How do I solve problem 2 using my answers from problem 1? I did this for Qt1.

a) let Fourier series of fon [-1, 1] as 2 f (+ ) = Qo + Sax cos (KTCt) + bx Sin (kx+) ) K =1 Where OK = f( +) cos ( KC+ ) at & bx = ) f ( + )Sin (kit ) dt , Where K 21. a. = / f(t) at ". a . = [f( that = J ( -) d+ + ) (1) 2+ + ) (-1) at 1 /2 = -1 (4+1)+ (2+2)-1(1-2) = -1 ( 2 ) + 1 - (2 ) = - 2-2+2=0 .. a. = 0 a x = J f ( + ) cos ( KTc + ) at = ) ( - 1) Cos ( Knit ) &+ + ) ( D cas ( Kx+ ) d+ + ) (-Dcos ( kit ) dt ax = - / Sin (kit ) ) 2 Sin ( K 7+ ) ) 2 * Sin (KT+) ) KT Kx + ( - ) Kx | 1/2 (- sin ( 2) Sin (KTC) Si (Kx ) Sin 2 + Sin (2) Kx KT SIn( KET ) - Sin ( KRC ) + 2 Sim $2 - Sin (KT) + Sin KTC Kx KR = 4Sin 2 Sin KTO kx ak = 4 Sinb) g.(+) = J f( + ) at = )/ 45m KT bx COS (KTIt) Cos (KTIt ) = 2 . Cos ( KTC+ ) + 0 + K= 1 KT KTC 1/ 2 4 . Sin En a(t ) = 2 IM8 KTE COS ( 2 ) - COS ( KT ) + COS ( KIT ) - Cos/ Cos ( KTC+ ) alt 45inKT Sin (kit ) = KT (0 - Cos( KIT ) + Cos( KIT ) - 0 ) = 0 2 K=I KI 4 Sin KIT ". by : 0 at Kz1 ( KTC ) 2 (Sin(KT+)) K= 1 = ) Fourier Series of f: f(+) = 5, a, Cos ( knit ) = 4 Sin ( 2). Cos (kit ) KT . " . coefficients of g (t ) decay *= ak = 4 Sin KTI - 0 and bK = 0 at ko . . Both ax & by decay2. (8 points) Using Fourier Series, compute the general solution for the following differential equation: y" - 8y' + 20y = f(t) with f(t) from problem #1. Careful: you need to use f(t), not g(t)!12 points) Consider the 2-periodic piecewise function 1, -1