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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y = x + e^{x}

y^{^{''''}} - 5 y^{^{''}} + 4 y = e^{x} (- 122 - 48 x + 72 x^{2})

y^{^{''}} + y^{^{'}} - 12 y = e^{t} + e^{2 t} - 1

y^{^{'''}} + 100 y^{^{'}} = tan (10 x)

y^{^{''}} - 6 y^{^{'}} + 8 y = 5 x^{2} sin (4 x)

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