The solution of the differential equation dy/dx = 1+y²/1+x²

The correct option is y – x = k(1 + xy).

To find the solution of the given differential equation, we start with:
$$\frac{dy}{dx}=\frac{1+y^2}{1+x^2}$$

We can solve this differential equation by using the method of separation of variables. By separating the variables, we group all terms containing y with dy on one side, and all terms containing x with dx on the other side:
$$\frac{dy}{1+y^2}=\frac{dx}{1+x^2}$$

Now, we integrate both sides of the equation:
$$\int \frac{dy}{1+y^2}=\int \frac{dx}{1+x^2}$$

Using the standard integration formula $\int \frac{du}{1+u^2}={\tan }^{-1}u+C$, we get:
$${\tan }^{-1}y={\tan }^{-1}x+c$$
where c is the constant of integration.

Rearranging the terms, we obtain:
$${\tan }^{-1}y-{\tan }^{-1}x=c$$

Next, we apply the trigonometric identity for the difference of arctangents:
${\tan }^{-1}A-{\tan }^{-1}B={\tan }^{-1}\left(\frac{A-B}{1+AB}\right)$
Applying this to our equation gives:
$${\tan }^{-1}\left(\frac{y-x}{1+xy}\right)=c$$

Taking the tangent of both sides to eliminate the inverse tangent:
$$\frac{y-x}{1+xy}=\tan \left(c\right)$$

Since c is a constant, $\tan \left(c\right)$ is also a constant. Let us define a new constant k such that k = tan(c).
This simplifies the expression to:
$$\frac{y-x}{1+xy}=k$$

Multiplying both sides by $(1+xy)$ yields the final solution:
$$y-x=k(1+xy)$$