What is the solution for y^{'''}-3y^{''}+3y^'-y=x-4e^x ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int sin (2 x - 1) d x

x^{2} y^{^{'}} - x y = y^{2}

\frac{\partial}{\partial y} (y)

\frac{4}{3} x y^{^{'}} - \frac{3}{8} y = 0

\int 2 x cos (3 x) d x

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