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

The solution for y^{'''}+y^{''}+y^'+y=xe^x is y=c_{1}e^{-x}+c_{2}cos(x)+c_{3}sin(x)+(e^xx)/4-3/8 e^x

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

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

y = c_{1} e^{- x} + c_{2} cos (x) + c_{3} sin (x) + \frac{e ^{x} x}{4} - \frac{3}{8} e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} cos (x) + c_{3} sin (x) + \frac{e ^{x} x}{4} - \frac{3}{8} e^{x}

y = c_{1} e^{- x} + c_{2} cos (x) + c_{3} sin (x) + \frac{e ^{x} x}{4} - \frac{3}{8} e^{x}

y^{^{''}} - 4 y = 4 e^{3 x}

y^{^{''}} + 2 y^{^{'}} + y = g (t)

y^{^{''}} - 4 y^{^{'}} = 3 x e^{x} + 2

y^{^{''}} + 25 y = 8 cos (5 x) + 12 sin (5 x)

y^{^{''}} - 6 y^{^{'}} + 25 y = 2 sin (\frac{x}{2}) - cos (\frac{x}{2})

What is the solution for y^{'''}+y^{''}+y^'+y=xe^x ?
The solution for y^{'''}+y^{''}+y^'+y=xe^x is y=c_{1}e^{-x}+c_{2}cos(x)+c_{3}sin(x)+(e^xx)/4-3/8 e^x