$$\cot y \, y^{''} = y^{'}{^2} +c$$ Substitute $y'=p \implies y''=p\frac {dp}{dy}$ $$pp'\cot y = p^2 +c$$ $$\int \frac {2p}{ p^2 +c}dp=2\int \tan(y) dy$$ $$\ln|p^2 +c|=-2\ln |\cos(y)|+K_1$$ $$p^2 =\frac {K_1} {\cos^2(y)}-c$$ $$y'=\sqrt{\frac {K_1} {\cos^2(y)}-c}$$ $$\int \frac {dy}{\sqrt{\frac {K_1} {\cos^2(y)}-c}}=x+K_2$$ With $u=\sin(y)$ $$\int \frac {du}{\sqrt{K_1+cu^2}}=x+K_2$$