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(C) Given the differential equation: $\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}$Separate the variables:$\frac{1}{1 + y^2} dy = \frac{1}{1 + x^2} dx$Inte

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$y - x = c(1 + xy)$

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(C) Given the differential equation: $\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}$Separate the variables:$\frac{1}{1 + y^2} dy = \frac{1}{1 + x^2} dx$Integrate both sides:$\int \frac{1}{1 + y^2} dy = \int \frac{1}{1 + x^2} dx$This gives:$	an^{-1} y = 	an^{-1} x + C$Let the constant $C = 	an^{-1} c$:$	an^{-1} y - 	an^{-1} x = 	an^{-1} c$Using the formula $	an^{-1} A - 	an^{-1} B = 	an^{-1} \left( \frac{A - B}{1 + AB} ight)$:$	an^{-1} \left( \frac{y - x}{1 + yx} ight) = 	an^{-1} c$Taking tangent on both sides:$\frac{y - x}{1 + xy} = c$Therefore,the solution is:$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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