Find the two power series solution of the differential equation $\ y''-3xy=0 \$

Find the two power series solution of the differential equation $\ y''-3xy=0 \$ at the ordinary point $\ x=0 \$.

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 ?

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