Uniqueness of initial value problem to ODE

Question

I was presented with an i.v.p.:$$y'=\frac{1+y^2}{1+x^2},y(0)=1$$Using separation of variables I obtained $$\arctan(y)=\arctan(x)+c$$ Substituting x=0 and y=1 gives $c=\frac{\pi}{4}$ but if I solve for y and get $$y=\tan(\arctan(x)+c)$$ then plugging in those values reduces to $$1=\tan(c)$$ which has a solution $c=\frac{\pi}{4}+k\pi$. Which approach is correct? Does this i.v.p. have only one solution or infinitely many of them

Answer 1

It doesn't really matter, because $\tan$ is periodic with period $\pi$. These differennt values of $c$ all correspond to the same solution.

BTW, you can also express the solution as $$ y = \frac{x+1}{1-x}$$ using the addition formula for $\tan$.