What is the solution for x^{''}+12x^'+27x=0 ?

The solution for x^{''}+12x^'+27x=0 is x=c_{1}e^{-3t}+c_{2}e^{-9t}

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x^{''}+12x^'+27x=0

x^{^{''}} + 12 x^{^{'}} + 27 x = 0

x = c_{1} e^{- 3 t} + c_{2} e^{- 9 t}

Solution steps

x^{''} (t) + 12 x^{'} (t) + 27 x = 0

Solve linear ODE:

x = c_{1} e^{- 3 t} + c_{2} e^{- 9 t}

x = c_{1} e^{- 3 t} + c_{2} e^{- 9 t}

(d^{5} - d^{3}) \cdot y = 0

y^{^{''}} - 16 y^{^{'}} + 64 y = 0, y (0) = 0, y^{^{'}} (0) = 9

y^{^{'''}} - 3 y^{^{''}} - y^{^{''}} + 3 y = 0

y^{^{''}} + (a^{2}) y = 0

y^{^{''}} = k y (x)

What is the solution for x^{''}+12x^'+27x=0 ?
The solution for x^{''}+12x^'+27x=0 is x=c_{1}e^{-3t}+c_{2}e^{-9t}