What is the solution for y^{''}-6y^'+9y=4e^{3x} ?

The solution for y^{''}-6y^'+9y=4e^{3x} is y=c_{1}e^{3x}+c_{2}xe^{3x}+2x^2e^{3x}

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y^{''}-6y^'+9y=4e^{3x}

y^{^{''}} - 6 y^{^{'}} + 9 y = 4 e^{3 x}

y = c_{1} e^{3 x} + c_{2} x e^{3 x} + 2 x^{2} e^{3 x}

Solution steps

y^{''} (x) - 6 y^{'} (x) + 9 y = 4 e^{3 x}

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} x e^{3 x} + 2 x^{2} e^{3 x}

y = c_{1} e^{3 x} + c_{2} x e^{3 x} + 2 x^{2} e^{3 x}

y^{^{''}} + 3 y^{^{'}} - 4 y = (6 x + 1) e^{2 x}

y^{^{''}} + y^{^{'}} + y = 13 e^{- 2 x} (cos (x) + 2 sin (x))

y^{^{''}} + y^{^{'}} = 7, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 10 y^{^{'}} + 25 y = \frac{e ^{- 10 x}}{x ^{2}}

y^{^{'''}} - 2 y^{^{''}} + y^{^{'}} = x^{3} + 2 e^{x}

What is the solution for y^{''}-6y^'+9y=4e^{3x} ?
The solution for y^{''}-6y^'+9y=4e^{3x} is y=c_{1}e^{3x}+c_{2}xe^{3x}+2x^2e^{3x}