What is the solution for y^{''}-6y^'+9y=x^{-3}e^{3x} ?

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

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y^{''}-6y^'+9y=x^{-3}e^{3x}

y^{^{''}} - 6 y^{^{'}} + 9 y = x^{- 3} e^{3 x}

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

Solution steps

y^{''} (x) - 6 y^{'} (x) + 9 y = x^{- 3} e^{3 x}

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} + 2 y = 12 x^{2}

d (n) = 6 - 4 (n - 1)

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

y^{^{''}} - 4 y^{^{'}} + 5 y = 125 t^{2}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 18 y^{^{'}} + 45 y = e^{2 x} + x

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