Solving a separable differential equation Solve the differential equation:
$$y'=\frac{1-y^2}{1-x^2}$$
My book says the solution is: $$y=\frac{x+c}{cx+1},$$ where $c$ is a constant.
It's been ten minutes I tried to verify if it was correct but I'm pretty sure the book is wrong. Can someone confirm it?

$$y'=\frac{cx+1-c(x+c)}{(cx+1)^2}=\frac{1-c^2}{(cx+1)^2},$$
$$\frac{1-y^2}{1-x^2}=\frac{\frac{(cx+1)^2-(x+c)^2}{(cx+1)^2}}{(1-x^2)}=\frac{(c^2-1)x^2+(1-c^2)}{(cx+1)^2(1-x^2)}=\frac{1-c^2}{(cx+1)^2}.$$
