Question

If $x\sqrt{1+y}+y\sqrt{1+x}=0$, prove that ${\left(1+x\right)}^2\frac{dy}{dx}+1=0$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},x\sqrt{1+y}+y\sqrt{1+x}=0 \\ \Rightarrow x\sqrt{1+y}=-y\sqrt{1+x} \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}, \\ \Rightarrow {\left(x\sqrt{1+y}\right)}^2={\left(-y\sqrt{1+x}\right)}^2 \\ \Rightarrow x^2\left(1+y\right)=y^2\left(1+x\right) \\ \Rightarrow x^2+x^{2}y=y^2+y^{2}x \\ \Rightarrow x^2-y^2=y^{2}x-x^{2}y \\ \Rightarrow \left(x-y\right)\left(x+y\right)=xy\left(y-x\right) \\ \Rightarrow \left(x+y\right)=-xy \\ \Rightarrow y+xy=-x \\ \Rightarrow y\left(1+x\right)=-x \\ \Rightarrow y=\frac{-x}{\left(1+x\right)} \end{gathered}$$
Differentiating with respect to x, we get,
$$\begin{gathered} \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\left[\frac{-\left(1+x\right){\displaystyle \frac{d}{dx}}\left(x\right)-\left(-x\right){\displaystyle \frac{d}{dx}}\left(x+1\right)}{{\left(1+x\right)}^2}\right] \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\left[\frac{-\left(1+x\right)\left(1\right)+x\left(1\right)}{{\left(1+x\right)}^2}\right] \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\left[\frac{-1-x+x}{{\left(1+x\right)}^2}\right] \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\frac{-1}{{\left(1+x\right)}^2} \\ \Rightarrow {\left(1+x\right)}^2\frac{\mathit{d}y}{\mathit{d}x}=-1 \\ \Rightarrow {\left(1+x\right)}^2\frac{\mathit{d}y}{\mathit{d}x}+1=0 \end{gathered}$$