What is the solution for (D^2+8D+16)y=0 ?

The solution for (D^2+8D+16)y=0 is y=c_{1}e^{-4x}+c_{2}xe^{-4x}

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(D^2+8D+16)y=0

(D^{2} + 8 D + 16) y = 0

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

Solution steps

(D^{2} + 8 D + 16) y = 0

Solve linear ODE:

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

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

y^{^{''}} - \frac{1}{9} y = 0

y^{^{'''}} + y^{^{''}} - 6 y^{^{'}} = 0 y

77 y^{^{''''}} + 156 y^{^{''}} + 64 y = 0

y^{^{''}} + 13 y^{^{'}} + 42 y = 0, y (0) = 8, y^{^{'}} (0) = - 51

y^{^{''}} + 4 y^{^{'}} + 5 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 2

What is the solution for (D^2+8D+16)y=0 ?
The solution for (D^2+8D+16)y=0 is y=c_{1}e^{-4x}+c_{2}xe^{-4x}