Recent content by TedMurphy

I'm going to write the simplification out, because it took me a while. \tan^{-1}y - \tan^{-1}x = C In order to take tan on the left side, this equation needs to be re-written as: \tan^{-1}\frac{y-x}{1+yx}=C then we can take the tangent of both sides, giving us: \frac{y-x}{1+yx}=C then we...

Obtain the solution to the differential equation: \frac{dy}{dx} = \frac{1+y^2}{1+x^2} Multiple choice answer: a) \frac{Cx}{1-Cx} b) \frac{Cx}{1+Cx} c) \frac{C-x}{1-Cx} d) \frac{1-Cx}{x+C} e) \frac{x+C}{1-Cx} Tried integrating two sides to arrive at arctan y = arctan x + C, but not sure...