What is the solution for y^{''}+2y^'+y=2x^2+4x+2 ?

The solution for y^{''}+2y^'+y=2x^2+4x+2 is y=c_{1}e^{-x}+c_{2}xe^{-x}+2x^2-4x+6

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

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

y = c_{1} e^{- x} + c_{2} x e^{- x} + 2 x^{2} - 4 x + 6

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} + 2 x^{2} - 4 x + 6

y = c_{1} e^{- x} + c_{2} x e^{- x} + 2 x^{2} - 4 x + 6

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

y^{^{''}} - 8 y^{^{'}} + 16 y = t^{- 5} e^{4 t}

x^{^{''}} + 7 x^{^{'}} + 12 x = e^{2 t} + cos (4 t)

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

y^{^{'''}} - y^{^{''}} - 16 y^{^{'}} + 16 y = 8 - e^{x} + e^{4 x}

What is the solution for y^{''}+2y^'+y=2x^2+4x+2 ?
The solution for y^{''}+2y^'+y=2x^2+4x+2 is y=c_{1}e^{-x}+c_{2}xe^{-x}+2x^2-4x+6