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

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

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y^{'''}-y^{''}-y^'+y=2(1+e^{-x})

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 2 (1 + e^{- x})

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

Solution steps

y^{'''} (x) - y^{''} (x) - y^{'} (x) + y = 2 (1 + e^{- x})

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} - 2 y = 18 x e^{- 2 x} - 12 e^{- 2 x}

y^{^{''}} + 5 y^{^{'}} + 4 y = 10 x e^{5 x}

y^{^{''}} + 2 y^{^{'}} - 3 y = t

y^{^{''}} - 6 y^{^{'}} + 9 y = x^{2} e^{3 x} - 8 e^{3 x}

y^{^{''}} + 5 y^{^{'}} + 6 y = 10

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