Solve the initial value problem $1+y\sinh(x)+(1+\cosh(x))y'=0,y(0)=\frac{1}{2}$

Question

The question asks to solve the initial value problem: $$1+y\sinh(x)+(1+\cosh(x))y'=0,y(0)=\frac{1}{2}$$ I have tried substituting $1=\frac{\sinh(x)}{\sinh(x)}$ and $1=\frac{\cosh(x)}{\cosh(x)}$ but can't seem to find a workable form to start finding $y_h$

Answer 1

hint

To find $y_h $ use hyperbolic identities

$$1+\cosh (x)=2\cosh^2 (x/2) $$

$$\sinh (x)=2\sinh (x/2)\cosh (x/2) $$ and $$\cosh'(x)=\sinh (x) $$

you will get

$$\frac {y'}{y}=-\frac{\sinh (x)}{1+\cosh (x)} $$ $$=-\frac {\sinh (x/2)}{\cosh (x/2)} $$

and $$y_h=\frac {\lambda}{\cosh^2 (x/2)}.$$