What is the solution for y^{''}+y^'-6y=6x^3-3x^2+12x ?

The solution for y^{''}+y^'-6y=6x^3-3x^2+12x is y=c_{1}e^{2x}+c_{2}e^{-3x}-x^3-3x-1/2

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y^{''}+y^'-6y=6x^3-3x^2+12x

y^{^{''}} + y^{^{'}} - 6 y = 6 x^{3} - 3 x^{2} + 12 x

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

Solution steps

y^{''} (x) + y^{'} (x) - 6 y = 6 x^{3} - 3 x^{2} + 12 x

Solve linear ODE:

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

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

y^{^{''}} + 6 y^{^{'}} + 12 y = 763 e^{2 t} cos (7 t)

y^{^{''}} + y = \frac{2}{cos ( x )}, y (0) = 2, y^{^{'}} (0) = 2

y^{^{''}} + y = 4 x + 10 sin (x)

y^{^{''}} - 5 y^{^{'}} + 6 y = t e^{t}

2 y^{^{''}} + 2 y^{^{'}} + y = 4 x

What is the solution for y^{''}+y^'-6y=6x^3-3x^2+12x ?
The solution for y^{''}+y^'-6y=6x^3-3x^2+12x is y=c_{1}e^{2x}+c_{2}e^{-3x}-x^3-3x-1/2