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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y = 4 sec^{2} (x)

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

y^{^{''}} - 3 y^{^{'}} - 4 y = 30 e^{4 x}

y^{^{''}} - y^{^{'}} - 12 y = 3 x - 2 + 5 e^{x}

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

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