What is the solution for y^{''}+2ky^'+k^2y=0 ?

The solution for y^{''}+2ky^'+k^2y=0 is y=c_{1}e^{-kt}+c_{2}te^{-kt}

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y^{''}+2ky^'+k^2y=0

y^{^{''}} + 2 k y^{^{'}} + k^{2} y = 0

y = c_{1} e^{- k t} + c_{2} t e^{- k t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- k t} + c_{2} t e^{- k t}

y = c_{1} e^{- k t} + c_{2} t e^{- k t}

y^{^{''}} + 2 y^{^{'}} + 63 y = 0

4 x^{^{''}} + 12 x^{^{'}} + 9 x = 0

y^{^{'''}} + 10 y^{^{''}} + 9 y^{^{'}} = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 7 y^{^{'}} + 10 y = 0, y^{^{'}} (0) = - \frac{1}{3}

What is the solution for y^{''}+2ky^'+k^2y=0 ?
The solution for y^{''}+2ky^'+k^2y=0 is y=c_{1}e^{-kt}+c_{2}te^{-kt}