What is the solution for y^{''}-k^2y=0,y(0)=0,y^'(0)=1 ?

The solution for y^{''}-k^2y=0,y(0)=0,y^'(0)=1 is y=(e^{kt})/(2k)-(e^{-kt})/(2k)

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y^{''}-k^2y=0,y(0)=0,y^'(0)=1

y^{^{''}} - k^{2} y = 0, y (0) = 0, y^{^{'}} (0) = 1

y = \frac{e ^{k t}}{2 k} - \frac{e ^{- k t}}{2 k}

Solution steps

y^{''} (t) - k^{2} y = 0

Solve linear ODE:

y = \frac{e ^{k t}}{2 k} - \frac{e ^{- k t}}{2 k}

y = \frac{e ^{k t}}{2 k} - \frac{e ^{- k t}}{2 k}

x^{^{'''}} + 5 x^{^{''}} - 25 x^{^{'}} - 125 x = 0

y^{^{''}} - 12 y^{^{'}} + 35 y = 0, y (0) = e^{k x}

\frac{d ^{3} y}{d x ^{3}} + 5 y = 0

y^{^{''}} + 5 y^{^{'}} - 14 y = 0, y^{^{'}} (0) = 56

y^{^{''}} - 2 y^{^{'}} + 5 y = 0, y^{^{'}} (0) = 2, y (0) = 1

What is the solution for y^{''}-k^2y=0,y(0)=0,y^'(0)=1 ?
The solution for y^{''}-k^2y=0,y(0)=0,y^'(0)=1 is y=(e^{kt})/(2k)-(e^{-kt})/(2k)