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प्रश्न
If \[x \sqrt{1 + y} + y \sqrt{1 + x} = 0\] , prove that \[\left( 1 + x \right)^2 \frac{dy}{dx} + 1 = 0\] ?
उत्तर
\[\text { We have }, x\sqrt{1 + y} + y\sqrt{1 + x} = 0\]
\[ \Rightarrow x\sqrt{1 + y} = - y\sqrt{1 + x}\]
\[\text{ Squaring both sides, we get } , \]
\[ \Rightarrow \left( x\sqrt{1 + y} \right)^2 = \left( - y\sqrt{1 + x} \right)^2 \]
\[ \Rightarrow x^2 \left( 1 + y \right) = y^2 \left( 1 + x \right)\]
\[ \Rightarrow x^2 + x^2 y = y^2 + y^2 x\]
\[ \Rightarrow x^2 - y^2 = y^2 x - x^2 y\]
\[ \Rightarrow \left( x - y \right)\left( x + y \right) = xy\left( y - x \right)\]
\[ \Rightarrow \left( x + y \right) = - xy\]
\[ \Rightarrow y + xy = - x\]
\[ \Rightarrow y\left( 1 + x \right) = - x\]
\[ \Rightarrow y = \frac{- x}{\left( 1 + x \right)}\]
Differentiating with respect to x, we get,
\[\Rightarrow \frac{d y}{d x} = \left[ \frac{- \left( 1 + x \right)\frac{d}{dx}\left( x \right) - \left( - x \right)\frac{d}{dx}\left( x + 1 \right)}{\left( 1 + x \right)^2} \right]\]
\[ \Rightarrow \frac{d y}{d x} = \left[ \frac{- \left( 1 + x \right)\left( 1 \right) + x\left( 1 \right)}{\left( 1 + x \right)^2} \right]\]
\[ \Rightarrow \frac{d y}{d x} = \left[ \frac{- 1 - x + x}{\left( 1 + x \right)^2} \right]\]
\[ \Rightarrow \frac{d y}{d x} = \frac{- 1}{\left( 1 + x \right)^2}\]
\[ \Rightarrow \left( 1 + x \right)^2 \frac{d y}{d x} = - 1\]
\[ \Rightarrow \left( 1 + x \right)^2 \frac{d y}{d x} + 1 = 0\]
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