What is the solution for y^{''}-2y^'-8y=3e^{3x} ?

The solution for y^{''}-2y^'-8y=3e^{3x} is y=c_{1}e^{4x}+c_{2}e^{-2x}-3/5 e^{3x}

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y^{''}-2y^'-8y=3e^{3x}

y^{^{''}} - 2 y^{^{'}} - 8 y = 3 e^{3 x}

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

Solution steps

y^{''} (x) - 2 y^{'} (x) - 8 y = 3 e^{3 x}

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} - 5 y = (4 x^{2} - 7 x + 0) e^{4 x}

y^{^{''}} - 2 y^{^{'}} + y = 4 x^{2} - 6

y^{^{''''}} - 2 y^{^{''}} + y = 5 sin (2 x)

y^{^{''}} + 3 y = e^{t}

y^{^{''}} - 9 y^{^{'}} = 9 x e^{- 3 x}

What is the solution for y^{''}-2y^'-8y=3e^{3x} ?
The solution for y^{''}-2y^'-8y=3e^{3x} is y=c_{1}e^{4x}+c_{2}e^{-2x}-3/5 e^{3x}