Power series method

II- Power series method
We consider the second-order linear homogeneous differential equation for y=y(x)
P(x) y^ +Q(x) y^ +R(x)y=0
where ??(??), ??(??) and ??(??) are polynomials
The idea of power series method is to assume that the unknown function y can be expanded into a power series:
y=?_(n=0)^??a_n x^n=a_0+a_1 x+a_2 x^2+a_3 x^3+a_4 x^4+a_5 x^5+?
y^ =a_1+2a_2 x+3a_3 x^2+4a_4 x^3+5a_5 x^4+6a_6 x^5+?
y^ =2a_2+6a_3 x+12a_4 x^2+20a_5 x^3+30a_6 x^4+42a_7 x^5+?
Substituting in the differential equation and requiring that the coefficients of each power of x sum to zero, we can find all the constants a_n ? n=2,3,4,? in terms of a_0 or a_1.
Example 1: Solve y^ +2xy^ +2y=0 by power series method
Solution :
y=a_0+a_1 x+a_2 x^2+a_3 x^3+a_4 x^4+a_5 x^5+?
y^ =a_1+2a_2 x+3a_3 x^2+4a_4 x^3+5a_5 x^4+6a_6 x^5+?
y^ =2a_2+6a_3 x+12a_4 x^2+20a_5 x^3+30a_6 x^4+42a_7 x^5+?
Now
y^ =?2a?_2+6a_3 x+12a_4 x^2+20a_5 x^3+30a_6 x^4+42a_7 x^5+?
2xy^ =0 +2a_1 x+4a_2 x^2+6a_3 x^3 +8a_4 x^4 +12a_5 x^5+?
2y=2a_0 +2a_1 x+2a_2 x^2+2a_3 x^3 +2a_4 x^4 +2a_5 x^5+?
2a_2+2a_0=0 ? ?(a_2=-a_0 ) and 6a_3+2a_1+2a_1=0 ? ?(a_3=-(2?3) a_1 )
12a_4+4a_2+2a_2=0 ? a_4=-(1?2) a_2 ? ?(a_4=(1?2) a_0 )
20a_5+6a_3+2a_3=0 ? a_5=-(2?5) a_3 ? a_5=(4?15) a_1
y=a_0+a_1 x-a_0 x^2-(2?3) a_1 x^3+(1?2) a_0 x^4+(4?15) a_1 x^5+?
y=a_0 (1-x^2+( 1 )/( 2 ) x^4+?)+a_1 (x-( 2 )/( 3 ) x^3+( 4 )/15 x^5+?)

Example 2: Solve y^ -xy^ +4y=0 by using power series method
Solution :
y=a_0+a_1 x+a_2 x^2+a_3 x^3+a_4 x^4+a_5 x^5+?
y^ =a_1+2a_2 x+3a_3 x^2+4a_4 x^3+5a_5 x^4+6a_6 x^5+?
y^ =2a_2+6a_3 x+12a_4 x^2+20a_5 x^3+30a_6 x^4+42a_7 x^5+?
Now
y^ =2a_2+6a_3 x+12a_4 x^2+20a_5 x^3 +30a_6 x^4+42a_7 x^5+?
-xy^ = 0 -a_1 x-2a_2 x^2 -3a_3 x^3 -4a_4 x^4 -5a_5 x^5-?
4y=4a_0 +4a_1 x +4a_2 x^2 +4a_3 x^3 +4a_4 x^4 +4a_5 x^5+?
2a_2+4a_0=0? ?(a_2=-2a_0 ) ,6a_3-a_1+4a_1=0 ? ?(a_3=-(1?2) a_1 )
12a_4-2a_2+4a_2=0 ? a_4=-(1?6) a_2 ? ?(a_4=(1?3) a_0 )
20a_5-3a_3+4a_3=0 ?a_5=-(1?20) a_3 ? ?(a_5=(1?40) a_1 )
30a_6-4a_4+4a_4=0 ? ?(a_6=0) and a_8=a_10=?=0
y=a_0+a_1 x-2a_0 x^2-(1?2) a_1 x^3+(1?3) a_0 x^4+(1?40) a_1 x^5+?
y=a_0 (1-2x^2+x^4/3)+a_1 (x-x^3/2+x^5/40+? )


Solve the differential equations by using power series method
(1) x y^ +y^ +xy=0
(2) x^2 y^ +y^ +x^2 y=0
(3) y^ +3xy^ -y=0
(4) (x-1) y^ +2y^ =0