What is the solution for (D^2-2D+1)y=xe^xsin(x) ?

The solution for (D^2-2D+1)y=xe^xsin(x) is y=c_{1}e^x+c_{2}xe^x-e^xxsin(x)-2e^xcos(x)

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(D^2-2D+1)y=xe^xsin(x)

(D^{2} - 2 D + 1) y = x e^{x} sin (x)

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

Solution steps

(D^{2} - 2 D + 1) y = x e^{x} sin (x)

Solve linear ODE:

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

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

y^{^{''}} + 100 y = 200 sin (10 x), y^{^{'}} (0) = 0, y^{^{''}} (0) = 0

y^{^{''}} + 9 y = cos (3 x) + sin (3 x), y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} - 6 y^{^{'}} + 9 y = t^{- 6} e^{3 t}

y^{^{''}} + 5 y^{^{'}} + 6 y = 12 e^{t}, y (0) = - 11, y (0) = 31

y^{^{''}} + y^{^{'}} - y = t^{3}

What is the solution for (D^2-2D+1)y=xe^xsin(x) ?
The solution for (D^2-2D+1)y=xe^xsin(x) is y=c_{1}e^x+c_{2}xe^x-e^xxsin(x)-2e^xcos(x)