What is the solution for y^{''}+8y^'+16y=(3e^{-4x})/(x^4) ?

The solution for y^{''}+8y^'+16y=(3e^{-4x})/(x^4) is y=c_{1}e^{-4x}+c_{2}xe^{-4x}+(e^{-4x})/(2x^2)

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y^{''}+8y^'+16y=(3e^{-4x})/(x^4)

y^{^{''}} + 8 y^{^{'}} + 16 y = \frac{3 e ^{- 4 x}}{x ^{4}}

y = c_{1} e^{- 4 x} + c_{2} x e^{- 4 x} + \frac{e ^{- 4 x}}{2 x ^{2}}

Solution steps

y^{''} (x) + 8 y^{'} (x) + 16 y = \frac{3 e ^{- 4 x}}{x ^{4}}

Solve linear ODE:

y = c_{1} e^{- 4 x} + c_{2} x e^{- 4 x} + \frac{e ^{- 4 x}}{2 x ^{2}}

y = c_{1} e^{- 4 x} + c_{2} x e^{- 4 x} + \frac{e ^{- 4 x}}{2 x ^{2}}

y^{^{''}} - 2 y^{^{'}} - 3 y = 4, y^{^{'}} (0) = - 1

y^{^{''}} + 9 y = cos (3 t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 9 y = - 18

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

1 q^{^{''}} + 4 q^{^{'}} + 5 q = 5 cos (40 t)

What is the solution for y^{''}+8y^'+16y=(3e^{-4x})/(x^4) ?
The solution for y^{''}+8y^'+16y=(3e^{-4x})/(x^4) is y=c_{1}e^{-4x}+c_{2}xe^{-4x}+(e^{-4x})/(2x^2)