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(B) Given the differential equation: $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$.By separating the variables,we get: $\frac{dy}{1+y^2} = \frac{dx}{1+x^2}$.I

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$y-x = c(1+xy)$,where $c$ is a constant of integration.

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(B) Given the differential equation: $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$.By separating the variables,we get: $\frac{dy}{1+y^2} = \frac{dx}{1+x^2}$.Integrating both sides: $\int \frac{dy}{1+y^2} = \int \frac{dx}{1+x^2}$.This yields: $	an^{-1}(y) = 	an^{-1}(x) + C$,where $C$ is a constant of integration.Rearranging the terms: $	an^{-1}(y) - 	an^{-1}(x) = C$.Using the formula $	an^{-1}(A) - 	an^{-1}(B) = 	an^{-1}(\frac{A-B}{1+AB})$,we get: $	an^{-1}(\frac{y-x}{1+xy}) = C$.Taking the tangent of both sides: $\frac{y-x}{1+xy} = 	an(C)$.Let $	an(C) = c$,where $c$ is a new constant.Thus,$y-x = c(1+xy)$.

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

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