Question 6 has been answered. The answer is correct. Please explain how to get said answer. The power series table has been provided in case you need it. 6. (6.2) Find two power series solutions of the differential equationy" - xy' +2y = o about the ordinary point x = 0.1.1Y1 = 1– r²and42 = r -r*120 IntervalMaclaurin Seriesof Convergencee* = 1 +1!Σ(-00, 00)2!3!-o n!x2cos x = 1(-1)"(-0, 0)....2!4!6!N=0 (2n)!(-1)"Σ(2n + 1)!sin x = x -3!(-0, 00)5!7!0x x7(-1)"tan x = x -[-1, 1](2)2n + 10= 1 +2!(-0, 0)cosh x+6!, (2n)!4!x71Σ(2n + 1)!sinh x = x ++5!3!(-0, 0)+... =7!=0x?In(1 + x) = x -(-1)"+1(-1, 1]341+ x+ x?(-1, 1)....n=0+++

Given ODE y"-xy'+2y=0 about ordinary point x=0. Assume that y=a0+a1x+a2x2+a3x3+........y=∑n=0∞anxnDifferentiating with respect to x,y'=∑nn=1∞anxn-1xy'=∑nn=0∞anxnAgain differentiating with respect to x,y"=∑n=2∞n(n-1)anxn-2Substituting in the equation we get,∑n=2∞n(n-1)anxn-2-∑nn=0∞anxn+2∑n=0∞anxn=0First series shift the index of summation by replacing n by n+2 and starting the sum at 0, the equation becomes,∑n=0∞n+2(n+1)an+2xn-∑nn=0∞anxn+2∑n=0∞anxn=0∑n=0∞n+2(n+1)an+2-nan+2anxn=0 Equating coefficients on both sides, n+2n+1an+2-n-2an=0 , n=0,1,2,.... an+2=n-2n+2n+1an, n=0,1,2,... Find first four terms ,Put n=0, a2=-2a02=-a0Put n=1, a3=-a16Put n=2, a4=0Put n=3, a5=a320=120-a16=-a1120Put n=4, a6=2a430=2300=0 Substituting these values in y we get, y=a0+a1x+a2x2+a3x3+.....y=a0+a1x-a0x2-a16x3-a1120x5+.....y=a01-x2+a1x-16x3-1120x5-..... The solution can be written as y=a0y1+a1y2. Comparing with above equation, y1=1-x2 and y2=x-16x3-1120x5-......