# power series approach to ODE

Consider the ODE $x^2y''(x)+xy'(x)-\lambda x^2y(x)=0, x\in[0,1], y(1)=0$. Determine the first 4 terms of a power series approach for $y(x)$ and compute the corresponding approximation to $\lambda$

Plugging the series into the ODE I get: $$\sum_{k=0}^{\infty}b_kk(k-1)x^k+\sum_{k=0}^{\infty}b_kkx^k-\lambda\sum_{k=0}^{\infty}b_kx^{k+2}.$$ Shifting indices I get $$\sum_{k=0}^{\infty}b_{k+2}(k+2)(k+1)x^{k+2}+\sum_{k=0}^{\infty}b_{k+2}(k+2)x^{k+2}+b_1x-\lambda\sum_{k=0}^{\infty}b_kx^{k+2}=\\ b_1x+\sum_{k=0}^{\infty}(b_{k+2}(k+2)(k+1)+b_{k+2}(k+2)-\lambda b_k)x^{k+2}=0.$$ From this I deduce that $b_1=0$. Now how do I compute the $b_k$ from the recurrence relation containing the unknown $\lambda$?

• Only one boundary condition? – Aretino Nov 11 '15 at 14:57
• yes, only one boundary condition – blst Nov 11 '15 at 16:39
• If you don't have a boundary condition for $y(0)$ then $b_0$ cannot be determined. From $b_0$ you could derive all the other even $b_k$, while the odd ones vanish, by the recurrence relation. You need anyway a second boundary condition to solve your equation. – Aretino Nov 11 '15 at 17:53

It turns out you don't actually need to know $b_0$. By the recurrence relation we have $b_0$ free to choose, $b_1=0, b_2=\lambda b_0 /4, b_3=0$. Using the initial value $B(1)=0$ we get $b_0+\lambda b_0/4\approx 0$ thus $\lambda\approx -4$.