What is the solution for y^{''}+6y^'+9y= 1/x ?

The solution for y^{''}+6y^'+9y= 1/x is y=c_{1}e^{-3x}+c_{2}xe^{-3x}-1/3+e^{-3x}x\Ei(3x)

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y^{''}+6y^'+9y= 1/x

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{1}{x}

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

Solution steps

y^{''} (x) + 6 y^{'} (x) + 9 y = \frac{1}{x}

Solve linear ODE:

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

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

x^{^{''}} + 4 x^{^{'}} + 4 x = 100 sin (4 t)

x^{^{''}} + 50 x^{^{'}} + 400 x = 100 cos (20 t)

y^{^{''}} + y^{^{'}} - 2 y = e^{t} sin (t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 22 y^{^{'}} + 121 y = 16.3 u

y^{^{''}} + y = 2 e^{- 8 t}, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}+6y^'+9y= 1/x ?
The solution for y^{''}+6y^'+9y= 1/x is y=c_{1}e^{-3x}+c_{2}xe^{-3x}-1/3+e^{-3x}x\Ei(3x)