Answer to Question #87056 in Differential Equations for MADHU

Question

Solve the following ODE using the power series method: (x^2-1) y′′ +3 xy′ + xy =0

Answer 1

We seek the solution of the equation in the form of a power series

$$y(x)= \textstyle\sum_{n=0}^{\infty} a_n x^n$$

Then ODE:

$$(x^2-1)\sum_{n=2}^{\infty} a_n n (n-1) x^{n-2} +3 x \sum_{n=1}^{\infty} a_n n x^{n-1} + x \sum_{n=0}^{\infty} a_n x^n = 0$$ $$\sum_{n=2}^{\infty} a_n n (n-1) x^n - 2a_2 -6a_3 x - \sum_{n=2}^{\infty} a_{n+2} (n+2) (n+1) x^n + 3a_1 x +$$ $$3 \sum_{n=2}^{\infty} a_n n x^n + a_0 x + \sum_{n=2}^{\infty} a_{n-1} x^n = 0$$

$$\begin{cases} -2 a_2 = 0 \\ -6 a_3 + 3 a_1 + a_0 = 0 \\ a_n n (n-1) - a_{n+2} (n+2) (n+1) + 3 a_n n + a_{n-1} = 0, n=2,3,4... \end{cases}$$ $$\begin{cases} a_2 = 0 \\ a_3 =\frac {3 a_1 + a_0} 6 \\ a_{n+2} = \frac {a_n n (n+2) + a_{n-1}} {(n+2) (n+1)}, n=2,3,4... \end{cases}$$

Answer:

$$y(x) = a_0 +a_1 x + \frac {3 a_1 + a_0} {6} x^3 + \sum_{n=2}^{\infty} \frac {a_n n (n+2) + a_{n-1}} {(n+2)(n+1)} x^{n+2}$$