Question

3. The second order variable coefficient differential equation Bxy" - ay = 0, (3) has a regular singular point at r = 0, where a > 0 and 8 >0 are given constants. Therefore, equation (3) has at least one solution of the form y(x) = ame" .m+r mao where r is chosen so that do 70. (a) Find the indicial equation and solve it for r. (b) For the larger value of r from part (a), find the corresponding recurrence relation for am

Answer 1

3. The end onder differential equation Bryan ay so regular singin ularr point at a=0 where X, B7o. y (2) = x ama where is chosen so that afo. has a 8 so, imun m=0 m+30-1 g (%) - Ž (m + r) am ze y"Ca) a § (mana) (min-dama meo man-2 . = mno

Now putting this value in 6 weget, mtnc) ß E (man) (m +0.1) am a a. I am x men no mao mco or, B. (no (3-1)) mol x mt 2-1 PS (m+rs) (m+30-1) am a mal mtn O +(-x) z ama mzo or, ß no (0-1) aox maro + ß s a mti (m+mo+1) (m+20) x X m20 mer as ama Co mco Now equating to zero the coefficient of the lowest porrer of a me have, mo (n-1) = 0 = 20 =0,1 this it we indicial solving x is the equation nation and noa0,1 are no ai (5) get. from (a) Now, equating to porners of a me have, see that, the larger value of zero coefficient of the higher ß amal Comitm +.1) (meno) = xam my,o. x am - amal ß (mArt) (m+0) So, for mai =1, amal x am BC m + 2)(m+1) xao So, a, - Il xao 2.B dao .ß. 2 . aao 2B ag = q2 do 17 ao १२ B. Con + 2)2 1) 2 2 p.2.3 4B2,3

ďao R. 1 az 4823 B. (2+2).(2+1) a ao 48².3 B. 4. 3. abão 4 $? 3 u x. A ao 3 ao aA ao K.: xta q3p3 325 2. 4² B² 3 B.5.4 3 p 325. m+l. so, E Om x I since zo si] mco 3 2 xt ar ta, a + .ag . + aga 4 2 3 ca. xao Lao a + - dort + + 2ß (23)3 3 (4M)? 32