What is the solution for (D^2+2D+17)y=0 ?

The solution for (D^2+2D+17)y=0 is y=e^{-x}(c_{1}cos(4x)+c_{2}sin(4x))

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(D^2+2D+17)y=0

(D^{2} + 2 D + 17) y = 0

y = e^{- x} (c_{1} cos (4 x) + c_{2} sin (4 x))

Solution steps

(D^{2} + 2 D + 17) y = 0

Solve linear ODE:

y = e^{- x} (c_{1} cos (4 x) + c_{2} sin (4 x))

y = e^{- x} (c_{1} cos (4 x) + c_{2} sin (4 x))

3 y^{^{'''}} - 5 y^{^{''}} - 16 y^{^{'}} + 12 y = 0

y^{^{''}} + y^{^{'}} + 4.25 y = 0, y (0) = 5

y^{^{''}} + 0 y^{^{'}} + 36 y = 0, y (0) = 3, y^{^{'}} (0) = 0

y^{^{''}} - 12 y^{^{'}} + 40 y = 0, y (0) = 6, y^{^{'}} (0) = - 2

y^{^{''}} + 8 y^{^{'}} + 4 y = 0, y^{^{'}} (0) = 1

What is the solution for (D^2+2D+17)y=0 ?
The solution for (D^2+2D+17)y=0 is y=e^{-x}(c_{1}cos(4x)+c_{2}sin(4x))