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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 ?

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  • $\begingroup$ +1 it seems correct to me $\endgroup$ – Isham Apr 11 '18 at 12:56

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