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 y^{^{'}} = 3 x y^{^{''}}

y^{^{''}} - (2 α - 12) y^{^{'}} + (α^{2} - 12 α + 32) y = 0

3 y^{^{''}} + 12 y = 0, y (0) = 5, y^{^{'}} (0) = 12

y = \frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x}

36 y^{^{''}} - 81 y = 0, y (0) = 0, y^{^{'}} (0) = 1

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))