$\displaystyle Let\ f(z)=\ u+iv\ be\ analytic\ \\[2ex] \displaystyle By\ Cauchy\ rehmans\ equations \\[2ex] Ux=Vy\ and\ Uy=-Vx\ \grave{a}(1)\\[2ex] \displaystyle Given\ a=\frac{\sin{2x}}{\cosh{2y}-\cos{2x}}\\[4ex] \displaystyle Let\ D=\ \cosh{2y}-\cos{2x} \\[2ex] $

$\displaystyle Differentiating\ wrt\ x, \\[2ex] \displaystyle \therefore{}Ux=(\cosh{2y}-\cos{2x})\bullet{}\cos{2x}-\sin{2x}\bullet{}(0+\sin{2x}\bullet{}2)/\ D^2 \\[2ex] \displaystyle =2(\cosh{2y}\cos{2x}\ {cos}^2{2y}-\ {sin}^2{2x}/D^2\ \\[2ex] =2(\cosh{2y}\cos{2x}-1)/\ D^2...(2)\\[2ex] $

$Similarly,\ differentiating\ ‘u’\ partially\ wrt\ ‘y’\\[2ex] \displaystyle \therefore{}Uy=(\cosh{2y}-\cos{2x})\bullet{}0-\ \sin{2x}\bullet{}(\sin{2y}\bullet{}2-0)/\ D^2 \\[2ex] \displaystyle =2(\cosh{2y}\cos{2x}\ {cos}^2{2x}-\ {sin}^2{2y}/D^2\ \\[2ex] $

$\displaystyle Now,\ {(D}^2-4)y=3e^t \\[2ex] \displaystyle \therefore{}{(D}^2y-4y)=\ 3e^t \\[2ex] \displaystyle \therefore{}L\left[D^2y\right]-4L\left[y\right]=3L[e^t] \\[2ex] \displaystyle \therefore{}s^2y-sy\left(0\right)-y^{'}\left(0\right)-4y=3L\left[e^t\right] \\[2ex] \displaystyle \therefore{}s^2y-s\left(0\right)-3-4y=\frac{3}{s-1} \\[2ex] \displaystyle \therefore{}s^2y-4y=\frac{3}{s-1}+3 \\[2ex] \displaystyle \therefore{}{y(s}^2-4)=\frac{3+3s-3}{s-1} \\[2ex] \displaystyle \therefore{}y\left(s-2\right)(s+2)=\frac{3s}{s-1} \\[2ex] \displaystyle \therefore{}y=\frac{3s}{(s-1)\left(s-2\right)(s+2)} \\[2ex] $

$\displaystyle \therefore{}y=L^{-1}\left[\frac{3s}{\left(s-1\right)\left(s-2\right)\left(s+2\right)}\right] \\[2ex] \displaystyle \therefore{}y=L^{-1}\left[\frac{-1}{s-1}+\frac{\frac{3}{2}}{s-2}-\frac{\frac{1}{2}}{s+2}\right]\ ..By\ partial\ fractions \\[2ex] \displaystyle =-e^t+\frac{3}{2}e^{2t}-\frac{1}{2}e^{-2t} \\[2ex] $