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

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

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

linear

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

y = c_{1} + c_{2} e^{- 8 t} - 2 t

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 8 t} - 2 t

y = c_{1} + c_{2} e^{- 8 t} - 2 t

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 e^{2 t}, y^{^{'}} (0) = 5, y (0) = - 3

y^{^{''}} - 4 y^{^{'}} + 8 y = 16 x

(d^{2} + 1) y = x^{3}

y^{^{''}} + 4 y = 8 x^{2} + 9 x

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = x e^{x}

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