Consider the differential equation
$ \ \ 4\frac{d^2x}{dt^2}+5x=f(t) \ $ ,
Find a particular solution of the above differential equation of the form
$ x_p(t)=\sum_{n=1}^{\infty} A_n \cos \frac{n \pi}{p} t=\sum_{n=1}^{\infty} g(t,n) \ $
Then find out the function $ \ g(t,n) \ $ ?
where $ \ f(t)=\begin{cases} 2 , \ \ 0 <t < \pi \\ -2, \ \ \pi <t<2 \pi \end{cases} \ $ and $ \ f(t+2 \pi)=f(t) \ $
Also given that when $ \ f(t) \ $ is extended to negative t-axis in a periodic manner, the resulting function is $ \ even \ $
Answer:
The given equation is
$ \ \ 4\frac{d^2x}{dt^2}+5x=f(t) \ $
The corresponding homogeneous part is
$ 4\frac{d^2x}{dt^2}+5x=0 \ $
The auxiliary equation is
$ 4m^2+5=0 \\ \Rightarrow m=\pm \frac{\sqrt 5}{2} \large i \ $
Thus the complementary function is
$ x(t)=A \cos (\frac{\sqrt 5}{2}t)+B \sin (\frac{\sqrt 5}{2} t) \ $ ,
where $ \ A , \ B \ $ are constants.
Now I can not find out the particular integral and the required function $ \ g(t,n) \ $
Actually I can not understand how to answer the question and in what process to approach .
I need help to get out of this.
You don't really need to find the complementary solution, just do what the questions asks.
You want the particular solution to match the periodicity of $f$, so $p = 2\pi$ and
$$ x_p(t) = \sum_{n=1}^{\infty} A_n\cos\left(\frac{n}{2}t\right) $$
To find the $A_n$, just plug $x_p$ into the ODE to get $$ 4{x_p}'' + 5x_p = \sum_{n=1}^{\infty} (5-n^2)A_n\cos\left(\frac{n}{2}t\right) = f(t) $$
You may recognize this as a Fourier representation of $f(t)$. Therefore
$$ (5-n^2)A_n = \frac{1}{\pi}\int_0^{2\pi} f(t) \cos \left(\frac{n}{2}t \right) dt $$