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In this problem you will solve the differential equation (x + 2)y" - (11-x)y' + y = 0. (1) By analyzing the singular points of the differential equation, we know that a series solution of the form y=0Ck ak for the differential equation will converge at least on the interval (-2,2) (2) Substituting y = kock into (a + 2)y" - (11-x)y' + y = 0, you get that C + с + M8 с + с x"=0

In this problem you will solve the differential equation (x + 2)y" - (11-x)y' + y = 0. (1) By analyzing the singular points of the differential equation, we know that a series solution of the form y=0Ck ak for the differential equation will converge at least on the interval (-2,2) (2) Substituting y = kock into (a + 2)y" - (11-x)y' + y = 0, you get that C + с + M8 с + с x"=0

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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In this problem you will solve the differential equation (x + 2)y" - (11-x)y' + y = 0. (1) By analyzing the singular points of the differential equation, we know that a series solution of the form y=0C * for the differential equation will converge at least on the interval (-2,2) (2) Substituting y = Σock into (+2)y" - (11-x)y' + y = 0, you get that ]=0 с + с + 8W n=1 C с + с
Transcribed Image Text:In this problem you will solve the differential equation (x + 2)y" - (11-x)y' + y = 0. (1) By analyzing the singular points of the differential equation, we know that a series solution of the form y=0C * for the differential equation will converge at least on the interval (-2,2) (2) Substituting y = Σock into (+2)y" - (11-x)y' + y = 0, you get that ]=0 с + с + 8W n=1 C с + с
Poblem 4) In this problem you will solve differerent equidio (x+2)y" - (11-8)y' ty=0 (1) By anayking the singular point of the equation differential, we know that a jerres solutions of the form y= {'redch For the differential equation will convere at least on the interval 1-2, 2 Po(x) = x+₂ (x+2)=0 X=-2 p=distance from 0 to -2=2 2 Interval (0 £2, (0-2) = (2,-2) you get that o f ≤ ск (2) Subusdituding y = { x= 4k xk into Cxtz)g" - (11->) y 'dy = 0, y ¹² k³0 to 08 83 y= 36x² = €₂ €6,0² +4₂ +² +6₂.23 fom GK₂ 12 13 ₁4 CAKX 个 K-2 y". ≤ (xki(k-1) x² d ½ C₁ fick x² no A • (K ´k ( ₂ ) ( k + 1) ²₂ соприва Y メラ+レーンメン(メックスタンプ/3(+) ран (K-1) X² = (11-X) 5 € / kx K=O (x+2) ≤ 4 = 0 C₁K 21 ARICA ( K-147 & 2 CK K (K-1) x4-²2 - ZK =201lc kx² +¹ +2 C₁ kx² + 2 KO B K=O D E C ++ 7/₂
Transcribed Image Text:Poblem 4) In this problem you will solve differerent equidio (x+2)y" - (11-8)y' ty=0 (1) By anayking the singular point of the equation differential, we know that a jerres solutions of the form y= {'redch For the differential equation will convere at least on the interval 1-2, 2 Po(x) = x+₂ (x+2)=0 X=-2 p=distance from 0 to -2=2 2 Interval (0 £2, (0-2) = (2,-2) you get that o f ≤ ск (2) Subusdituding y = { x= 4k xk into Cxtz)g" - (11->) y 'dy = 0, y ¹² k³0 to 08 83 y= 36x² = €₂ €6,0² +4₂ +² +6₂.23 fom GK₂ 12 13 ₁4 CAKX 个 K-2 y". ≤ (xki(k-1) x² d ½ C₁ fick x² no A • (K ´k ( ₂ ) ( k + 1) ²₂ соприва Y メラ+レーンメン(メックスタンプ/3(+) ран (K-1) X² = (11-X) 5 € / kx K=O (x+2) ≤ 4 = 0 C₁K 21 ARICA ( K-147 & 2 CK K (K-1) x4-²2 - ZK =201lc kx² +¹ +2 C₁ kx² + 2 KO B K=O D E C ++ 7/₂
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