If \(x\sqrt{1+y} + y\sqrt{1+x} = 0\); \(x \ne y\), then value of \(\displaystyle\frac{d^{2}y}{dx^{2}} + 2\frac{dy}{dx}\) at \(x = 1\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(-\displaystyle\frac{1}{4}\)
Moving one term across and squaring gives y in terms of x:
\(x\sqrt{1+y} = -y\sqrt{1+x}\)
\(x^{2}(1+y) = y^{2}(1+x)\)
\(x^{2} - y^{2} + x^{2}y - xy^{2} = 0\)
\((x - y)(x + y) + xy(x - y) = 0\)
\((x - y)(x + y + xy) = 0\)
Since \(x \ne y\), the factor \(x - y\) can be cancelled:
\(x + y + xy = 0\)
\( \displaystyle y = -\frac{x}{1+x} = -1 + \frac{1}{1+x} \)
Differentiating twice:
\( \displaystyle \frac{dy}{dx} = -\frac{1}{(1+x)^{2}} \)
\( \displaystyle \frac{d^{2}y}{dx^{2}} = \frac{2}{(1+x)^{3}} \)
At \(x = 1\):
\( \displaystyle \frac{dy}{dx} = -\frac{1}{4}, \quad \frac{d^{2}y}{dx^{2}} = \frac{2}{8} = \frac{1}{4} \)
\( \displaystyle \frac{d^{2}y}{dx^{2}} + 2\frac{dy}{dx} = \frac{1}{4} - \frac{2}{4} = -\frac{1}{4} \)