What is the solution for y^{'''}-3y^{''}+3y^'-y=x-8e^x ?

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

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

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

y = c_{1} e^{x} + c_{2} x e^{x} + c_{3} x^{2} e^{x} - x - 3 - \frac{4 e ^{x} x ^{3}}{3}

Solution steps

y^{'''} (x) - 3 y^{''} (x) + 3 y^{'} (x) - y = x - 8 e^{x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + c_{3} x^{2} e^{x} - x - 3 - \frac{4 e ^{x} x ^{3}}{3}

y = c_{1} e^{x} + c_{2} x e^{x} + c_{3} x^{2} e^{x} - x - 3 - \frac{4 e ^{x} x ^{3}}{3}

\frac{d ^{2} y}{d t ^{2}} = 0.5 t - y

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

y^{^{'''}} - y^{^{'}} = 7 t

y^{^{'''}} + 4 y^{^{'}} = t

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

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