What is the solution for y^{'''}-4y^{''}-y^'+4y=e^{6x} ?

The solution for y^{'''}-4y^{''}-y^'+4y=e^{6x} is y=c_{1}e^{4x}+c_{2}e^{-x}+c_{3}e^x+1/70 e^{6x}

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

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

y = c_{1} e^{4 x} + c_{2} e^{- x} + c_{3} e^{x} + \frac{1}{70} e^{6 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{- x} + c_{3} e^{x} + \frac{1}{70} e^{6 x}

y = c_{1} e^{4 x} + c_{2} e^{- x} + c_{3} e^{x} + \frac{1}{70} e^{6 x}

What is the solution for y^{'''}-4y^{''}-y^'+4y=e^{6x} ?
The solution for y^{'''}-4y^{''}-y^'+4y=e^{6x} is y=c_{1}e^{4x}+c_{2}e^{-x}+c_{3}e^x+1/70 e^{6x}