Home page for accesible maths Math 101 Chapter 3: Differentiation

Style control - access keys in brackets

Font (2 3) - + Letter spacing (4 5) - + Word spacing (6 7) - + Line spacing (8 9) - +

3.37 Solution of differential equation for hyperbolic functions

We have

f(x)=βAsinhβx+Bcoshβxf^{\prime}(x)=\beta A\sinh\beta x+B\cosh\beta x
f′′(x)=β2Acoshβx+Bβsinhβxf^{\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)=β2f(x)=kmf(x).f^{\prime\prime}(x)=\beta^{2}f(x)={{k}\over{m}}f(x).