1. (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?

  • $\begingroup$ MathJax hint: Use \sin and \cos for proper spacing. Also the homogenous solution should contain $\sin(\sqrt 2 t)$ and $\cos(\sqrt2 t)$ terms. The ones you wrote are constant w.r.t. $t$ so they make no sense. $\endgroup$ – AlexR Mar 12 '15 at 16:06
  • $\begingroup$ I made the edits. Any idea how to solve? $\endgroup$ – marcus Mar 12 '15 at 16:13
  • $\begingroup$ My guess would be $\omega = 0$ will definately be aperiodic because it blows up as $t\to\infty$. This may in fact be the only critical value. Try to find a particular solution assuming $\omega\ne 0$, maybe? $\endgroup$ – AlexR Mar 12 '15 at 16:14

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.

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