From above, the general solution of 2y" sinh x - 2 cosh xy' + y sinh³ x = 0 is 1 cosh x) + D sin(. √2 which can also be written as y = C cos(- so that as required. 1 √2 1 y = Esin(√√2 cosh x + F). This latter form is better when applying the boundary conditions; y(1) = 0 immediately gives Then applying y(2) = 2 gives y = Esin(- (cosh x-cosh 1)). E = y = cosh x), 2 9 sin((cosh 2-cosh 1)) how Come ?? 2 sin( (cosh x - cosh 1)) - sin((cosh 2 - cosh 1))

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