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}-1/4 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}{4} 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}{4} e^{- x} - 2 x + \frac{1}{3}

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{1}{4} e^{- x} - 2 x + \frac{1}{3}

5 y^{^{''}} - 15 y = 40 e^{x}

y^{^{''}} + 6 y^{^{'}} + 10 y = e^{- 3 t}, y (0) = 4, y^{^{'}} (0) = 5

y^{^{'''}} = 3 y^{^{''}} + 9 (e^{(5 x)}) + 2 (cos (x))

y^{^{''}} - y^{^{'}} - y = x^{2} - 2

y^{^{''}} - 16 y^{^{'}} = 32 + 45 e^{x}

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}-1/4 e^{-x}-2x+1/3