What is the solution for (D^3-2D^2+4D-8)y=0 ?

The solution for (D^3-2D^2+4D-8)y=0 is y=c_{1}e^{2x}+c_{2}cos(2x)+c_{3}sin(2x)

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(D^3-2D^2+4D-8)y=0

(D^{3} - 2 D^{2} + 4 D - 8) y = 0

y = c_{1} e^{2 x} + c_{2} cos (2 x) + c_{3} sin (2 x)

Solution steps

(D^{3} - 2 D^{2} + 4 D - 8) y = 0

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} cos (2 x) + c_{3} sin (2 x)

y = c_{1} e^{2 x} + c_{2} cos (2 x) + c_{3} sin (2 x)

y^{^{''}} + 16 y^{^{'}} + 100 y = 0, y (0) = 6, y^{^{'}} (0) = 3

θ^{^{''}} (t) = 3 θ (t)

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

4 y^{^{''}} + \frac{1}{4} y^{^{'}} + y = 0

y^{^{''}} + 2 y^{^{'}} - 15 y = 0, y (0) = x, y^{^{'}} (0) = 20

What is the solution for (D^3-2D^2+4D-8)y=0 ?
The solution for (D^3-2D^2+4D-8)y=0 is y=c_{1}e^{2x}+c_{2}cos(2x)+c_{3}sin(2x)