Question

The solution of the differential equation $\frac{dy}{dx}=\frac{1+y^2}{1+x^2}$ is

[SCRA 1986]

• A. $1+xy+c(y+x)=0$
• B. $x+y=c(1-xy)$
• C. $y-x=c(1+xy)$
• D. $1+xy=c(x+y)$

Answer 1

The correct option is C $y-x=c(1+xy)$

$\frac{dy}{dx}=\frac{1+y^2}{1+x^2}\Rightarrow \frac{1}{1+y^2}dy=\frac{1}{1+x^2}dx$
Now on integrating both sides, we get
$ta{n}^{-1}y=ta{n}^{-1}x+ta{n}^{-1}c\Rightarrow ta{n}^{-1}y=ta{n}^{-1}\left(\frac{x+c}{1-cx}\right)\Rightarrow y=\frac{x+c}{1-cx}\Rightarrow y-x=c(1+xy).$