5th order ODE question 


*(a) Find the general solution of the fifth order ODE
$$y^{(5)}+4 y^{\prime \prime \prime}+4 y^{\prime}=0$$
(b) Consider the fifth order ODE
$$y^{(5)}+4 y^{\prime \prime \prime}+4 y^{\prime}=\cos \omega t$$
For which real numbers $\omega$ equation (2) does NOT have any periodic or constant solutions?



Yesterday I wrote my 2nd differential equations midterm and that was one of the questions. To solve part a) I wrote the characteristic equation: $$r^5+4r^3+4r=r(r^4+4r^2+4)=r(r^2+2)^2$$
and found that the homogeneous solution is $$y=c_1+c_2\cos\sqrt{2}t + c_3\sin\sqrt{2}t+c_4t\cos\sqrt{2}t + c_5t\sin\sqrt{2}t$$
I need help with part b). I do not know how to solve it. Using the method of undetermined coefficients I guessed a solution to be in the form of $Y=Acos\omega t+Bsin \omega t$. I then differentiated it 5 times and subbed everything into the equation. I still didn't understand what to do or how to do it so for my final answer I just wrote $\omega=0$ Could someone explain how to solve it?

You guessed the solution Y correctly but didn't use the solution from part a. The solution for part b is the sum of the solution from part A and your guess solution. Then you go through differentiation and equate corresponding coefficients for sin(t) and cos(t) to get values for A and B.
