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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} = 8 x e^{3 x}

y^{^{''}} - 6 y^{^{'}} + 9 y = 4 e^{x}

y^{^{''}} - 4 y^{^{'}} + 7 y = t e^{t}

y^{^{''}} + 9 y = sin (2 x), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 4 y = 8 sin (2 x) + 5 e^{x}

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