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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (5 x - 8)

X^{\frac{9}{6}}, 7

\frac{d}{d x} (tan (3^{x}))

\frac{\partial}{\partial x} (\frac{2}{x + y})

\frac{x}{x + 9}

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