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

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

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y^{''}-6y^'+9y=(e^{3x})/(x^2)

y^{^{''}} - 6 y^{^{'}} + 9 y = \frac{e ^{3 x}}{x ^{2}}

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

Solution steps

y^{''} (x) - 6 y^{'} (x) + 9 y = \frac{e ^{3 x}}{x ^{2}}

Solve linear ODE:

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

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

\frac{d y}{d x} = \frac{10 x ^{2}}{1 + x ^{3}}

x = 15 - y^{2}, x = y^{2} - 3

\int_{0}^{\frac{π}{8}} arctan (2 x) d x

\frac{\partial}{\partial w} (w arctan (uv))

y^{^{''}} + 2 y - 1 (y^{^{'}}) 2 = y^{^{'}}

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