A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
hi, i already got the fourier series for f(x) = x where pi/2 =< x =< pi/2
which is f(x) = sigma, n=1 to infinity ( (1)^n+1*sin (2nx) / n )
in order to find particular solution for y'' + 4y = f(x)
i have to equate with with y(x)_p = A0 + sigma, n=1 to infinity (An*cos(2nx) + Bn*sin(2nx))
and i get y(x)_p = sigma, n=1 to infinity ( (1)^n+1 (sin(2nx) ) / 4n(1n^2) ) which is the correct answer. but this is not valid if n = 1.
so is anyone can show me how to get particular solution for n = 1. i already equate with
y_p = axcos 2x + bxsin x but i didnt get the answer
the correct answer for n=
anonymous
 5 years ago
hi, i already got the fourier series for f(x) = x where pi/2 =< x =< pi/2 which is f(x) = sigma, n=1 to infinity ( (1)^n+1*sin (2nx) / n ) in order to find particular solution for y'' + 4y = f(x) i have to equate with with y(x)_p = A0 + sigma, n=1 to infinity (An*cos(2nx) + Bn*sin(2nx)) and i get y(x)_p = sigma, n=1 to infinity ( (1)^n+1 (sin(2nx) ) / 4n(1n^2) ) which is the correct answer. but this is not valid if n = 1. so is anyone can show me how to get particular solution for n = 1. i already equate with y_p = axcos 2x + bxsin x but i didnt get the answer the correct answer for n=

This Question is Closed
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.