Question

3. Find the eigenvalues and eigenfunctions for the given boundary-value problem. There are 3 cases to consider. g" + Ag = 0 y(0) = 0, y'(%) = 0
and

Answer 1

3) Now. the given differential equation ist y" + ay = 0.76), y (0) = 0, $'(0%) = 0 If x=(-4²) <0. then the auxiliary equation is> mp_k=0 → m = +K N then y=Gekx + ekx where are are Now, y'=qk e kx + G.tk e ka arbitary constant ... Now, yO) =0 = 4 +C =0 -->lii) y'[") -- > k[gewice"] ce Theo (171) Now, the coefficient deterriminant for C,& iss 1 -1 OTK TIK ! Пk so, G = G = 0. then y=o.le we get toivine thus aco. is not possible Tf 2 = 0. then ci) equation Redures to → y"=0 y =dxtde where d, is arbitary Constant ... Now, y'=di talvelun ... Now, yo =0 → dy = o y ' (%) = 0 = dy=0. thus pros y=o. For a=0 we get a trivial solution. so, the case 1=0. is not possible

Il 2-k?>0. then the auxiliary equation of li) differential equation is to delive r m+K2=0 = m = t ik therefore, y = a loskx + a sinkx. : Now y'=- ak sin kx +qk.coskx Now, y(0) = 0 + 0 = 0.1 ... poste y'(W) -0. → Ask.cos (km) = 0, [=0=0] . Now, a to. If > Ls (187) = 0 = (04{ent)}} we get frivial 7 sluition ] → ķT - (20+1) → K = (2n+1) => K?- (2n+1)2 - 9 = (2n+)2 , N=12... Then, y = ag sin [(2n+1)x]./ M Therefore, the eigen Values is → an = (21+12 where nolana and the eigen function is An (x) = An sin {(2n+1)x} Where n=1,2,...

den holi) Nole consider the differential equation 20(t). do + ... + ani (L) dhe u an(t).y = F(!) where as, ap..., and an and Fare continuo us on the interval aLtab and ao (t) to on aLLLL Let to be a point of the interval alteb and let Co, G..., Cnro be a sel of Real constants. Then I a Unique solution & of (i) such that Ø (to) = Co. $ 'lto) = 9,---, gnil to) =(n- and this solution is defined over the entire interval actcb.

8) Now, the given differential equation iss (x-4). y" + 2ay' +17y - Inx, y(3) =1, y(3) =0 :. Here 20 (x) = (2-4). 04 (X) = 2X, Az (X) = 17, Flx) = ln 2 .. Now, 20 (), 4(), ag(x) are continue ony and F(X) is continuous for xyo. The coefficient of you is vanished at x=4. l. e ao (x) = 0 at a=4.. Thus, the given problem has a unique solution in the interval a LXLb. where Ocak 2=3 < b <4.