What is the solution for y^{''}+4y^'+4y=(6e^{-2x})/(x^5) ?

The solution for y^{''}+4y^'+4y=(6e^{-2x})/(x^5) is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(e^{-2x})/(2x^3)

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y^{''}+4y^'+4y=(6e^{-2x})/(x^5)

y^{^{''}} + 4 y^{^{'}} + 4 y = \frac{6 e ^{- 2 x}}{x ^{5}}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{e ^{- 2 x}}{2 x ^{3}}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = \frac{6 e ^{- 2 x}}{x ^{5}}

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{e ^{- 2 x}}{2 x ^{3}}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{e ^{- 2 x}}{2 x ^{3}}

2 y^{^{''}} + 7 y = - \frac{8}{3} x

y^{^{''}} + y = 5 e^{- 3 x}, y (0) = 0, y^{^{'}} (0) = 0

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

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

x^{^{''}} + 8 x^{^{'}} + 16 x = 6 sin (4 t)

What is the solution for y^{''}+4y^'+4y=(6e^{-2x})/(x^5) ?
The solution for y^{''}+4y^'+4y=(6e^{-2x})/(x^5) is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(e^{-2x})/(2x^3)