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

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

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y^{''}-y^'-6y=-e^x+12x

y^{^{''}} - y^{^{'}} - 6 y = - e^{x} + 12 x

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} + \frac{1}{6} e^{x} - 2 x + \frac{1}{3}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} + \frac{1}{6} e^{x} - 2 x + \frac{1}{3}

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} + \frac{1}{6} e^{x} - 2 x + \frac{1}{3}

y^{^{''}} - 9 y = sinh (3 x)

y^{^{''''}} - 2 y^{^{'''}} + y^{^{''}} = e^{x} + 1

y^{^{''}} + 6 y^{^{'}} - 7 y = 3 e^{2 x} - e^{- x}

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

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

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