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

The solution for y^{'''}-3y^{''}+3y^'-y=x is y=c_{1}e^x+c_{2}xe^x+c_{3}x^2e^x-x-3

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (cos (ln (x)) + sin (ln (x)))

\int \frac{x}{2} + \frac{2}{x} d x

\frac{\partial}{\partial x} (\frac{x + y}{x y - 1})

\frac{d y}{d x} = sin (y) + 1

\frac{d}{d x} (x^{3 c o s (x)})

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