Question

The solution of the differential equation (x² + 1) $\frac{dy}{dx}$ + (y² + 1) = 0, is
(a) y = 2 + x²

(b) $y=\frac{1+x}{1-x}$

(c) y = x (x − 1)

(d) $y=\frac{1-x}{1+x}$

Answer 1

$\left(\mathrm{d}\right)y=\frac{1-x}{1+x}$


$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \left(x^2+1\right)\frac{dy}{dx}+(y^2+1)=0 \\ \Rightarrow \left(x^2+1\right)\frac{dy}{dx}=-\left(y^2+1\right) \\ \Rightarrow \frac{1}{\left(y^2+1\right)}dy=-\frac{1}{\left(x^2+1\right)}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{\left(y^2+1\right)}dy=-\int \frac{1}{\left(x^2+1\right)}dx \\ \Rightarrow {\tan }^{-1}y=-{\tan }^{-1}x+{\tan }^{-1}C \\ \Rightarrow {\tan }^{-1}y+{\tan }^{-1}x={\tan }^{-1}C \\ \Rightarrow {\tan }^{-1}\left(\frac{x+y}{1-xy}\right)={\tan }^{-1}C \\ \Rightarrow \frac{x+y}{1-xy}=C \\ Disclaimer:Theinitialvalueconditionsarenotgiven,sothefinalanswerwillbeobatinedonlyif \\ C=1.So, \\ \Rightarrow x+y=1-xy \\ \Rightarrow y+xy=1-x \\ \Rightarrow y\left(1+x\right)=1-x \\ \Rightarrow y=\frac{1-x}{1+x} \end{gathered}$$