What is the solution for y^{''}-y^'-2y=16xe^{3x} ?

The solution for y^{''}-y^'-2y=16xe^{3x} is y=c_{1}e^{2x}+c_{2}e^{-x}+4xe^{3x}-5e^{3x}

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y^{''}-y^'-2y=16xe^{3x}

y^{^{''}} - y^{^{'}} - 2 y = 16 x e^{3 x}

y = c_{1} e^{2 x} + c_{2} e^{- x} + 4 x e^{3 x} - 5 e^{3 x}

Solution steps

y^{''} (x) - y^{'} (x) - 2 y = 16 x e^{3 x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- x} + 4 x e^{3 x} - 5 e^{3 x}

y = c_{1} e^{2 x} + c_{2} e^{- x} + 4 x e^{3 x} - 5 e^{3 x}

y^{^{''}} - 4 y = 8 x

y^{^{''}} + 9 y = cos (x)

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} - 12 y = x e^{2 x}

y^{^{''}} + 3 y^{^{'}} - 4 y = (x^{3} + x) e^{x}

y^{^{''}} - 5 y^{^{'}} + 6 y = 3 e^{3 x} + 36 x^{2}

What is the solution for y^{''}-y^'-2y=16xe^{3x} ?
The solution for y^{''}-y^'-2y=16xe^{3x} is y=c_{1}e^{2x}+c_{2}e^{-x}+4xe^{3x}-5e^{3x}