Third order non-homogeneous differential equation I have no idea on how to work this out.  I've tried variation of parameters, undetermined coefficients, making it into a system, etc.
$$y'''+2y''+5y'+20e^{-x}\cos(2x)=0$$
 A: Hint
If the differential equation is $$y'''+2y''+5y'+20e^{-x}\cos(2x)=0$$ first reduce the order using $y'=z$ and it becomes $$z''+2z'+5z+20e^{-x}\cos(2x)=0$$ The characteristic equation is $m^2+2m+5=0$ and its roots are $m_{\pm}=-1\pm2i$,so you know that the solution will contain the factor $e^{-x}$; so, define $z=e^{-x}u$ and the equation becomes $$u''+4u +20\cos(2x)=0$$ $$u_c=c_1\cos(2x)+c_2\sin(2x)$$ $$u_p=x\big(a\cos(2x)+b\sin(2x)\big)$$ which make the derivatives much simpler $$u'_p=\sin (2 x) (b-2 a x)+\cos (2 x) (a+2 b x)$$ $$u''_p=4 \cos (2 x) (b-a x)-4 \sin (2 x) (a+b x)$$ which lead to $$u''+4u +20\cos(2x)=4 (b+5) \cos (2 x)-4 a \sin (2 x)=0$$
I am sure that you can take from here.
A: you have two ways to solve this O.D.E as follow:
1- find the characteristic equation 
$$m^3+2m^2+5m=0$$
$$m(m^2+2m+5)=0$$
$$m_1=0$$
$$m_2=-1+2i$$
$$m_3=-1-2i$$
the pcomplementry  solution is
$$y_c=C_1e^0+e^{-x}[C_2\cos(2x)+C_3\sin(2x)]$$
now we will find the particular solution 
$$y_p=xe^{-x}[A\cos(2x)+B\sin(2x)]$$
and then complete the solution
2-take integration for both sides to get
$$y''+2y'+5y=\int -20e^{-x}\cos(2x)dx$$
$$y''+2y'+5y=-1/5e^{-x}[\cos(2x)-2\sin(2x)]+C$$
you can get the $C$ by using the boundary condition and then solve the second-order D.E by finding the $y_c$ firstly and $y_p$ secondly
