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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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