What is the solution for y^{''}+8y^'+16y=0 ?

The solution for y^{''}+8y^'+16y=0 is y=c_{1}e^{-4t}+c_{2}te^{-4t}

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y^{''}+8y^'+16y=0

y^{^{''}} + 8 y^{^{'}} + 16 y = 0

y = c_{1} e^{- 4 t} + c_{2} t e^{- 4 t}

Solution steps

y^{''} (t) + 8 y^{'} (t) + 16 y = 0

Solve linear ODE:

y = c_{1} e^{- 4 t} + c_{2} t e^{- 4 t}

y = c_{1} e^{- 4 t} + c_{2} t e^{- 4 t}

\frac{d y}{d x} = x^{5} y

\frac{d}{d t} (\frac{t ^{2} + 2}{t ^{4} - 3 t ^{2} + 1})

\frac{d}{d x} (4 sin^{2} (2 x))

\frac{\partial}{\partial y} (4 x^{3} y^{3} - 4)

y^{^{''}} - 2 y^{^{'}} - 4 y = 0

What is the solution for y^{''}+8y^'+16y=0 ?
The solution for y^{''}+8y^'+16y=0 is y=c_{1}e^{-4t}+c_{2}te^{-4t}