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

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

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

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

y = c_{1} e^{2 x} + c_{2} e^{- x} + c_{3} e^{x} + \frac{1}{8} e^{3 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- x} + c_{3} e^{x} + \frac{1}{8} e^{3 x}

y = c_{1} e^{2 x} + c_{2} e^{- x} + c_{3} e^{x} + \frac{1}{8} e^{3 x}

y^{^{''}} + y = x e^{x}

y^{^{''}} + 5 y^{^{'}} + 6 y = 24 e^{t} + 36 t^{2}

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

y^{^{''}} - 2 y^{^{'}} + 2 = 0

y^{^{''}} - 2 y^{^{'}} + 82 y = e^{x} cos (9 x)

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