1st order linear differential equation

Question

Can anyone help me solve this 1st order linear differential equation $$(1+x^2)y'=2 \cosh(y)$$ The answer on the back of the book is $$y=\ln\left(\frac{x+c}{1-cx}\right).$$ I found $\arctan(\exp y)= \arctan(x)+c$, but I don't know how to proceed.

Answer 1

The equation is separable : $$(1+x^2)y'=2 \cosh(y)$$ $$\int \frac {dy}{\cosh(y)}=2 \int \frac {dx}{(1+x^2)}$$ $$\int \frac {e^y}{e^{2y}+1}dy=\int \frac {dx}{(1+x^2)}$$ With $u=e^y \implies du=e^ydy$ $$\int \frac {du}{u^2+1}= \arctan(x)+K $$ $$\arctan(e^y)=\arctan (x)+K$$ Take tangent of both side $$e^y=\tan (\arctan(x)+K)$$ use the formula $\tan (A+B)=\frac {\tan A+\tan B}{1-\tan A \tan B}$ $$\implies e^y=\dfrac {x+C}{1-xC}$$ $$ \boxed {y(x)= \ln \left| \frac {x+C}{1-xC}\right |}$$