TITLE: y'''+4y"+4y'=2 solution of non-homogenous differential equation
QUESTION [6 upvotes]: This is non-homogenous differential equation :
$$y'''+4y''+4y'=2.$$
Of course, I started with characteristic polynomial of homogenous case:
$$t^3+4t^2+4t=0$$ then $$t(t^2+4t+4)=0$$ we have:
$$t_1=0; t_{2,3}=-2.$$ So, solution of homogenous case is:
$$y_s(x)=c_1 + c_2e^{-2x}+c_3xe^{-2x}$$
Now, I want to continue from this point to solution of non-homogenous differential equation. Please give any hint or general solution!! Thanks in advance.

REPLY [3 votes]: we can make the D.E as a homogeneous D.E by taking the derivative 
$$y''''+4y'''+4y''=0$$
the characteristics equation
$$r^2(r^2+4r+4)=0$$
so
$$y=c_1+c_2x+c_3e^{-2x}+c_4xe^{-2x}$$