find the general solution of the differential equation (x+1)y\'\'+y\'+xy=0 by me
ID: 2939289 • Letter: F
Question
find the general solution of the differential equation (x+1)y''+y'+xy=0 by means of a power series about the base pointx0=0. express your answer in the form y =a0y1(x)+a1y2(x), where{y1(x),y2(x)} is a fundamental set ofsolutions, and each of y1and y2 are given aspower series. find the general solution of the differential equation (x+1)y''+y'+xy=0 by means of a power series about the base pointx0=0. express your answer in the form y =a0y1(x)+a1y2(x), where{y1(x),y2(x)} is a fundamental set ofsolutions, and each of y1and y2 are given aspower series.Explanation / Answer
QuestionDetails: find the general solution of the differential equation (x+1)y''+y'+xy=0 by means of a power series about the base pointx0=0. express your answer in the form y =a0y1(x)+a1y2(x), where{y1(x),y2(x)} is a fundamental set ofsolutions, and each of y1and y2 are given aspower series.LET
Y=A0+A1X+A2X2+A3X3+................+ANXN+..........................1
Y'=A1+2A2X+3A3X2+.........................+(N+1)AN+1XN+..........................2
Y''=2A2+6A3X+....................................+(N+2)(N+1)AN+2XN+...........................3
XY''=2A2X+6A3X2+....................................+(N+1)(N)AN+1XN+...........................4
XY=A0X+A1X2+A2X3+A3X4+................+AN-1XN+..........................5
(X+1)Y''=2A2+(6A3+2A2)X+....................................+[(N+2)(N+1)AN+2+(N+1)(N)AN+1]XN+................6
EQNS.2+5+6 GIVES
[A1+2A2]+[A0+4A2+6A3]X+......+[AN-1+(N+1)2AN+1+(N+2)(N+1)AN+2]XN+.......=0
[A1+2A2] =0......................7
[A0+4A2+6A3]=0..........................8
IN GENERAL
[AN-1+(N+1)2AN+1+(N+2)(N+1)AN+2]XN=0......................................9
TAKING P=A0 AND Q=A1, WE GET
A0 =P
A1=Q
A2= -(1/2)Q........FROM EQN.7
A3= -[P-2Q]/6................FROM EQN.8
PUTTING N=2 IN THE GENERAL EQN.9
A1+9A3+12A4=0...SO....A4=-[Q+{3(P-2Q)/2}]/12
THUS YOU CAN GET ALL VALUES OF COEFFICIENTSA4,A5,...ETC...
BY PUTTING SUCCESSIVELY N=3,4...ETC..
WE FIND ALL COEFFICIENTS CAN BE EVALUATED IN TERMS OF P AND Q
SO FINALLY COMBINING THEM ALL USING EQN.1WE GET
Y=P[Y1(X)]+Q[Y2(X)]=A0[Y1(X)]+A1[Y2(X)] ASDESIRED ...
HOPE YOU CAN COMPLETE THE LEG WORK
IF STILL IN DIFFICULTY PLEASE COME BACK
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