What is the solution for 2y^{''}+4y^'+ky=0,y(0)=2,y^'(0)=0 ?

The solution for 2y^{''}+4y^'+ky=0,y(0)=2,y^'(0)=0 is

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2y^{''}+4y^'+ky=0,y(0)=2,y^'(0)=0

2 y^{^{''}} + 4 y^{^{'}} + k y = 0, y (0) = 2, y^{^{'}} (0) = 0

y = e^{\frac{t ( - 4 + 16 - 8 k )}{4}} (2 + \frac{- - k + 2 + 2}{2 - k}) - \frac{e ^{\frac{t ( - 4 - 16 - 8 k )}{4}} ( - - k + 2 + 2 )}{2 - k}

Solution steps

2 y^{''} (t) + 4 y^{'} (t) + k y (t) = 0

Solve linear ODE:

y = e^{\frac{t ( - 4 + 16 - 8 k )}{4}} (2 + \frac{- - k + 2 + 2}{2 - k}) - \frac{e ^{\frac{t ( - 4 - 16 - 8 k )}{4}} ( - - k + 2 + 2 )}{2 - k}

y = e^{\frac{t ( - 4 + 16 - 8 k )}{4}} (2 + \frac{- - k + 2 + 2}{2 - k}) - \frac{e ^{\frac{t ( - 4 - 16 - 8 k )}{4}} ( - - k + 2 + 2 )}{2 - k}

0.1 \cdot y^{^{''}} + y^{^{'}} + y = 0

y^{^{''}} - v^{2} y = 0

x^{^{''}} + 60 x^{^{'}} + 500 x = 0

10 y^{^{''''}} + 53 y^{^{''}} + 36 y = 0

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

What is the solution for 2y^{''}+4y^'+ky=0,y(0)=2,y^'(0)=0 ?
The solution for 2y^{''}+4y^'+ky=0,y(0)=2,y^'(0)=0 is