Answer: (a) \(2\cosh^2 x=\cosh 2x+1\).
(b) \(y=\operatorname{sech}^2 x\left(\frac14\sinh 2x+\frac12 x+1\right)\).
(a) Using the definition
\(\cosh x=\dfrac{e^x+e^{-x}}{2}\).
Then
\(2\cosh^2 x=2\left(\dfrac{e^x+e^{-x}}{2}\right)^2=\dfrac12(e^x+e^{-x})^2\).
Expanding,
\(\dfrac12(e^{2x}+2+e^{-2x})=\dfrac12(e^{2x}+e^{-2x})+1\).
Since \(\cosh 2x=\dfrac{e^{2x}+e^{-2x}}{2}\), it follows that
\(2\cosh^2 x=\cosh 2x+1\).
(b) The equation is
\(\dfrac{dy}{dx}+2y\tanh x=1\).
This is linear, with integrating factor
\(\exp\left(\int 2\tanh x\,dx\right)=\exp\left(2\ln(\cosh x)\right)=\cosh^2 x\).
Multiplying through by \(\cosh^2 x\),
\(\cosh^2 x\,\dfrac{dy}{dx}+2y\tanh x\cosh^2 x=\cosh^2 x\),
so
\(\dfrac{d}{dx}(y\cosh^2 x)=\cosh^2 x\).
Integrating,
\(y\cosh^2 x=\int \cosh^2 x\,dx+C\).
Using part (a),
\(\cosh^2 x=\dfrac12(\cosh 2x+1)\).
Hence
\(\int \cosh^2 x\,dx=\dfrac12\int (\cosh 2x+1)\,dx=\dfrac12\left(\dfrac12\sinh 2x+x\right)+C\)
\(=\dfrac14\sinh 2x+\dfrac12x+C\).
Therefore
\(y\cosh^2 x=\dfrac14\sinh 2x+\dfrac12x+C\).
Using \(y=1\) when \(x=0\),
\(1\cdot \cosh^2 0=\dfrac14\sinh 0+\dfrac12\cdot 0+C\),
so \(1=C\).
Thus
\(y\cosh^2 x=\dfrac14\sinh 2x+\dfrac12x+1\),
and therefore
\(y=\operatorname{sech}^2 x\left(\frac14\sinh 2x+\frac12x+1\right).\)
