Find the general solution to the ODE:$$y''+9y=3t+e^{3t}$$I have the answer but I don't know how to do this one

Question

anonymous · Answer

the answer i have in my exam review is $$y=\frac{t}{3}+\frac{e^{3t}}{18}+c_{1}sin(3t)+c_{2}cos(3t)$$and I'm just wondering how to do it because my professor only gave us the answer and I don't know how to start

accessdenied · Answer

Do you know how to solve the equation for a single function:
 Like solving each of these: $ y" + 9y = 3t $,  $y"+ 9y = e^{3t} $

anonymous · Answer

ok so all I have to do is solve it separately and then combine them in the end?

accessdenied · Answer

Essentially, yes.  There is something called the Principle of Superposition that states that, to find the particular solution of
  $y" + p(x) y' + q(x) y  = f_1 (x) + f_2 (x) + ... + f_N (x) $
If $Y_j$ is a solution to $y" + p(x) y'+ q(x) y = f_J(x) $, 
    then $Y_1 + Y_2 + Y_3 + ... + Y_N$ is a particular solution to the entire equation.

anonymous · Answer

ok this makes a lot of sense now. thank you so much. i know how to do the first part, but i am not too familiar with the $e^{3t}$ portion. can you guide me as to how to start?

accessdenied · Answer

Hmm..  admittedly, I am rusty on these second order DE.  So, I gotta use a book.  :P

Well, I believe there are a few methods that we could use to find the particular solution here.  Do you use Variation of Parameters or Undetermined Coefficients typically?

$ y"+ 9y = e^{3t} $

accessdenied · Answer

And I think undetermined coefficients would be easier here.

If we consider that the derivative of e^(3t) is always some Ae^(3t), we could try y=Ae^(3t) and solve for A...

anonymous · Answer

OK if I do that then I would end up with $y=\frac{e^{3t}}{18}$ and combining this and the $y=\frac{t}{3}$ from the first part I still don't know how to get the cos and sin part in the last half of the answer. Do I solve $y''+9y=0$ in order to get them?

accessdenied · Answer

We get the particular solution out of these methods, so the general solution is just tacking on the Homogenous DE's solution.  Yes.  :)

anonymous · Answer

so I need to solve 3 ODE's to get this answer, in short?
$$y''+9y=3t$$$$y''+9y=e^{3t}$$and $$y''+9y=0$$and then I will get the answer I want?
Thanks and sorry for keeping you so long.

accessdenied · Answer

Yep.  If we think about it, by combining all those together:
  $ y = Y_{p1} + Y_{p2} + y_H $    y_(p1) is solution to 1 DE, y_(p2) to another, and Y_H the homogenous DE solution
The original DE splits up :
  $y" + p(x) y'+ q(x) y = f(x) \  \to  (Y_{p1} + Y_{p2} + y_H)" + p(x) (Y_{p1} + Y_{p2} + y_H)' + q(x) (Y_{p1} + Y_{p2} + y_H) = f(x) \
\to (y_H" + p(x)y_H'  + q(x) y_H) + (Y_{p1}" + p(x) Y_{p1}' + q(x) Y_{p1}) \ \qquad + (Y_{p2}" + q(x) Y_{p2}' + p(x)Y_{p2}) $
The homogenous case will eliminate to 0, and those other DE we solved earlier will sum up to the other side's f(x)

Sorry this took a while to write.  I thought it was a strong point to understand though.

anonymous · Answer

that's ok and thank you so much this helps a lot. Let me attempt some more problems.

accessdenied · Answer

You're welcome!  :)