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

The solution for (D^2-2D+5)^2y=0 is y=e^x(c_{1}cos(2x)+c_{2}sin(2x))+xe^x(c_{3}cos(2x)+c_{4}sin(2x))

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

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

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + x e^{x} (c_{3} cos (2 x) + c_{4} sin (2 x))

Solution steps

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

Solve linear ODE:

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + x e^{x} (c_{3} cos (2 x) + c_{4} sin (2 x))

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + x e^{x} (c_{3} cos (2 x) + c_{4} sin (2 x))

3 x^{^{''}} + 25 x = 0, x (0) = - 0.5, x^{^{'}} (0) = 2

y^{^{''}} + 6 y^{^{'}} + 8.96 y = 0

y^{^{''}} + 2 y^{^{'}} + 1.25 y = 0

y^{^{''}} + 20 y^{^{'}} + 36 y = 0, y (0) = 3, y^{^{'}} (0) = 0

3 \frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + 3 y = 0

What is the solution for (D^2-2D+5)^2y=0 ?
The solution for (D^2-2D+5)^2y=0 is y=e^x(c_{1}cos(2x)+c_{2}sin(2x))+xe^x(c_{3}cos(2x)+c_{4}sin(2x))