What is the solution for y^{''}+10y^'+25y=(e^{-5x})/(x^6) ?

The solution for y^{''}+10y^'+25y=(e^{-5x})/(x^6) is y=c_{1}e^{-5x}+c_{2}xe^{-5x}+(e^{-5x})/(20x^4)

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y^{''}+10y^'+25y=(e^{-5x})/(x^6)

y^{^{''}} + 10 y^{^{'}} + 25 y = \frac{e ^{- 5 x}}{x ^{6}}

y = c_{1} e^{- 5 x} + c_{2} x e^{- 5 x} + \frac{e ^{- 5 x}}{20 x ^{4}}

Solution steps

y^{''} (x) + 10 y^{'} (x) + 25 y = \frac{e ^{- 5 x}}{x ^{6}}

Solve linear ODE:

y = c_{1} e^{- 5 x} + c_{2} x e^{- 5 x} + \frac{e ^{- 5 x}}{20 x ^{4}}

y = c_{1} e^{- 5 x} + c_{2} x e^{- 5 x} + \frac{e ^{- 5 x}}{20 x ^{4}}

y^{^{''}} - 4 y^{^{'}} + 8 y = 12 x e^{2 x} + 6 e^{2 x}

y^{^{''}} + y = 8 x cos (x)

(D^{2} - 6 D + 9) y = x^{2} e^{2 x}

y^{^{''}} + 2 y^{^{'}} + 10 = 0, y (0) = - 2, y^{^{'}} (0) = 0

4 y^{^{''}} + y = 10 cos (x)

What is the solution for y^{''}+10y^'+25y=(e^{-5x})/(x^6) ?
The solution for y^{''}+10y^'+25y=(e^{-5x})/(x^6) is y=c_{1}e^{-5x}+c_{2}xe^{-5x}+(e^{-5x})/(20x^4)