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zidane
 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=
zidane
 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=

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