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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(d^{2} - 2 d + 2) y = 3 e^{x} sin (x)

y^{^{''}} + 5 y^{^{'}} + 6 y = 5 e^{3 t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = 2 e^{2 x}

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

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

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