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

The solution for y^{''}-2y^'+y=4x^2-4+x^{-1}e^x is y=c_{1}e^x+c_{2}xe^x+4x^2+16x-e^xx+e^xxln(x)+20

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

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

y = c_{1} e^{x} + c_{2} x e^{x} + 4 x^{2} + 16 x - e^{x} x + e^{x} x ln (x) + 20

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + 4 x^{2} + 16 x - e^{x} x + e^{x} x ln (x) + 20

y = c_{1} e^{x} + c_{2} x e^{x} + 4 x^{2} + 16 x - e^{x} x + e^{x} x ln (x) + 20

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

y^{^{''}} + 6 y^{^{'}} + 8 y = 2 t + e^{t}

x^{^{''}} + 25 x = 2 cos (5.1 t)

y^{^{'''}} + 9 y^{^{''}} + 27 y^{^{'}} + 27 y = 9 e^{- 3 x}

y^{^{''}} + y = sin (t), y (0) = 0, y^{^{'}} (0) = 2

What is the solution for y^{''}-2y^'+y=4x^2-4+x^{-1}e^x ?
The solution for y^{''}-2y^'+y=4x^2-4+x^{-1}e^x is y=c_{1}e^x+c_{2}xe^x+4x^2+16x-e^xx+e^xxln(x)+20