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

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

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

y^{^{''}} + 2 y^{^{'}} + y = \frac{e ^{- x}}{x ^{2}}

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{'''}} - y^{^{'}} = 10 sin (3 x)

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

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

y^{^{''}} - 8 y^{^{'}} + 7 y = x

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