Here's what I might try

First $\dfrac{d}{dx} \left(\dfrac{y'}{\sinh x}\right) = \dfrac{\sinh x \,y '' - \cosh x \,y'}{\sinh ^2 x}$

So your DE becomes

$2 \dfrac{d}{dx} \left(\dfrac{y'}{\sinh x}\right) +\sinh x \,y = 0$

or

$2 \dfrac{1}{\sinh x } \,\dfrac{d}{dx} \left(\dfrac{y'}{\sinh x}\right) +y = 0$

Now try a new substitution. Let $ t = \cosh x$.

See how that goes.