(a) Starting from the definitions of sinh and cosh in terms of exponentials, prove that
\[2 \sinh ^{2} x=\cosh 2 x-1\]
(b) Find the solution to the differential equation
\[\frac{\mathrm{d} y}{\mathrm{~d} x}+y \operatorname{coth} x=4 \sinh x\]
for which \(y=1\) when \(x=\ln 3\).