What is the solution for (D^2-2D+5)y=0 ?

The solution for (D^2-2D+5)y=0 is y=e^x(c_{1}cos(2x)+c_{2}sin(2x))

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(D^2-2D+5)y=0

(D^{2} - 2 D + 5) y = 0

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

Solution steps

(D^{2} - 2 D + 5) y = 0

Solve linear ODE:

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

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

y^{^{''}} + 10 y^{^{'}} + 16 y = 0

y^{^{''''}} + 2 y^{^{'''}} + 4 y^{^{''}} - 2 y^{^{'}} - 5 y = 0

x^{^{''}} + a x = 0

y^{^{''}} + 3 y^{^{'}} + 5 y = 0

y^{^{''''}} - 10 y^{^{'''}} + 25 y^{^{''}} = 0

What is the solution for (D^2-2D+5)y=0 ?
The solution for (D^2-2D+5)y=0 is y=e^x(c_{1}cos(2x)+c_{2}sin(2x))