Question

0 Find the Fourier Coefficients a, for the periodic function f(x) = {: for-2<2<0 O for 0 < x < 2 f(x + 4) = f(x)

section is fourier series and first order differential equations

Answer 1

0 osaca L 2 f(x) = {x for 2 sugo for. f*+4) fi) Fourier collecient an for find ! " An = 1 f f (2) Cos(nuk) da L) an & S flas cos (nich) dit forse cestning) +* ... Continente) da? Preto sa naging (ng) + bcs (n) [0+ (6 - sen ce mai) - les San Enr) Penge cos cono Florence (27(-)" ] p. 1l-6-19] In = (1-(-5] :(-19] i n=124367 5 -2 2 ht) neka n22

for -35460 OSH<3 ③ fow) = fal 3 f (x+6)=f(x) To for Fourier coefficients by for flood - bn = L L T bn = 1 I fox) sin(n) de 3 3 9 buat ff a sh(m2) dx = fo.sindrome , alors a bn = Sih ( 1 ) dox علا 3 3 [ * ) Cetags) + * sin(nam) 2.0 5f0.70-6343) ($(=nx) – jesint-nos] 하 -3(-) (-)” 3 nK 3 ПА - bn=-3 (-1)" ; n = 1, 2, 3 .... ht 3 frana In ocnck half i f(a+25) = f(x) range Cosine series of flor): - f(2)= 9ot Ean (man) ☺ 0<xL Cos L I Where do = ť s finida and on=2 & S few calinan dan

9. 2 . da el- a (na R3 Gm X 6 k 2 & ana txe.cos(nu) da X Cesnuda 2 0 x² Sin (nn) + au cas (nm) h² o tax cos(hr) - 0 h2 to ] २ (-1) b 2 TE [fon (-0!? 2 { cas (na) as the half range cosme 6 n2 n=1 Series of f(20) 4

Solution: Given y' = (1 + xy) -1, y(1) = 2, h = 0.1,y(1.1) = ? Forth order R-K method kq = nf(xo»yo) = (0.1171, 2) = (0.1). (0.333333) = 0.033333 = (0.1)|(1.05, 2.016667) = (0.1).(0.32077) = 0.032077 Ks = 1060 2. + ); -- nofoot h k2 kg Xa +20+ = (0.1)/(1.05, 2.016038) = (0.1) . (0.320838) = 0.032084 2 k4 = hf(xo + h, yg + kg) = (0.1)|1.1, 2.032084) = (0.1) - (0.309091) = 0.030909 YA = yo *a (ky + 2xy + 2kg +ka) 1 y, = 2 + 2 [0.033333 + 2(0.032077) + 2(0.032084) + (0.030909)] Y1 = 2.032094 :. y(1.1) = 2.032094 ::y(1.1) = 2.032094