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

y^{^{''}} - 5 y = 3 x^{3} + 2 x^{2} - x - 3

y^{^{''}} + 2 y^{^{'}} + y = (\frac{1}{6}) e^{- x}

d (x) = 2 x^{2} - 3

y^{^{'''}} + 2 y^{^{''}} - y^{^{'}} - 2 y = 5 H (t)

y^{^{''''}} + 3 y^{^{'''}} - 4 y^{^{''}} - 4 y^{^{'}} + \frac{1}{2} = 0

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