What is the solution for (D^3+D^2+D+1)y=xe^x ?

The solution for (D^3+D^2+D+1)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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(D^3+D^2+D+1)y=xe^x

(D^{3} + D^{2} + D + 1) 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

(D^{3} + D^{2} + D + 1) 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}

30 y^{^{''}} + 25 y^{^{'}} + 4 y = 40

y^{^{''}} + y^{^{'}} - 12 y = 4

y^{^{''}} - y = sec^{3} (x) - sec (x)

y^{^{''''}} - 5 y^{^{''}} + 4 y = 8 cos (2 x) - 12 e^{- x}

x^{^{''}} + x = 9 cos (8 t)

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