Question

PLEASE JUST DO QUESTION 2

be Kapwing - Where Cox 13 Upload Assignment: PX S PowerSeries.pdf х Le PT ds/PowerSeries.pdf Differential Equations // Math 2680 Section OL63 // Summer 2020 Using Power Series For each differential equation P(x)"+ Q(x)) + R(x)y=0, the solution will be a power series y(x) = n(3 - 20)" = 40 +212 + a2z? +... where to satisfies P(x0) +0, and ao = A and a1 = B are constants that determine all other coefficients. Example: y" + y = 0. Let 10 = 0. Plug in y(x) = and" into the differential equation "+y = (§an")" + (Žan=") = (§ (n = 1)ayx+-2) + (8") ((n + 2)(n + 1)ant2") + (auz") = }(n + 2)(n + 1)ent2 + cm) = -6 00 Therefore 0 = (n + 2)(n+1)an+2 +ana". Rearranging we get Gn+2 = (m+2)(n+1), which says 20,24 = wis = %, and as = mit = 70. Also, a3 = 3 =, as = (57 and ay = Tits = cod. Therefore y(x) = 40 + a2 + ()x2 + (x + 3) + (%)+33 + (b)x+ ... |y(z) = as (1 : +...) + az (: - 22 + + 2 24 720 6 120 5040 For the following differential equations, find az, 23, 24, 25, 26, and ay in terms of co and a and write the answer y(x) = 60 sum of terms + sum of terms 1. y" - y = expanding about 10 =0. y' - xy - y = 0 expanding about to 0. 2.

Answer 1

y" - ny ryzo Let us suppose 6 hay the solution of the form y=Ea Eauren nco a) y = nananny nco Y" Encn-ilan xn-2 substituate ② o. in o x nanana anzen -o Zn(n-1) anan-2 nco co nco nanzen -5 anach zo n(n-1) anan-2 no nco nzo L N-2 5 auraken (M-2) Gunder2 - Encn-1) and -2 hiz noa n=2 {{n(n-1) an (n=2+1) Ga-2721-2 = 0 n=2 En(n-1) an- (n-1) an-2} 2 -2 = n=2 CS Scanned with CamScanner

aura =) n Now n=2 a o Q2= ao 2 ar 3 series solution is Equating the coefficient of lowest power of a to zero, we get n (n-1) an - 6m-1) an-2=0 ana na 2,3,000 ass aus 4 a6 = a 5 y (n) = auan = as far faza? tagn?t aage facutare axt al x² + ar tas ne tot =hollt + y + 10 too) fai (x+33 + 1 ...) 8 A3 F al al 15 as 5 la sala au 6 12 a o 48 al at = 105 no - Go tarnt CS Scanned with CamScanner