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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} + 2 y - x - 3 = 0

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

y^{^{''}} - y = 8 e^{9 t} + t e^{9 t} + 5 t^{2} e^{9 t}

14 y^{^{''}} + y^{^{'}} + y = x^{2} - 4 x

y^{^{'''}} - y^{^{''}} = cos (2 t)

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