y'''+4y"+4y'=2 solution of non-homogenous differential equation 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.
 A: Using the Method of Undetermined Coefficients, the particular solution is of the form $Ax$.
Note that $y_p(x)=Ax\implies y'_p(x)=A\implies y''_p(x)=0\implies y'''_x(p)=0$
Now, substituting into our original differential equation, we get that $$0+4\cdot 0+4A=2$$
$$\therefore A=\frac 12\implies y_p(x)=\frac x2$$
$$\therefore y=\frac x2+c_1 + c_2e^{-2x}+c_3xe^{-2x}$$
A: Let $w=y'$, so that the differential equation becomes
$$w''+4w'+4w=2.$$
Then let $w_p=A$, and work with
$$w_p''+4w_p'+4w_p=2$$
Solve for $w_p$, then integrate to find $y_p$ (since $w_p=y_p'$).
A: HINT:
$$4y'(x)+4y''(x)+y'''(x)=2\Longleftrightarrow$$

The general solution will be the sum of the complementary solution and particular solution.
Find the complementary solution by solving:

$$4y'(x)+4y''(x)+y'''(x)=0\Longleftrightarrow$$

Assume a solution will be proportional to $e^{\lambda x}$ for some constant $\lambda$.
Substitute $y(x)=e^{\lambda x}$ into the differential equation:

$$\frac{\text{d}^3}{\text{d}x^3}\left(e^{\lambda x}\right)+4\frac{\text{d}^2}{\text{d}x^2}\left(e^{\lambda x}\right)+4\frac{\text{d}}{\text{d}x}\left(e^{\lambda x}\right)=0\Longleftrightarrow$$

Substitute $\frac{\text{d}^3}{\text{d}x^3}\left(e^{\lambda x}\right)=\lambda^3e^{\lambda x}$:

$$\lambda^3e^{\lambda x}+4\lambda^2e^{\lambda x}+4\lambda e^{\lambda x}=0\Longleftrightarrow$$
$$e^{\lambda x}\left(\lambda^3+4\lambda^2+4\lambda\right)=0\Longleftrightarrow$$

Since $e^{\lambda x}\ne0$ for any finite $\lambda$, the zeros must come from the polynomial:

$$\lambda^3+4\lambda^2+4\lambda=0\Longleftrightarrow$$
$$\lambda\left(\lambda+2\right)^2=0$$
A: 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
$$y=c_1+c_2x+c_3e^{-2x}+c_4xe^{-2x}$$ 
A: Hint:
$$y'''+4y''+4y'=2$$
the particular solution should be (according to the undetermined Coefficient Method)
$$y_p=A$$
but because there is a similirity with the complementary solution, so we should multiply by x
$$y_p=Ax$$
