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

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

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

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

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 (1 + x^{2})

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 12 e^{x} - 9

\frac{d ^{2} y}{d x ^{2}} + 4 y = sin (x)

y^{^{''}} + 16 y = cos (4 t)

4 y^{^{''}} - y = e^{\frac{x}{2}} + 4

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