Find the two power series solution of the differential equation $ \ y''-3xy=0 \ $ at the ordinary point $ \ x=0 \ $.
Answer:
Let $ \ y=\sum_{n=0}^{\infty} a_n x^n \ $ be the power series solution .
Substituting $ \ y, y'' \ $ in the above equation , we get
$ \sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}-\sum_{n=0}^{\infty} 3a_n x^{n+1}=0 \\ \Rightarrow \sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n}-\sum_{n=1}^{\infty} 3a_{n-1} x^{n}=0 , $ Comparing both sides , we get
$ a_2=0, \\ a_{n+2}=\frac{3}{(n+2)(n+1)} a_{n-1} \ , n \geq 1 $
solving , we get
$ a_2=a_5=a_8=....=0 \\ a_3=\frac{1}{2} a_0, \ a_6= \frac{1}{20}a_0 , ....... \\ a_4=\frac{1}{4} a_1, \ a_7=\frac{1}{56} a_1, .... $
Thus the power seies solution is
$ y(x)=a_0 (1+\frac{1}{2} x^3+\frac{1}{20} x^6+........)+a_1(x+\frac{1}{4} x^4+\frac{1}{56} x^7+.......) \ $
Thus the the two power series solutions are
$ y_1(x)= 1+\frac{1}{2} x^3+\frac{1}{20} x^6+........, \\ y_2(x)=x+\frac{1}{4} x^4+\frac{1}{56} x^7+....... \ $
Am I right ?