How can I solve \begin{equation} y'''+6y''+y'-34y=0 \end{equation} (I) by order reduction method, knowing that \begin{equation} y_1(x)=e^{-4x} cos(x) \end{equation} is a solution of (I)?
I did: \begin{equation} y(x)=v(x)e^{-4x} cos(x) \end{equation} And I arrived on \begin{equation} p''-p'(6+3tan(x))+p(12tan(x)-2)=0 \end{equation} where \begin{equation} p=v'(x) \end{equation} But I couln't move on from there.
It seems that you forgot the basic stuff when facing this kind of differential equations with constant coefficients.
Given $$\begin{equation} y'''+6y''+y'-34y=0 \end{equation}$$ the characteristic equation is $$r^3+6 r^2+r-34=0\implies (r-2)(r^2+8 r+17)=0$$ So, the roots are $$r_1=2\qquad r_2=-4-i\qquad r_3=-4+i$$ making the solution to be $$y=c_1 e^{2 x}+c_2 e^{-4 x} \sin (x)+c_3 e^{-4 x} \cos (x)$$
End of the story.
