0
$\begingroup$

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 $\lambda$?

$\endgroup$
  • $\begingroup$ Only one boundary condition? $\endgroup$ – Aretino Nov 11 '15 at 14:57
  • $\begingroup$ yes, only one boundary condition $\endgroup$ – blst Nov 11 '15 at 16:39
  • 1
    $\begingroup$ 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. $\endgroup$ – Aretino Nov 11 '15 at 17:53
0
$\begingroup$

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 $\lambda\approx -4$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.