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

The solution for y^{''}+2y^'+y=x^{-4}e^{-x} is y=c_{1}e^{-x}+c_{2}xe^{-x}+(e^{-x})/(6x^2)

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

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

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{e ^{- x}}{6 x ^{2}}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{e ^{- x}}{6 x ^{2}}

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{e ^{- x}}{6 x ^{2}}

y = \frac{x}{x ^{2} + 9}

x \to \infty lim (1 + \frac{2}{x})

\int (5 x - 2) d x

\int 4 x - 7 d x

y = 0, y = x^{2}, y^{2} = x

What is the solution for y^{''}+2y^'+y=x^{-4}e^{-x} ?
The solution for y^{''}+2y^'+y=x^{-4}e^{-x} is y=c_{1}e^{-x}+c_{2}xe^{-x}+(e^{-x})/(6x^2)