tan⁻¹ x + tan⁻¹ y = c is the general solution of the differential equation

The correct option is: (1 + x²)dy + (1 + y²)dx = 0

To find the differential equation corresponding to the given general solution, we need to eliminate the arbitrary constant $c$ by differentiating the equation with respect to $x$.

The given general solution is:
${\tan }^{-1}x+{\tan }^{-1}y=c$

Differentiating both sides of the equation with respect to $x$, we use the chain rule for the term involving $y$:
$\frac{d}{dx}\left({\tan }^{-1}x\right)+\frac{d}{dx}\left({\tan }^{-1}y\right)=\frac{d}{dx}\left(c\right)$

Applying the standard derivative formula $\frac{d}{dt}\left({\tan }^{-1}t\right)=\frac{1}{1+t^2}$ and knowing that the derivative of a constant is zero, we get:
$\frac{1}{1+x^2}+\frac{1}{1+y^2}·\frac{dy}{dx}=0$

To write this differential equation in differential form, we multiply the entire equation by $dx$:
$\frac{1}{1+x^2}dx+\frac{1}{1+y^2}dy=0$

Now, to clear the denominators, we multiply the entire equation by $(1+x^2)(1+y^2)$:
$(1+y^2)dx+(1+x^2)dy=0$

Rearranging the terms, we get:
$(1+x^2)dy+(1+y^2)dx=0$

This matches the correct option.