What is the solution for (d^2x)/(dt^2)+8(dx)/(dt)+16x=0 ?

The solution for (d^2x)/(dt^2)+8(dx)/(dt)+16x=0 is x=c_{1}e^{-4t}+c_{2}te^{-4t}

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(d^2x)/(dt^2)+8(dx)/(dt)+16x=0

\frac{d ^{2} x}{d t ^{2}} + 8 \frac{d x}{d t} + 16 x = 0

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

Solution steps

\frac{d ^{2} x}{d t ^{2}} + 8 \frac{d x}{d 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}

y^{^{'''}} + 2 y^{^{''}} - 10 y^{^{'}} + 4 y = 0

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

y^{^{''}} - 14 y^{^{'}} + 65 y = 0, y (0) = 4, y^{^{'}} (0) = - 4

- a y^{^{''}} + b y^{^{'}} + cy = 0

y^{^{''}} - 3 y^{^{'}} - 18 y = 0, y (0) = 24

What is the solution for (d^2x)/(dt^2)+8(dx)/(dt)+16x=0 ?
The solution for (d^2x)/(dt^2)+8(dx)/(dt)+16x=0 is x=c_{1}e^{-4t}+c_{2}te^{-4t}