d y}{d x} = - y^2 \]
\[ \Rightarrow 4\frac{d y}{d x} = 3 y^2 - 4 y^2 \]
\[ \Rightarrow 4\frac{d y}{d x} + 4 y^2 = 3 y^2 \]
\[ \Rightarrow 4\left( \frac{d y}{d x} + y^2 \right) = 3 y^2 \]
Dividing both side by x,
\[\Rightarrow \frac{4}{x}\left( \frac{d y}{d x} + y^2 \right) = \frac{3 y^2}{x}\]
\[ \Rightarrow y\left( \frac{d y}{d x} + y^2 \right) = \frac{3 y^2}{x} \]
\[ \Rightarrow x\left( \frac{d y}{d x} + y^2 \right) = \frac{3 y^2}{y}\]
\[ \Rightarrow x\left( \frac{d y}{d x} + y^2 \right) = 3y\]
