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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 6 y^{^{'}} + 9 y = t^{2} e^{3}

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

y^{^{''}} + 4 y = cos (2 t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + y^{^{'}} + 5 y = 4 e^{- t} cos (2 t)

y^{^{''}} - 5 y^{^{'}} + 4 y = e^{2 t}

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