What is the solution for (D^4-2D^3+2D^2-2D+1)y=0 ?

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

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(D^4-2D^3+2D^2-2D+1)y=0

(D^{4} - 2 D^{3} + 2 D^{2} - 2 D + 1) y = 0

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

Solution steps

(D^{4} - 2 D^{3} + 2 D^{2} - 2 D + 1) y = 0

Solve linear ODE:

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

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

y^{^{'''}} + 5 y^{^{''}} + 7 y^{^{'}} + 3 y = 0

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

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

y^{^{''}} = 6 y

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = - 4, y^{^{'}} (0) = 14

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