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Differential Equations                                        Quiz #9                                                      May 6, 2005

Name____________________                   R.  Hammack                                                 Score ______
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(1)    Find find two linearly independent power series solutions about the ordinary point x=0 of the differential equation  y''-2x y'+ y=0.


Try solution y=Underoverscript[∑ , n = 0, arg3]c_nx^n. Then y'=Underoverscript[∑ , n = 1, arg3]c_nn x^(n - 1), and y''=Underoverscript[∑ , n = 2, arg3]c_nn(n-1)x^(n - 2)
Plugging this in:
y''+x ^2y'+y=0
  Underoverscript[∑ , n = 2, arg3]c_nn(n-1)x^(n - 2)-2x Underoverscript[∑ , n = 1, arg3]c_nn x^(n - 1)+Underoverscript[∑ , n = 0, arg3]c_nx^n= 0
  Underoverscript[∑ , n = 2, arg3]c_nn(n-1)x^(n - 2)- Underoverscript[∑ 2, n = 1, arg3]c_nn x^n+Underoverscript[∑ , n = 0, arg3]c_nx^n= 0
  Underoverscript[∑ , k = 0, arg3]c_ (k + 2)(k+2)(k+1)x^k- Underoverscript[∑ , k = 0, arg3]2c_kk x^k+Underoverscript[∑ , k = 0, arg3]c_kx^k= 0
  Underoverscript[∑ , k = 0, arg3](c_ (k + 2)(k+2)(k+1)+(1-2k)c_k)x^k= 0
  
  Thus:  c_ (k + 2)(k+2)(k+1)+(1-2k)c_k= 0
  So c_ (k + 2)=(2k - 1)/((k + 2) (k + 1))c_k
  
  c_0=c_0
  c_2=-c_0/2
  c_4=3/(4  3)c_2=-3/4 !c_0
  c_6=7/(6  5)c_4=-(7  3)/6 !c_0
  c_8=11/(8  7)c_6=-(11  7  3)/8 !c_0
  c_ (2n)=-(3  7  11. .. (2n - 5))/(2n) !c_0
  Thus y_0=c_0(1-x^2/2 !-(3x^4)/4 !-(21 x^6)/6 !-(321x^8)/8 ! ... )
  
   
  c_1=c_1
  c_3=c_1/(3  2)
  c_5=5/(5  4)c_3=5/5 !c_1
  c_7=9/(7  6)c_5=(5  9)/7 !c_1
  c_9=13/(9  8)c_7=(5  9  13)/9 !c_1
  c_ (2n + 1)=(5  9  13. .. (2n - 5))/(2n + 1) !c_1
  Thus y_1=c_1(x+x^3/3 !+(5x^5)/5 !+(45 x^7)/7 !+(585x^9)/9 ! ... )