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

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

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

y^{^{''}} + 8 y^{^{'}} + 16 y = 3 \frac{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 = 3 \frac{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 \cdot y^{^{'}} + y = e^{- t}

y^{^{''}} - y = 6 e^{2 x}, y (0) = 4, y^{^{'}} (0) = 4

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

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{2 t}

(D^{4} - 1) \cdot y = sin (2 x)

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