What is the solution for y^{'''}-2y^{''}-y^'+2y=6e^{2x} ?

The solution for y^{'''}-2y^{''}-y^'+2y=6e^{2x} is y=c_{1}e^{2x}+c_{2}e^{-x}+c_{3}e^x+2xe^{2x}

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

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

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

Solution steps

y^{'''} (x) - 2 y^{''} (x) - y^{'} (x) + 2 y = 6 e^{2 x}

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} + y = sin (2 x)

y^{^{''}} + 9 y = cos (3 t), y (0) = 2, y^{^{'}} (0) = 5

y^{^{''}} - y^{^{'}} - 6 y = (x^{2} + 3 x) e^{2 x} + cos (2 x)

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} + 12 y = 20 cos (3 x)

y^{^{''}} + 6 y^{^{'}} + 13 y = e^{- 3 x} \cdot cos (2 x)

What is the solution for y^{'''}-2y^{''}-y^'+2y=6e^{2x} ?
The solution for y^{'''}-2y^{''}-y^'+2y=6e^{2x} is y=c_{1}e^{2x}+c_{2}e^{-x}+c_{3}e^x+2xe^{2x}