What is the solution for y^{''}-2y^'+y=x-2-2cos(8) ?

The solution for y^{''}-2y^'+y=x-2-2cos(8) is y=c_{1}e^x+c_{2}xe^x+x-2cos(8)

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y^{''}-2y^'+y=x-2-2cos(8)

y^{^{''}} - 2 y^{^{'}} + y = x - 2 - 2 cos (8)

y = c_{1} e^{x} + c_{2} x e^{x} + x - 2 cos (8)

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = x - 2 - 2 cos (8)

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + x - 2 cos (8)

y = c_{1} e^{x} + c_{2} x e^{x} + x - 2 cos (8)

y^{^{''}} + 4 y^{^{'}} = 4 x + 1

(d^{2} - 4 d + 3) y = e^{- 2 t}

y^{^{''}} + y^{^{'}} - 12 = 0

y^{^{''}} + 15 y^{^{'}} + 50 y = 1296 t e^{- t}

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

What is the solution for y^{''}-2y^'+y=x-2-2cos(8) ?
The solution for y^{''}-2y^'+y=x-2-2cos(8) is y=c_{1}e^x+c_{2}xe^x+x-2cos(8)