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-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 - 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 - sin (x)

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

y^{^{''}} + y^{^{'}} + y = 5 x + 3

y^{^{''}} + 5 y^{^{'}} + 6 y = 3 e^{- 2 t}

y^{^{'''}} - 2 y^{^{''}} - 21 y^{^{'}} - 18 y = 3 + 4 e^{- t}

y^{^{''}} + 9 y = 18 t, y (\frac{π}{2}) = 0, y (0) = 0

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