What is the solution for y^{''}-4y^'-12y=8e^{-2x} ?

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

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y^{''}-4y^'-12y=8e^{-2x}

y^{^{''}} - 4 y^{^{'}} - 12 y = 8 e^{- 2 x}

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

Solution steps

y^{''} (x) - 4 y^{'} (x) - 12 y = 8 e^{- 2 x}

Solve linear ODE:

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

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

\frac{d}{d x} (x e^{x} cos (x))

\int (5 x^{2} + 2 x^{3}) d x

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

\int (3 x + 5)^{6} d x

\frac{\partial}{\partial x} (5 cos (5 x))

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