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

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

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

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

y = c_{1} e^{- x} + c_{2} cos (x) + c_{3} sin (x) + \frac{e ^{- x} x}{2}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} cos (x) + c_{3} sin (x) + \frac{e ^{- x} x}{2}

y = c_{1} e^{- x} + c_{2} cos (x) + c_{3} sin (x) + \frac{e ^{- x} x}{2}

y^{^{''}} - 8 y^{^{'}} + 16 y = 16 e^{4 x}

\frac{d ^{3} y}{d x ^{3}} + y = 3 + e^{- x} + 5 e^{2 x}

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = 2 e^{3 x} + 4 x

y^{^{''}} - 3 y^{^{'}} = e^{3 x} + 2 x - 1

y^{^{''}} + 2 y^{^{'}} + y = 7 + 75 sin (2 x)

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