Using Dz and D'z to denote partial derivatives of z with respect to x and y respectively, the equation becomes (D^2 -1)z=0 or (D-1)(D+1)z=0. Put (D+1)z = u(x,y). Then we have (D-1)u =0. The general solution of this equation is u(x,y)=A(y).e^x, where A(y) is an arbitrary function of y, because we have integrated a partial derivative of u with respect to x, and we have to take an arbitrary function of y instead of an arbitrary constant.
Now the original equation becomes
(D+1)z=A(y)e^x and we integrate it with respect to x, giving ze^x = Int[e^x.A(y).e^x.dx] + B(y), (for another arbitrary function B(y)) = (1/2)e^(2x).A(y) +B(y) or
z(x,y) = B(y)e^(-x)+[A(y)e^x]/2 or
z = B(y)e^(-x) + C(y)e^x. Hence putting x=0,
e^y = B(y)+C(y) and Dz= -e^(-x)B(y) + e^x.C(y). Again putting x=0, we get 1 = -B(y) + C(y). Solving the two equations in B(y) and C(y), we have C(y) = (1+e^y)/2, and B(y) = (e^y -1)/2. Hence the required solution is
z = [(e^y -1).e^(-x)]/2 + [(e^y +1).e^x]/2 =
z = (e^y)cosh(x) + sinh(x).