What is the solution for y^{''}+2y^'+y=x^2-5 ?

The solution for y^{''}+2y^'+y=x^2-5 is y=c_{1}e^{-x}+c_{2}xe^{-x}+x^2-4x+1

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y^{''}+2y^'+y=x^2-5

y^{^{''}} + 2 y^{^{'}} + y = x^{2} - 5

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

Solution steps

y^{''} (x) + 2 y^{'} (x) + y = x^{2} - 5

Solve linear ODE:

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

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

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

y^{^{''}} - 6 y^{^{'}} + 9 y = t^{2} e^{3 t}

y^{^{''}} - 25 y = cosh (5 x)

y^{^{''}} - 3 y^{^{'}} - 4 y = e^{- x}

3 y^{^{''}} + 27 y = 2 cos (3 t)

What is the solution for y^{''}+2y^'+y=x^2-5 ?
The solution for y^{''}+2y^'+y=x^2-5 is y=c_{1}e^{-x}+c_{2}xe^{-x}+x^2-4x+1