Fourier series for particular solution when n=1

[dynezidane]

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(1-n^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=1 is y(x)_p = -1/4* (x*cos2x)
tq
Can anyone help me..

 

[mighty2000]

Hello there,

Thank you for sharing your progress with us. It seems like you are on the right track in finding the particular solution for y'' + 4y = f(x). In order to find the particular solution for n=1, you can follow these steps:

1. Start with the equation y_p = axcos 2x + bxsin x, as you have mentioned.

2. Substitute this equation into the original differential equation, y'' + 4y = f(x).

3. Simplify the equation and equate the coefficients of cos 2x and sin x to the corresponding coefficients in f(x).

4. Since f(x) = x, the coefficient of cos 2x should be 0 and the coefficient of sin x should be 1.

5. Solve for a and b by equating the coefficients to 0 and 1 respectively.

6. Once you have found the values of a and b, substitute them back into the equation y_p = axcos 2x + bxsin x to get the particular solution for n=1.

In this case, you should get y(x)_p = -1/4*x*cos 2x as the correct answer.

I hope this helps and clarifies any confusion. Keep up the good work!