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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y = (x^{2} - 3) \cdot sin (2 x)

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

y^{^{''}} + 8 y^{^{'}} + 17 y = 2 e^{- 3 x}

x^{^{''}} + 2 x^{^{'}} + 41 x = 8 cos (6 t)

y^{^{'''}} + 64 y^{^{'}} = tan (8 x)

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