The solution of the differential equation $\frac{dy}{dx}=\frac{1+y^2}{1+x^2}, $ is
Answer & explanation
Correct answer: option 2
The correct answer i soption (2) : $dy-x=C(1+xy)$
We have,
$\frac{dy}{dx}=\frac{1+y^2}{1+x^2}$
$⇒\frac{1}{1+y^2}dy =\frac{1}{1+x^2}dx$
On integrating, we get
$tan^{-1} y = tan^{-1} x + tan^{-1} C$
$⇒tan^{-1} y - tan^{-1}x + tan^{-1}C$
$⇒tan^{-1} \left(\frac{y-x}{1+xy}\right) = tan^{-1}C$
$⇒\frac{y-x}{1+xy}=C$
$⇒y - x = C (1+ xy)$