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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 5 y^{^{'}} = 2 x^{3} - 4 x^{2} - x + 6

(D^{2} + 4 D + 5) y = 50 x + 13 e^{3 x}

y^{^{''}} + y = 4 x cos (x)

(D^{2} - 1) y = \frac{2}{1 + e ^{x}}

y^{^{''}} + 4 y^{^{'}} + 3 y = 6 x^{2} + x + 9

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