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

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

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

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

y = c_{1} e^{3 x} + c_{2} x e^{3 x} - e^{3 x} x + e^{3 x} x ln (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} x e^{3 x} - e^{3 x} x + e^{3 x} x ln (x)

y = c_{1} e^{3 x} + c_{2} x e^{3 x} - e^{3 x} x + e^{3 x} x ln (x)

y^{^{''}} - 2 y^{^{'}} + 10 y = sin (3 x) + cos (3 x)

y^{^{''}} - 4 y^{^{'}} + 13 y = e^{x} cos (x)

d (x) = 5 (\frac{1}{3})^{x}

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

y^{^{''}} - 2 y^{^{'}} + y = e^{2 x} + x + 1

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