Question

Individual task 6 Fredholm integral equations. Freholm alternat Case 7 1. Test for solubility at different values of the parameter 1 following integral equation (cosh x sinh s.O SXSS 1.1. y(x) - 2 SK(x,s)y(s) ds = 1, where K(x,s) = cosh s sinhxis SXS 1.2. y(x) -15, * sin(275)y(s) ds = x.

Answer 1

.Y (2) - l'uc k (0,8) $(3)d8=1 K(0,8)= eashx sinhs o bass cash & sinha stne) (1 -) y (2) d (918) ds y(t) = 1 1 *«cm, 3) 3 (4) ds and 37 y (aj - She eashos sinhin y(-s) ds & ၊ ဟု cosha Sinhs (1) ds -1 = y(x) - A Sinha] Bosh (a) y ) ds en a (sinke) ye ) eshx 21 2) ve get. ylla 2) 1 eoshre ds I Sinha/eos ha y(2) Sind Differentiating wanita feesh (18) 3 (8) ha feriha _ olsinha J'ainhla y ceny ds + d cosha tinha J'essh (3996) di 20 » yl(a) - d eos ha J'aime Sinh (B) y (8) as =0 cue get, Again differentiating }" (+)- A Sinha fes | cahwa Gide ersh (8) 4 (8) ds - ers hañ g (2) deosha sinhs y(3) ds tid sinhañ g (2) 20 2 >> y"(*) – ( yl(a) – 1) - 1 y(x) = 0 yil (3) CHA) y (21) -1. (3)

Haccum with the y (0)= a (constant) y los O) = o The complementary famton foro (3) is m2 (d) = 0 m= ired. let ledyo. reda y (3) = Be-Vikin Be A e yllo) = Vied (A - B) =0 Vie CA-B)=0.9) ADB y(0) = A + B = à à .=) f = 9/2 = B. y (x)=eos havied) x + +tes (1) Let redzo y" (a) = -1 y can -ne A y ľn) = An-22% + B yllo) A= =0 y (0)= B = a so so y(x) = constant : (3) let it co m = 1 I Mi antar |(x) = C, eosune la Sinuina1 s) C2-0 yl(o)= C2M=0 s) y(0)C, - Your - Yara =) CIF Vart.a)

the Solution is y (a) = Care tor Care 1/me) cos puxe Can this) te) easpa + 1 1 - 14 1.2. y (20) = n sin (20 s) y(s) ds pr. y(x)=xa Sinets) y (S) Is é a let a= t's sin (28) y(s) ds. = s' Sinzas Ganen) ds. = alla+1) Sinan 8 ds = alaal). eos at 8 2 DT = . y (2) ski It is solvable for ale values of d.