Solving Higher-Order Non-Homogeneous ODE $y'''-3y''+3y'-y=24(x+1)e^{x}$ Given equation:

$$y'''-3y''+3y'-y=24(x+1)e^x$$ with initial values $y(0)=1$ and $y'(0)=2$ and $y''(0)=7$

I get the homogeneous equation as $(c_1+c_2x+c_3x^{2})e^{x}$

But I don't get how to get the particular solution as it is written $24(x+1)e^{x}$
Easy way:
$$y'''-3y''+3y'-y=24(x+1)e^x$$
$$(ye^{-x})'''=24(x+1)$$
Integrate three times.



Another way:
$$y'''-3y''+3y'-y=24(x+1)e^x$$
The poynomial characteristic is:
$$r^3-3r^2+3r-1=0 \implies (r-1)^3$$
The solution to the homogeneous equation is
$$y_h=c_1e^x+c_2xe^x+c_3x^2e^x$$
For the particular solution try:
$$y_p=x^3(Ax+B)e^x$$
$$\frac{1}{f(D)} \left ( e^{ax} \phi(x) \right ) = e^{ax} \left ( \frac{1}{f(D+a)} \phi(x) \right )$$

The formula above would be of help to solve the particular integral $$\int 24 x(e^x)\,dx.$$