What is the solution for y^{'''}-3y^{''}+3y^'-y=e^x-x+16 ?

The solution for y^{'''}-3y^{''}+3y^'-y=e^x-x+16 is y=c_{1}e^x+c_{2}xe^x+c_{3}x^2e^x+(e^xx^3)/6+x-13

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y^{'''}-3y^{''}+3y^'-y=e^x-x+16

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = e^{x} - x + 16

y = c_{1} e^{x} + c_{2} x e^{x} + c_{3} x^{2} e^{x} + \frac{e ^{x} x ^{3}}{6} + x - 13

Solution steps

y^{'''} (x) - 3 y^{''} (x) + 3 y^{'} (x) - y = e^{x} - x + 16

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + c_{3} x^{2} e^{x} + \frac{e ^{x} x ^{3}}{6} + x - 13

y = c_{1} e^{x} + c_{2} x e^{x} + c_{3} x^{2} e^{x} + \frac{e ^{x} x ^{3}}{6} + x - 13

(d^{2} + 4 d + 5) \cdot y = - 2 cos (x)

(d^{3} - 3 d + 2) \cdot y = 540 x^{3} \cdot e^{- x}

(d^{3} + 8) \cdot y = x^{4} + 2 x + 1 + 3 sin (x)

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

(d^{3} - d^{2} - d + 1) \cdot y = x (e^{- x} + 1)

What is the solution for y^{'''}-3y^{''}+3y^'-y=e^x-x+16 ?
The solution for y^{'''}-3y^{''}+3y^'-y=e^x-x+16 is y=c_{1}e^x+c_{2}xe^x+c_{3}x^2e^x+(e^xx^3)/6+x-13