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

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

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y^{''}-6y^'+9y= 1/(x^2)e^{3x}

y^{^{''}} - 6 y^{^{'}} + 9 y = \frac{1}{x ^{2}} e^{3 x}

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{1}{x ^{2}} e^{3 x}

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}

y^{^{''}} - 2 y^{^{'}} + y = \frac{15 e ^{t}}{1 + t ^{2}}

x^{^{''}} + 4 x^{^{'}} + 29 = 0, x (0) = 5, x^{^{'}} (0) = 5

y^{^{''}} - 4 y^{^{'}} + 3 y = 3 e^{- 2 t}

\frac{d ^{2} x}{d t ^{2}} + 2 \frac{d x}{d t} + x = 1

y^{^{''}} - y = e^{a t}

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