What is the solution for y^{''}+4y^'-12y=3x^2+x ?

The solution for y^{''}+4y^'-12y=3x^2+x is y=c_{1}e^{2x}+c_{2}e^{-6x}-(x^2)/4-x/4-1/8

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y^{''}+4y^'-12y=3x^2+x

y^{^{''}} + 4 y^{^{'}} - 12 y = 3 x^{2} + x

y = c_{1} e^{2 x} + c_{2} e^{- 6 x} - \frac{x ^{2}}{4} - \frac{x}{4} - \frac{1}{8}

Solution steps

y^{''} (x) + 4 y^{'} (x) - 12 y = 3 x^{2} + x

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 6 x} - \frac{x ^{2}}{4} - \frac{x}{4} - \frac{1}{8}

y = c_{1} e^{2 x} + c_{2} e^{- 6 x} - \frac{x ^{2}}{4} - \frac{x}{4} - \frac{1}{8}

2 y^{^{''}} + y^{^{'}} - 10 y = x e^{3 x} cos (π x)

y^{^{'''}} + 3 y^{^{''}} - 10 y^{^{'}} = x - 3

y^{^{'''}} - 24 y^{^{''}} + 2048 y = e^{16 x}

y^{^{''}} + y = sec (2) [t]

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

What is the solution for y^{''}+4y^'-12y=3x^2+x ?
The solution for y^{''}+4y^'-12y=3x^2+x is y=c_{1}e^{2x}+c_{2}e^{-6x}-(x^2)/4-x/4-1/8