1
$\begingroup$
  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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.