1
$\begingroup$

I am having trouble figuring out how to solve this problem using the Fourier series method. I'm not allowed to use methods of undetermined coefficient or variation of parameter. $$ y'' + 2y' + y = 25\cos(2t) $$ The general solution is obvious but I'm unable to find the particular solution. I tried to use the Fourier series representation of $25\cos(2t)$ but I get 0 for my $a_0$, $a_n$, and $b_n$. The only value that's not 0 is when my $n=2$, then my value becomes $25$.

In the book I use, this is called the steady state solution and Theorem 1 is called Forced Oscillation. I'm unable to apply the theorem because I am unable to obtain the Fourier coefficients.

Solution: $$ ce^{-t} + dte^{-t} -3\cos(2t)+4\sin(2t) $$

$\endgroup$
1
  • $\begingroup$ If $a_n$ and $b_n$ are the coefficients of the $\sin$ and $\cos$ terms in the Fourier series expansion, it makes sense that $a_n = b_n = 0$ for all $n\ne 2$, does it not? $\endgroup$
    – Xander Henderson
    Commented Sep 24, 2017 at 21:54

1 Answer 1

1
$\begingroup$

HINT

The idea is to assume that $$ y(x) = \sum_{n=0}^\infty \left[ a_n \sin(nx) + b_n \cos(nx)\right] $$ Now compute $y', y''$ and equate it to the right-hand side to get a recurrence relationship for $a_n$ and $b_n$ which you can solve.

$\endgroup$
3
  • $\begingroup$ I'm not sure that's how I'm supposed to solve the problem. books.google.com/… Look at page 70. The formula there is the one I'm supposed to use to solve the problem. $\endgroup$
    – Help Me
    Commented Sep 24, 2017 at 22:53
  • $\begingroup$ @HelpMe on pp.69-70 it seems to describe exactly the approach i advised you to take and Theorem 1 is really the final result of that approach. Knowing that will save you the arithmetic. Why don't you simply plug in? $\endgroup$
    – gt6989b
    Commented Sep 25, 2017 at 0:33
  • $\begingroup$ Thanks for clarifying. I just wanted to make sure I wasn't practicing the wrong method. $\endgroup$
    – Help Me
    Commented Sep 25, 2017 at 4:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .