Question

solve the equation
$x \sqrt{1 + y} + y \sqrt{1 + x} = 0 \: and \: x \: not \: equal \: to \: y \: \: \\ then \: prove \: that \: \frac{dy}{dx} = \frac{1}{ {x + 1}^ 2}$
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Answer 1

${\large{\textsf{\underline{\underline{Given :}}}}}$

$x\sqrt{1+y} +y\sqrt{1+x} =0$

${\large{\underline{\underline{\textsf{To Prove :}}}}}$

$\frac{dy}{dx}=\frac{1}{(x+1)^2}$

${\large{\textsf{\underline{\underline{Solution :}}}}}$

$\implies x\sqrt{1+y} + y\sqrt{1+x}=0$

$\implies x\sqrt{1+y}=-y\sqrt{1+x}$

$\text{Squaring on both sides}$

$\implies x^2(1+y)=-y^2(1+x)$

$\implies x^2+x^2y=y^2+xy^2$

$\implies x^2-y^2=xy^2-x^2y$

$\implies (x+y)(x-y) = -xy(x+y)$

$\implies (x-y) = -xy\implies x-y+xy=0$

$\implies x+y(1+x)=0\implies y(1+x)=-x$

$\implies y=\frac{-x}{(x+1)^2}$

$\implies \frac{dy}{dx}=\frac{(1+x)(-1)-(-x)(1)}{(1+x)^2}$

$\implies \frac{-1-x-(-x)}{(1+x)^2} \implies \frac{-1+x-x}{(1+x)^2}$

$\implies {\bf {\frac{-1}{(1+x)^2}}}$

$\text{Hence Proved}$