Question

Exercise 7.3.8. In the following equations classify the point x = 0 as ordinary, regular singular, or singular but not regular singular. a. 2²(1+x?)y" + xy = 0 b. x+y" + y' +y=0 C. ng” +cº+y= 0 d. xy" + xy' - e"y=0 e. a’y" + xạy' + xạy=0

Answer 1

SOLUTIONO Note : • А point point of If both x = lo is called ordinary y" + pomy' + Qoy=0 px) 4 (x) are analytic atx=x a If x=xo is not ordinary point and if (x-xo) pox) 4 x-xoa x) are analytic at x = ko then given differential equation is heung regular singu ular pont at x = xo. otherwise x= xo is not reqular singular. singular but Now ; a 22 (1+x2) " + xy =0 y" + -Yeo 22 (1+x2) Y" + 0.4't y co Xll+x²) here p(x)=0 Q(x) = / x (1+x2) lim pel = tim X- O-O lim Hence Q(x) = lom = DNE (do not exist) xo X (1+x2) is not ordinary point Cx-xo) poe) = (-o).0=0 Xc0 Now (x-xo) ². Qxz son x² X0 & Cita2) Hence x=0 is REGULAR SINGULAR point

22y"+y'ty=0 yhy' p(x) = 4 22 Q = 1/2 lim P(x) = D.N.E. lim QW= DNE Hence No point ordmary = D.N.E. im be? 1-1 x2 > Hence SINGULAR BUT NOT REGULAR SINGULAR xy" + x3y'+y=0 y"+x2y'tly=0 lim x2 = 0 xt = DNE 2 >0 Hence No ordinary point x.x²=0 Im x21 = 2 Hence REGULAR SINGULAR

xy" + xy'-exy=0 y' + y - oxy=0 Je lim 1=1 - en ze =lonit do not exist Hence No ordmay pont Now Im xel= 0 X>0 xto Hence REGULAR SINSULAR x²y" + x²y'+x²y=0 y"+y+y=0 Im Im 1=1 bolal Hence ORDINARY POINT

Answers

a) REGULAR SINGULAR

b) SINGULAR BUT NOT REGULAR SINGULAR

c) REGULAR SINGULAR

d) REGULAR SINGULAR

e) ORDINARY

NOTE- in part b) I wrote DNE (do not exist in first validation) it's values are infinity hence not analytic at x=0 .