What is the solution for y^{''}-4y^'-12y=x-6 ?

The solution for y^{''}-4y^'-12y=x-6 is y=c_{1}e^{6x}+c_{2}e^{-2x}-x/(12)+19/36

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y^{''}-4y^'-12y=x-6

y^{^{''}} - 4 y^{^{'}} - 12 y = x - 6

y = c_{1} e^{6 x} + c_{2} e^{- 2 x} - \frac{x}{12} + \frac{19}{36}

Solution steps

y^{''} (x) - 4 y^{'} (x) - 12 y = x - 6

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} e^{- 2 x} - \frac{x}{12} + \frac{19}{36}

y = c_{1} e^{6 x} + c_{2} e^{- 2 x} - \frac{x}{12} + \frac{19}{36}

x^{^{''}} + x = cos (3 t)

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 5 + 4 sin (x)

y^{^{''}} + 2 y^{^{'}} - 4 y = 8 x^{2}

8 y^{^{''}} - 32 y = 16 cos (x)

y^{^{''}} - 6 y^{^{'}} + 5 y = x, y (0) = e^{x}

What is the solution for y^{''}-4y^'-12y=x-6 ?
The solution for y^{''}-4y^'-12y=x-6 is y=c_{1}e^{6x}+c_{2}e^{-2x}-x/(12)+19/36