What is the solution for 2*y^{''}+8*y^'+8y=0 ?

The solution for 2*y^{''}+8*y^'+8y=0 is y=c_{1}e^{-2t}+c_{2}te^{-2t}

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2y^{''}+8y^'+8y=0

2 \cdot y^{^{''}} + 8 \cdot y^{^{'}} + 8 y = 0

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

Solution steps

2 \cdot y^{''} (t) + 8 \cdot y^{'} (t) + 8 y = 0

Solve linear ODE:

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

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

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 1

c^{2} \cdot f^{^{''}} (t) = a f^{^{''}} (t) - b f (t)

(y^{^{''}} + 4 y^{^{'}} + 20 y)^{2} = 0

y^{^{'''}} + 36 y^{^{'}} = 0, y^{^{'}} (0) = 6

y^{^{'''}} - 5 y^{^{''}} - 2 y^{^{'}} + 24 y = 0

What is the solution for 2*y^{''}+8*y^'+8y=0 ?
The solution for 2*y^{''}+8*y^'+8y=0 is y=c_{1}e^{-2t}+c_{2}te^{-2t}