What is the solution for y^{''}-y^'-12y=98(x+2)e^{-3x} ?

The solution for y^{''}-y^'-12y=98(x+2)e^{-3x} is y=c_{1}e^{4x}+c_{2}e^{-3x}-7x^2e^{-3x}-30e^{-3x}x

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y^{''}-y^'-12y=98(x+2)e^{-3x}

y^{^{''}} - y^{^{'}} - 12 y = 98 (x + 2) e^{- 3 x}

y = c_{1} e^{4 x} + c_{2} e^{- 3 x} - 7 x^{2} e^{- 3 x} - 30 e^{- 3 x} x

Solution steps

y^{''} (x) - y^{'} (x) - 12 y = 98 (x + 2) e^{- 3 x}

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{- 3 x} - 7 x^{2} e^{- 3 x} - 30 e^{- 3 x} x

y = c_{1} e^{4 x} + c_{2} e^{- 3 x} - 7 x^{2} e^{- 3 x} - 30 e^{- 3 x} x

What is the solution for y^{''}-y^'-12y=98(x+2)e^{-3x} ?
The solution for y^{''}-y^'-12y=98(x+2)e^{-3x} is y=c_{1}e^{4x}+c_{2}e^{-3x}-7x^2e^{-3x}-30e^{-3x}x