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

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

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

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

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

Solution steps

2 y^{''} (t) + 8 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}

5 x^{4} e^{x} + e^{x} x^{5}

\frac{d}{d x} (x^{4} - 2 x^{2} - 8)

\frac{d y}{d x}, y = 2^{x}

x \to 0 lim (\frac{2 x}{3 x})

\int_{0}^{1} 5^{x} d x

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