Question

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

Answer 1

$\mathrm{We}\mathrm{have},y\sqrt{x^2+1}=\log \left(\sqrt{x^2+1}-x\right)$
Differentiating with respect to x, we get,
$$\begin{gathered} \Rightarrow \frac{d}{dx}\left(y\sqrt{x^2+1}\right)=\frac{d}{dx}\log \left(\sqrt{x^2+1}-x\right)\left[\mathrm{using}\mathrm{product}\mathrm{rule}\mathrm{and}\mathrm{chain}\mathrm{rule}\right] \\ \Rightarrow y\frac{d}{dx}\left(\sqrt{x^2+1}\right)+\sqrt{x^2+1}\frac{dy}{dx}=\frac{1}{\left(\sqrt{x^2+1}-x\right)}\times \frac{d}{dx}\left(\sqrt{x^2+1}-x\right) \\ \Rightarrow \frac{y}{2\sqrt{x^2+1}}\times \frac{d}{dx}\left(x^2+1\right)+\sqrt{x^2+1}\frac{dy}{dx}=\frac{1}{\left(\sqrt{x^2+1}-x\right)}\times \left[\frac{1}{2\sqrt{x^2+1}}\frac{d}{dx}\left(x^2+1\right)-1\right] \\ \Rightarrow \frac{2xy}{2\sqrt{x^2+1}}+\sqrt{x^2+1}\frac{dy}{dx}=\frac{1}{\left(\sqrt{x^2+1}-x\right)}\left[\frac{2x}{2\sqrt{x^2+1}}-1\right] \\ \Rightarrow \sqrt{x^2+1}\frac{dy}{dx}=\left[\frac{1}{\sqrt{x^2+1}-x}\right]\left[\frac{x-\sqrt{x^2+1}}{\sqrt{x^2+1}}\right]-\frac{xy}{\sqrt{x^2+1}} \\ \Rightarrow \sqrt{x^2+1}\frac{dy}{dx}=\frac{-1}{\sqrt{x^2+1}}-\frac{xy}{\sqrt{x^2+1}} \\ \Rightarrow \sqrt{x^2+1}\frac{dy}{dx}=\frac{-\left(1+xy\right)}{\sqrt{x^2+1}} \\ \Rightarrow \left(x^2+1\right)\frac{dy}{dx}=-\left(1+xy\right) \\ \Rightarrow \left(x^2+1\right)\frac{dy}{dx}+1+xy=0 \end{gathered}$$