What is the solution for y^{'''}-5y^{''}-y^'+5y=e^{4x} ?

The solution for y^{'''}-5y^{''}-y^'+5y=e^{4x} is y=c_{1}e^{5x}+c_{2}e^{-x}+c_{3}e^x-1/15 e^{4x}

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y^{'''}-5y^{''}-y^'+5y=e^{4x}

y^{^{'''}} - 5 y^{^{''}} - y^{^{'}} + 5 y = e^{4 x}

y = c_{1} e^{5 x} + c_{2} e^{- x} + c_{3} e^{x} - \frac{1}{15} e^{4 x}

Solution steps

y^{'''} (x) - 5 y^{''} (x) - y^{'} (x) + 5 y = e^{4 x}

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} e^{- x} + c_{3} e^{x} - \frac{1}{15} e^{4 x}

y = c_{1} e^{5 x} + c_{2} e^{- x} + c_{3} e^{x} - \frac{1}{15} e^{4 x}

(D^{2} - D) y = 2 - 2 x

y^{^{''}} - 6 y^{^{'}} + 8 y = x, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 2 y^{^{'}} - 63 y = x^{2}

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{t}, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + y = cos (a x)

What is the solution for y^{'''}-5y^{''}-y^'+5y=e^{4x} ?
The solution for y^{'''}-5y^{''}-y^'+5y=e^{4x} is y=c_{1}e^{5x}+c_{2}e^{-x}+c_{3}e^x-1/15 e^{4x}