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3.37 Solution of differential equation for hyperbolic functions

We have

f′⁢(x)=β⁢A⁢sinh⁡β⁢x+B⁢cosh⁡β⁢xf^{\prime}(x)=\beta A\sinh\beta x+B\cosh\beta x
f′′⁢(x)=β2⁢A⁢cosh⁡β⁢x+B⁢β⁢sinh⁡β⁢xf^{\prime\prime}(x)=\beta^{2}A\cosh\beta x+{{B}{\beta}}\sinh\beta x

so f⁢(0)=Af(0)=A, f′⁢(0)=Bf^{\prime}(0)=B and

f′′⁢(x)=β2⁢f⁢(x)=km⁢f⁢(x).f^{\prime\prime}(x)=\beta^{2}f(x)={{k}\over{m}}f(x).