Question

- y(x) 41. Use the method of Frobenius to solve the equation: x y"(x) + I Solution, Tanerad..to students and extra credit for undergraduates:

Answer 1

I zey" + (n - Ž) ' + £x 3 122 200 By Frobenius method, assume a solution of the form y = xk 2 anx , so that ya all Putting и+-) — 7 an (h+k) x an Ch+k) (ntke- 1 x tez these two in egn 6 , we ntkel ntk-/ htk-/ Zan (n+k) (h+k-1) x + Zan (nta) anthe - Ž Z an (nth) anth + Ź {an (th=0 2 an (nt k) ant - 1 s nt k-1-12} + 2 an (nt k) Michat + z zan x 0

the indicial eqh i obtained by setting Anika n=0, we get Date Page ask (K-₂) + 1 = 0 Since as is defined as the first non- zero term, we have k(K-32) ork or, k² - 3* = -² or, 2K-3k+1=0 or 24² - 2k-kt150 or 2K (k-1)-1(K-1)=0 C or, (k-1) (2k-1) - 0 K=1, Y If k=1, n=1 - a 2 (1+1-X-1/2) + 2 60+ 2a0 = 0 - a *3% =0 ) q o =-2 ao 79 2-4 for no 2, 3 ... and so I coeff of x 76-7 an/ (140 (n+/- v- ) + 1] + 2-1 (0*) = 0 or an [ {n+1) (n = 42) + 4] = - an-1 (n+1) an nytn-& - to an 1 or an = - (n+1) - 26n+1) ani (2011) th

An Ef az£ no a [136) (X+4-X-1) + 2] + ao » q + 3a020 - q = -390 an [(n+ b )(n+8-1-96) +47 + [n++]4.430 ov, an fwa fa-1 +452 - (024) 60-1 ov, an in-4») - - kme nt For, an fheb nten-1) -- (2011) :-/ and so putting these to get the on. values of solution. an in