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

The solution for y^{''}+2y^'+y=x^2+2 is y=c_{1}e^{-x}+c_{2}xe^{-x}+x^2-4x+8

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

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

y = c_{1} e^{- x} + c_{2} x e^{- x} + x^{2} - 4 x + 8

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} + x^{2} - 4 x + 8

y = c_{1} e^{- x} + c_{2} x e^{- x} + x^{2} - 4 x + 8

y^{^{''}} + y = sin (3 t)

y^{^{''}} + 2 y^{^{'}} + 2 y = 1 + x

y^{^{''}} - 8 y^{^{'}} + 16 y = t, y (0) = 0, y^{^{'}} (0) = 1

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

y^{^{''}} + 4 y = 4 cos (x) + 3 sin (x) - 8

What is the solution for y^{''}+2y^'+y=x^2+2 ?
The solution for y^{''}+2y^'+y=x^2+2 is y=c_{1}e^{-x}+c_{2}xe^{-x}+x^2-4x+8