What is the solution for (d^3-3d^2+3d-1)y=4cos(x) ?

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

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(d^3-3d^2+3d-1)y=4cos(x)

(d^{3} - 3 d^{2} + 3 d - 1) y = 4 cos (x)

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

Solution steps

(d^{3} - 3 d^{2} + 3 d - 1) y = 4 cos (x)

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} = 8 x^{2}

2 y^{^{''}} = - 2 y + cos^{2} (x)

(d^{2} - 2 d + 1) \cdot y = e^{x} \cdot x^{2}

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = e^{x} - x + 16

(d^{2} + 4 d + 5) \cdot y = - 2 cos (x)

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