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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} - y^{^{'}} = e^{t} cos (t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 10 y^{^{'}} + 25 y = e^{5 x} ln (4 + x^{2})

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e^{x} - 9

y^{^{''}} - 10 y^{^{'}} + 9 y = 5 t, y (0) = - 1, y^{^{'}} (0) = 2

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