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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y = 0, y^{^{'}} (\frac{π}{4}) = 2, y^{^{'}} (0) = 1

y^{^{''}} + 0 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} + 2 y^{^{'}} - 15 y = 0, y (0) = 1

5 y^{^{''}} + 20 y = 0, y (\frac{π}{4}) = 10, y (0) = 15

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

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